<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:wfw="http://wellformedweb.org/CommentAPI/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
	xmlns:slash="http://purl.org/rss/1.0/modules/slash/"
	>

<channel>
	<title>Partial pressure Archives - The Fact Factor</title>
	<atom:link href="https://thefactfactor.com/tag/partial-pressure/feed/" rel="self" type="application/rss+xml" />
	<link>https://thefactfactor.com/tag/partial-pressure/</link>
	<description>Uncover the Facts</description>
	<lastBuildDate>Thu, 01 Oct 2020 15:29:52 +0000</lastBuildDate>
	<language>en-US</language>
	<sy:updatePeriod>
	hourly	</sy:updatePeriod>
	<sy:updateFrequency>
	1	</sy:updateFrequency>
	<generator>https://wordpress.org/?v=6.9</generator>
	<item>
		<title>Equilibrium Constant</title>
		<link>https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/equilibrium-constant/11136/</link>
					<comments>https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/equilibrium-constant/11136/#respond</comments>
		
		<dc:creator><![CDATA[Hemant More]]></dc:creator>
		<pubDate>Thu, 02 Apr 2020 13:06:59 +0000</pubDate>
				<category><![CDATA[Physical Chemistry]]></category>
		<category><![CDATA[Active mass]]></category>
		<category><![CDATA[Chemical equilibrium]]></category>
		<category><![CDATA[Chemical reaction]]></category>
		<category><![CDATA[Chemistry]]></category>
		<category><![CDATA[Concentration quotient]]></category>
		<category><![CDATA[Direction of reaction]]></category>
		<category><![CDATA[Dynamic equilibrium]]></category>
		<category><![CDATA[Endothermic reaction]]></category>
		<category><![CDATA[Equilibrium]]></category>
		<category><![CDATA[Equilibrium concentration]]></category>
		<category><![CDATA[Equilibrium constant.]]></category>
		<category><![CDATA[Equilibrium mixture]]></category>
		<category><![CDATA[Exothermic reaction]]></category>
		<category><![CDATA[Extent of reaction]]></category>
		<category><![CDATA[Heterogeneous equilibrium]]></category>
		<category><![CDATA[Heterogeneous reaction]]></category>
		<category><![CDATA[Homogeneous equilibrium]]></category>
		<category><![CDATA[Homogeneous reaction]]></category>
		<category><![CDATA[Irreversible reaction]]></category>
		<category><![CDATA[Law of chemical equilibrium]]></category>
		<category><![CDATA[Law of mass action]]></category>
		<category><![CDATA[Partial pressure]]></category>
		<category><![CDATA[Physical equilibrium]]></category>
		<category><![CDATA[Products]]></category>
		<category><![CDATA[Rate of chemical reaction]]></category>
		<category><![CDATA[Reactants]]></category>
		<category><![CDATA[Reversible reaction]]></category>
		<guid isPermaLink="false">https://thefactfactor.com/?p=11136</guid>

					<description><![CDATA[<p>Science &#62; Chemistry &#62; Chemical Equilibrium &#62; Equilibrium Constant Characteristics of Equilibrium Constant: It has a definite value for every chemical reaction at a particular temperature. It is independent of the initial concentrations of the reacting species. It changes with the change in the temperature. It depends on the nature of the reaction. It is [&#8230;]</p>
<p>The post <a href="https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/equilibrium-constant/11136/">Equilibrium Constant</a> appeared first on <a href="https://thefactfactor.com">The Fact Factor</a>.</p>
]]></description>
										<content:encoded><![CDATA[
<h4 class="wp-block-heading"><strong>Science &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/" target="_blank">Chemistry</a> &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/chemical-equilibrium/" target="_blank">Chemical Equilibrium</a> &gt; Equilibrium Constant</strong></h4>



<p><strong>Characteristics of Equilibrium Constant:</strong></p>



<ul class="wp-block-list"><li>It has a definite value for every chemical reaction at a particular temperature.</li><li>It is independent of the initial concentrations of the reacting species.</li><li>It changes with the change in the temperature.</li><li>It depends on the nature of the reaction.</li><li>It is independent of the change of pressure, volume and concentrations of the reactants and products.</li><li>It is not affected by the introduction of the catalyst.</li><li>The expression for it may contain the concentrations of gases or molecules and ions in solution but not of pure solids or pure liquids.</li><li>The expression for it and its magnitude depends on the stoichiometric form of the balanced chemical equation.</li><li>When the equation for equilibrium is multiplied by a factor, then the equilibrium constant must be raised to the power equal to the factor.</li></ul>



<p class="has-text-align-center">If K<sub>c</sub> is equilibrium constant for reaction&nbsp;aA + bB&nbsp;<strong>⇌</strong>&nbsp; cC + dD</p>



<p class="has-text-align-center">Then its value for reaction naA + nbB&nbsp;<strong>⇌</strong>&nbsp; ncC + ndD is given by</p>



<p class="has-text-align-center">K&#8217;<sub>c</sub>&nbsp;= (K<sub>c</sub>)<sup>n</sup></p>



<ul class="wp-block-list"><li>When the addition of two equilibria leads to another equilibrium then the product of their equilibrium constants gives the value of K<sub>C</sub> of the resultant equilibrium.</li></ul>



<p class="has-text-align-center">K<sub>(resultant)</sub> = K<sub>(Reaction 1)&nbsp;&nbsp;</sub>x K<sub>(Reaction 2)</sub></p>



<ul class="wp-block-list"><li>If K<sub>1</sub>, k<sub>2</sub>, K<sub>3</sub>, &#8230;. are equilibrium constant for recation<sub>1</sub>, reaction<sub>2</sub>, reaction<sub>3</sub>, &#8230;&#8230;. Then the value of K<sub>C</sub> for reaction, a x recation<sub>1</sub>+ b x reaction<sub>2</sub>, c x reaction<sub>3</sub>, &#8230;&#8230;.&nbsp;is given by</li></ul>



<p class="has-text-align-center">K<sub>C</sub> = (K<sub>1</sub>)<sup>a</sup>(K<sub>2</sub>)<sup>b</sup>(K<sub>3</sub>)<sup>c</sup>&#8230;&#8230;..</p>



<ul class="wp-block-list"><li>For a reversible reaction, the value of K<sub>C</sub> for the backward reaction is inverse of the equilibrium constant for the forward reaction</li></ul>



<p class="has-text-align-center">K<sub>(backward)</sub> = 1 / K<sub>(forward)</sub></p>



<ul class="wp-block-list"><li>If it is expressed in terms of concentration, it has different units for different reactions.</li></ul>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>K<sub>C</sub> is Dimensionless Unitless Quantity:</strong></p>



<p>Depending upon the stoichiometric coefficients of a chemical reaction, the equilibrium constant should have a unit. Its unit should be </p>



<p class="has-text-align-center">(mol dm<sup>-3</sup>)<sup>(∑n products &#8211;&nbsp;∑n reactants)</sup></p>



<p>Note that 1 dm<sup>3</sup> = 1 L. Hence concentration can be expressed as (mol L<sup>-1</sup>)</p>



<p>These days we specify equilibrium constant in terms of dimensionless quantities by specifying the standard state of reactants and products. The standard state pressure for pure gas is 1 bar and the partial pressures of the gases are measured with respect to this standard state.</p>



<p>If a gas has a partial pressure of 1.5 bar, then in terms of standard state its pressure would be equal to 1.5 bar/1bar = 1.5, a dimensionless number. Similarly, for concentrations, the standard state is 1 M ( 1mol dm<sup>-3</sup>). If the concentration of the substance is 2.0 M. Then in terms of standard state it will be expressed as 2.0 M/1 M = 2. Thus using partial pressures and concentrations K<sub>P</sub> and K<sub>C</sub> obtained are dimensionless. Hence equilibrium constant is considered as dimensionless, unitless quantity.</p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>K<sub>C</sub> of a General Reaction and its Multiple:</strong></p>



<p>Consider a hypothetical reversible reaction</p>



<p class="has-text-align-center">aA + bB&nbsp;<strong>⇌</strong>&nbsp; cC + dD</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-45.png" alt="" class="wp-image-11141" width="96" height="50"/></figure></div>



<p class="has-text-align-center">The equilibrium constant of the reaction given by</p>



<p class="has-text-align-center">Consider a multiple of above reaction</p>



<p class="has-text-align-center">naA + nbB&nbsp;<strong>⇌</strong>&nbsp; ncC + ndD</p>



<p class="has-text-align-center">The equilibrium constant of the reaction given by</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-46.png" alt="" class="wp-image-11143" width="227" height="93"/></figure></div>



<p>Thus when the equation for an equilibrium is multiplied by a factor, then the equilibrium constant must be raised to the power equal to the factor.</p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong><strong>K<sub>C</sub></strong> for the addition of Two Chemical Equilibria:</strong></p>



<p>Consider following equilibria</p>



<p>1) N<sub>2(g)</sub> +&nbsp; &nbsp;O<sub>2(g)</sub> &nbsp;⇌&nbsp; &nbsp; &nbsp;2NO<sub>(g)</sub></p>



<p class="has-text-align-center">The equilibrium constant for the reaction is</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-47.png" alt="" class="wp-image-11145" width="227" height="50"/></figure></div>



<p>2)&nbsp; &nbsp;2NO<sub>(g)</sub> +&nbsp; &nbsp;O<sub>2(g)</sub> &nbsp;⇌&nbsp; &nbsp; &nbsp;2NO<sub>2(g)&nbsp;</sub></p>



<p class="has-text-align-center">The equilibrium constant for the reaction is</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-48.png" alt="" class="wp-image-11147" width="227" height="51"/></figure></div>



<p>3)&nbsp; N<sub>2(g)</sub> +&nbsp; 2O<sub>2(g)</sub> &nbsp; ⇌&nbsp; &nbsp;2NO<sub>2(g)</sub></p>



<p class="has-text-align-center">The equilibrium constant for the reaction is</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-49.png" alt="" class="wp-image-11149" width="231" height="50"/></figure></div>



<p class="has-text-align-center">Multiplying equation (1) by (2) we get</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-50.png" alt="" class="wp-image-11151" width="227" height="121"/></figure></div>



<p>Thus When the addition of two equilibria leads to another equilibrium then the product of their equilibrium constants gives the equilibrium constant of the resultant equilibrium.</p>



<p class="has-text-color has-background has-medium-font-size has-luminous-vivid-orange-color has-very-light-gray-background-color"><strong>Temperature Dependence of Equilibrium&nbsp;Constant:</strong></p>



<p>At chemical equilibrium, the rate of the forward reaction is equal to the rate of backward reaction. When the temperature is increased, in general, the rate of both the forward reaction and the backward reaction increases. As the energy of activation of the&nbsp;forward reaction and the backward reaction are different, the extent of the increase of the&nbsp;forward reaction and the backward reaction is different. Thus the value of K<sub>f</sub> (rate constant for the&nbsp;forward reaction) and K<sub>b</sub> (rate constant for the backward reaction) changes to a different extent. Thus the ratio K<sub>f</sub> / K<sub>b</sub> changes. i.e. the value of equilibrium constant changes with the change in temperature.</p>



<p>The value of the equilibrium constant for endothermic reaction increases with an increase in temperature, while the value of the equilibrium constant for exothermic reaction decreases with an increase in temperature. The temperature dependence of equilibrium constant can be written mathematically as</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-51.png" alt="" class="wp-image-11153" width="210" height="43"/></figure></div>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Uses of K<sub>C</sub>:</strong></p>



<h4 class="wp-block-heading"><strong>It helps in the prediction of the extent of reaction:</strong></h4>



<p>The magnitude of the K<sub>C</sub> tells us about the extent in which the reactants are converted into the products before the equilibrium is attained. Larger values of K indicates that the extent of reactants converting into products is greater. The generalization is</p>



<ul class="wp-block-list"><li>If K<sub>C</sub> &gt; 10<sup>3</sup> Products predominates the reactants. i.e. the concentration of products is very high compared to that of reactants t equilibrium and the reaction proceeds nearly to completion</li><li>If K<sub>C</sub> &lt; 10<sup>-3</sup> Reactants predominates the products. i.e. the concentration of products is very less compared to that of reactants t equilibrium and the reaction hardly proceeds.</li><li>10<sup>-3</sup> &lt; K<sub>C</sub> &lt; 10<sup>-3</sup>&nbsp;, appreciable concentrations of both the reactants and products are present at equilibrium</li></ul>



<div class="wp-block-image"><figure class="aligncenter size-large"><img loading="lazy" decoding="async" width="300" height="72" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-52.png" alt="" class="wp-image-11155"/></figure></div>



<p>It gives us an idea of relative stabilities of reactants and products:</p>



<p>If the K<sub>C</sub> is large products are more stable than the reactants. Whereas, If the value of the equilibrium constant is small reactants are more stable than the products.&nbsp; The generalisation is</p>



<p>If K<sub>C</sub> &gt; 10<sup>3</sup> Products are stable than reactants.</p>



<p>If K<sub>C</sub> &lt; 10<sup>-3</sup> Reactants are stable than products.</p>



<h4 class="wp-block-heading"><strong>It helps in the prediction of the direction of a net reaction.</strong></h4>



<p>The value of K<sub>C</sub> helps in the prediction of the direction in which the net reaction is proceeding at given concentrations or partial pressures of reactants and products. If Q<sub>c</sub> is the concentration quotient and K<sub>c</sub> be the equilibrium constant of a chemical reaction.</p>



<p class="has-text-align-center">For reaction aA + bB&nbsp;<strong>⇌</strong>&nbsp; cC + dD</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-53.png" alt="" class="wp-image-11157" width="106" height="50"/></figure></div>



<p class="has-text-align-center">When computing Q<sub>C</sub> the concentration at that instant are used</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-45.png" alt="" class="wp-image-11141" width="109" height="57"/></figure></div>



<p class="has-text-align-center">When computing K<sub>C</sub> the concentration at equilibrium are used. The generalization is</p>



<ul class="wp-block-list"><li>If Q<sub>C</sub> &gt; K<sub>C</sub> , the reaction is taking place in a backward direction i.e. in the direction of reactants</li><li>If Q<sub>C</sub> &lt; &nbsp;K<sub>C</sub> , the reaction is taking place in a forward direction i.e. in the direction of products.</li><li>If Q<sub>C</sub> = &nbsp;K<sub>C</sub> The reaction is in the equilibrium state and hence no net reaction is taking place.</li></ul>



<h4 class="wp-block-heading"><strong>It helps in the calculation of equilibrium constants and equilibrium pressures.&nbsp;</strong></h4>



<p>If the equilibrium concentrations of various reactants and products are known for a reaction, the equilibrium constant can be calculated. On the other hand, if the equilibrium constant is known, the equilibrium concentrations can be calculated.</p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Steps Involved in the Calculation of K<sub>C</sub> and Equilibrium Pressures:</strong></p>



<ol class="wp-block-list" type="1"><li>Write the chemical equation for the equilibrium</li><li>Write expression for K<sub>C</sub> or K<sub>P</sub> for the reaction.</li><li>Express all unknown concentrations or partial pressures in terms of a single variable x.</li><li>Substitute equilibrium concentrations or partial pressures in terms of x in the expression for K<sub>C</sub> or K<sub>P</sub>.</li><li>Solve the equation for x</li><li>Substitute the value obtained for x in the expression in step 3 to calculate equilibrium concentrations or equilibrium partial pressures</li></ol>



<h4 class="wp-block-heading"><strong>Science &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/" target="_blank">Chemistry</a> &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/chemical-equilibrium/" target="_blank">Chemical Equilibrium</a> &gt; Equilibrium Constant</strong></h4>
<p>The post <a href="https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/equilibrium-constant/11136/">Equilibrium Constant</a> appeared first on <a href="https://thefactfactor.com">The Fact Factor</a>.</p>
]]></content:encoded>
					
					<wfw:commentRss>https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/equilibrium-constant/11136/feed/</wfw:commentRss>
			<slash:comments>0</slash:comments>
		
		
			</item>
		<item>
		<title>Writing Expression for Equilibrium Constant</title>
		<link>https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/expression-for-equilibrium-constant/11052/</link>
					<comments>https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/expression-for-equilibrium-constant/11052/#respond</comments>
		
		<dc:creator><![CDATA[Hemant More]]></dc:creator>
		<pubDate>Thu, 02 Apr 2020 12:04:11 +0000</pubDate>
				<category><![CDATA[Physical Chemistry]]></category>
		<category><![CDATA[Active mass]]></category>
		<category><![CDATA[Chemical equilibrium]]></category>
		<category><![CDATA[Chemical reaction]]></category>
		<category><![CDATA[Chemistry]]></category>
		<category><![CDATA[Concentration quotient]]></category>
		<category><![CDATA[Dynamic equilibrium]]></category>
		<category><![CDATA[Endothermic reaction]]></category>
		<category><![CDATA[Equilibrium]]></category>
		<category><![CDATA[Equilibrium concentration]]></category>
		<category><![CDATA[Equilibrium constant.]]></category>
		<category><![CDATA[Equilibrium mixture]]></category>
		<category><![CDATA[Exothermic reaction]]></category>
		<category><![CDATA[Heterogeneous equilibrium]]></category>
		<category><![CDATA[Heterogeneous reaction]]></category>
		<category><![CDATA[Homogeneous equilibrium]]></category>
		<category><![CDATA[Homogeneous reaction]]></category>
		<category><![CDATA[Irreversible reaction]]></category>
		<category><![CDATA[Law of chemical equilibrium]]></category>
		<category><![CDATA[Law of mass action]]></category>
		<category><![CDATA[Partial pressure]]></category>
		<category><![CDATA[Physical equilibrium]]></category>
		<category><![CDATA[Products]]></category>
		<category><![CDATA[Rate of chemical reaction]]></category>
		<category><![CDATA[Reactants]]></category>
		<category><![CDATA[Reversible reaction]]></category>
		<guid isPermaLink="false">https://thefactfactor.com/?p=11052</guid>

					<description><![CDATA[<p>Science &#62; Chemistry &#62; Chemical Equilibrium &#62; Writing Expression for Equilibrium Constant In this article, we shall stuudy to write expression for equilibrium constant. Steps Involved in Writing Expression for Equilibrium Constant of a Reaction: Write the balanced chemical equation for the reaction. Write the products of equilibrium concentrations of the products in the numerator. [&#8230;]</p>
<p>The post <a href="https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/expression-for-equilibrium-constant/11052/">Writing Expression for Equilibrium Constant</a> appeared first on <a href="https://thefactfactor.com">The Fact Factor</a>.</p>
]]></description>
										<content:encoded><![CDATA[
<h4 class="wp-block-heading"><strong>Science &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/" target="_blank">Chemistry</a> &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/chemical-equilibrium/" target="_blank">Chemical Equilibrium</a> &gt; Writing Expression for Equilibrium Constant</strong></h4>



<p>In this article, we shall stuudy to write expression for equilibrium constant.</p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Steps Involved in Writing Expression for Equilibrium Constant of a Reaction:</strong></p>



<ul class="wp-block-list"><li>Write the balanced chemical equation for the reaction.</li><li>Write the products of equilibrium concentrations of the products in the numerator. Omit pure solids, pure liquids and the solvents in dilute solutions.</li><li>Write the products of equilibrium concentrations of the reactants in the denominator. Omit pure solids, pure liquids and the solvents in dilute solutions.</li><li>Raise each concentration term to the power equal to stoichiometric coefficients of the species in the equation.</li></ul>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Homogeneous Equilibrium:</strong></p>



<p>The equilibrium in which all the substances involved exist in a single homogeneous phase is called homogeneous equilibrium.</p>



<p><strong>Examples</strong>:</p>



<p class="has-text-align-center">N<sub>2(g)</sub> +&nbsp; 3 H<sub>2(g)</sub> &nbsp;⇌&nbsp; &nbsp; &nbsp;2NH<sub>3(g)</sub></p>



<p class="has-text-align-center">H<sub>2(g)</sub> +&nbsp; I<sub>2(g)</sub> ⇌&nbsp;&nbsp;&nbsp;&nbsp; 2HI<sub>(g)</sub></p>



<p class="has-text-align-center">2SO<sub>2(g)</sub> +&nbsp; O<sub>2(g)</sub> &nbsp;⇌&nbsp;&nbsp; 2SO<sub>3(g)</sub></p>



<p class="has-text-align-center">NH<sub>3(aq)</sub>&nbsp;+ H<sub>2</sub>O<sub>(l)</sub> ⇌&nbsp;&nbsp;&nbsp;&nbsp; 2NH<sub>4</sub>+<sub>(aq)</sub> +&nbsp; OH-<sub>(aq)</sub></p>



<p class="has-text-align-center">2N<sub>2</sub>O<sub>(g)</sub> ⇌ 2N<sub>2(g)</sub>&nbsp; +&nbsp;&nbsp; O<sub>2(g)</sub></p>



<p class="has-text-align-center">CH<sub>3</sub>COOC<sub>2</sub>H<sub>5(aq)</sub> + H2O<sub>(l)</sub> ⇌&nbsp;CH<sub>3</sub>COOH<sub>(aq)</sub> +&nbsp;&nbsp; C2H5OH<sub>(aq)</sub></p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Heterogeneous Equilibrium:</strong></p>



<p>The equilibrium in which the substance involved are present in different phases is called heterogeneous equilibrium.</p>



<p><strong>Examples:</strong></p>



<p class="has-text-align-center">CaCO<sub>3(s)</sub> ⇌ &nbsp;CaO<sub>(s)</sub> + . CO<sub>2(g)</sub></p>



<p class="has-text-align-center">CaCO<sub>3(s)</sub>+H<sub>2</sub>O(l)+CO<sub>2</sub>(g) ⇌ Ca<sup>2+</sup><sub>(aq)</sub>+ 2HCO<sub>3</sub><sup>&#8211;</sup><sub>(aq)</sub></p>



<p class="has-text-align-center">(NH<sub>4</sub>)<sub>2</sub>CO<sub>3(s)</sub> ⇌ 2NH<sub>3(g)</sub> + CO<sub>2(g)</sub>+&nbsp; H<sub>2</sub>O<sub>(g)</sub></p>



<p class="has-text-align-center">2Mg<sub>(s)</sub> + O2<sub>(g)</sub> &nbsp;⇌ &nbsp; &nbsp;2MgO<sub>(s)</sub></p>



<p class="has-text-color has-text-align-center has-vivid-red-color"><strong>Note: Pure liquids and solids are ignored while writing the equilibrium constant expression.</strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img loading="lazy" decoding="async" width="300" height="131" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-09.png" alt="" class="wp-image-11062"/></figure></div>



<p>Now, for pure solids and pure liquids, the molar mass M and density ρ&nbsp;are constant. Hence the concentration remains constant.</p>



<ul class="wp-block-list"><li>When pure solids are involved in equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentrations of all solids involved in equilibrium are taken as unity.</li><li>When pure liquids are involved in equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentrations of all liquids involved in equilibrium are taken as unity.</li><li>For equilibria in the aqueous medium, the concentration of solvent (water) will not change appreciably because it is present in large excess. Hence by convention, the concentration of solvent (water) is taken as unity.</li><li>Hence in general, pure liquids, pure solids, and solvents can be ignored while writing the equilibrium constant expression.</li></ul>



<p class="has-text-color has-background has-medium-font-size has-luminous-vivid-orange-color has-very-light-gray-background-color"><strong>Writing the Expression for K<sub>c</sub> and K<sub>p</sub>:</strong></p>



<p class="has-text-align-center">BaCO<sub>3(s)&nbsp;</sub>&nbsp;⇌&nbsp; BaO<sub>(s)&nbsp;</sub>+ CO<sub>2(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-10.png" alt="" class="wp-image-11064" width="224" height="30"/></figure></div>



<p class="has-text-align-center">4NH<sub>3(g)</sub> + 5O<sub>2(g)</sub> ⇌&nbsp; &nbsp;4NO<sub>(g)</sub>+ 6H<sub>2</sub>O<sub>(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-11-1.png" alt="Expression for equilibrium constant" class="wp-image-11066"/></figure></div>



<p class="has-text-align-center">NH<sub>3(g)</sub> + HCl<sub>(g)</sub>&nbsp; ⇌&nbsp; NH<sub>4</sub>Cl<sub>(s)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-12-1.png" alt="Expression for equilibrium constant" class="wp-image-11068" width="224" height="51"/></figure></div>



<p class="has-text-align-center">3Cl<sub>2(g)</sub> + 2NO<sub>2(g)</sub>&nbsp; ⇌&nbsp; 2NO<sub>2</sub>Cl<sub>3(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-13-1.png" alt="Expression for equilibrium constant" class="wp-image-11070"/></figure></div>



<p class="has-text-align-center">Fe<sup>3+</sup><sub>(aq)</sub> + SCN<sup>&#8211;</sup><sub>(aq)</sub> ⇌ FeNCS<sup>2+</sup><sub>(aq)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-14-1.png" alt="Expression for equilibrium constant" class="wp-image-11072" width="170" height="65"/></figure></div>



<p class="has-text-align-center">CH<sub>4(g)</sub> + 2H<sub>2</sub>S<sub>(g)&nbsp;</sub> ⇌&nbsp; CS<sub>2(g)</sub>+ 4H<sub>2(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-15-1.png" alt="Expression for equilibrium constant" class="wp-image-11075"/></figure></div>



<p class="has-text-align-center">MgCO<sub>3(s)&nbsp;&nbsp;</sub>⇌&nbsp; MgO<sub>(s)&nbsp;&nbsp;</sub>+ CO<sub>2(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-16-1.png" alt="" class="wp-image-11077" width="217" height="33"/></figure></div>



<p class="has-text-align-center">AgBr<sub>(s)</sub> ⇌ Ag<sup>+</sup><sub>(aq)</sub>+ Br<sup>&#8211;</sup><sub>(aq)</sub></p>



<p class="has-text-align-center">K<sub>c</sub> = [Ag<sup>+</sup>][Br&nbsp;<sup>&#8211;</sup>]</p>



<p class="has-text-align-center">CH<sub>3</sub>COCH<sub>3(l)</sub>&nbsp;⇌CH<sub>3</sub>COCH<sub>3(g)</sub></p>



<p class="has-text-align-center">K<sub>c</sub> = [CH<sub>3</sub>COCH<sub>3(g)</sub>]</p>



<p class="has-text-align-center">CH<sub>4(g)</sub> + 2O<sub>2(g)</sub> ⇌ CO<sub>2(g)</sub>+ 2H<sub>2</sub>O<sub>(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-17-1.png" alt="Expression for equilibrium constant" class="wp-image-11079"/></figure></div>



<p class="has-text-align-center">Al<sub>(s)</sub> + 3H<sup>+</sup><sub>(aq)&nbsp;</sub>⇌ Al<sup>3+</sup><sub>(aq)</sub>+ 3/2H<sub>2(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-18-1.png" alt="Expression for equilibrium constant" class="wp-image-11081" width="181" height="65"/></figure></div>



<p class="has-text-align-center">HPO<sub>4</sub><sup>2-</sup><sub>(aq)</sub> + H<sub>2</sub>O<sub>(l)</sub> ⇌ H<sub>3</sub>O<sup>+</sup><sub>(aq)</sub>+ PO<sub>4</sub><sup>3-</sup><sub>(aq)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-19.png" alt="Expression for equilibrium constant" class="wp-image-11082" width="166" height="60"/></figure></div>



<p class="has-text-align-center">Ag<sub>2</sub>O<sub>(s)</sub> + 2HNO<sub>3(aq)</sub> ⇌&nbsp; 2AgNO<sub>3(aq)</sub>+ H<sub>2</sub>O<sub>(l)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-20-1.png" alt="Expression for equilibrium constant" class="wp-image-11084" width="110" height="51"/></figure></div>



<p class="has-text-align-center">Ni<sub>(s)</sub> + 4CO<sub>(g)</sub> ⇌&nbsp; Ni(CO)<sub>4(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-21.png" alt="Expression for equilibrium constant" class="wp-image-11086" width="135" height="45"/></figure></div>



<p class="has-text-align-center">CuO<sub>(s)</sub> + H<sub>2(g)</sub> ⇌&nbsp; Cu<sub>(s)</sub>+ H<sub>2</sub>O<sub>(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-22-1.png" alt="" class="wp-image-11090" width="124" height="51"/></figure></div>



<p class="has-text-align-center">N<sub>2(g)</sub> + 3H<sub>2(g)</sub> ⇌&nbsp; 2NH<sub>3(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img loading="lazy" decoding="async" width="300" height="64" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-23-1.png" alt="Expression for equilibrium constant" class="wp-image-11092"/></figure></div>



<p class="has-text-align-center">Fe<sup>3+</sup><sub>(aq)</sub> + 3OH<sup>&#8211;</sup><sub>(aq)</sub> ⇌&nbsp; Fe(OH)<sub>3(s)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-24.png" alt="Expression for equilibrium constant" class="wp-image-11093" width="177" height="67"/></figure></div>



<p class="has-text-align-center">2N<sub>2</sub>O<sub>(g)</sub> ⇌ 2N<sub>2(g)</sub> + O<sub>2(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-25-1.png" alt="Expression for equilibrium constant" class="wp-image-11095"/></figure></div>



<p class="has-text-align-center">C<sub>(s)</sub>+&nbsp; CO<sub>2(g)</sub> ⇌&nbsp; 2CO<sub>(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-26-1.png" alt="Expression for equilibrium constant" class="wp-image-11097" width="226" height="59"/></figure></div>



<p>Consider a hypothetical reversible reaction involving homogeneous gaseous phase</p>



<p class="has-text-align-center">aA(g) + bB(g)&nbsp;⇌&nbsp; cC(g) + dD(g)</p>



<p>The equilibrium constant in terms of the partial pressure is given by</p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img loading="lazy" decoding="async" width="300" height="144" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-27.png" alt="Expression for equilibrium constant" class="wp-image-11099"/></figure></div>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-28-1.png" alt="Expression for equilibrium constant" class="wp-image-11102" width="220" height="172"/></figure></div>



<p class="has-text-align-center">This is the relation between K<sub>c</sub> and K<sub>p</sub>.</p>



<p>Where Δn = (Sum of the exponents in the numerator of concentration quotient) &#8211; (Sum of the exponents in the denominator of concentration quotient)</p>



<p class="has-text-color has-background has-medium-font-size has-luminous-vivid-orange-color has-very-light-gray-background-color"><strong>Steps Involved in Finding Relation Between K<sub>c</sub> and K<sub>p</sub> :</strong></p>



<ul class="wp-block-list"><li>Write the balanced chemical equation for the reaction.</li><li>Find the change in the number of moles of gaseous species by the following formula.</li></ul>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-29-1.png" alt="Expression for equilibrium constant" class="wp-image-11104"/></figure></div>



<ul class="wp-block-list"><li>Now, use the relation</li></ul>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-30-1.png" alt="" class="wp-image-11106" width="144" height="35"/></figure></div>



<p class="has-text-color has-background has-medium-font-size has-luminous-vivid-orange-color has-very-light-gray-background-color"><strong>Examples to Find Relation between&nbsp;Between K<sub>c</sub> and K<sub>p</sub> :</strong></p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Example &#8211; 01:</strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img loading="lazy" decoding="async" width="300" height="178" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-31.png" alt="" class="wp-image-11107"/></figure></div>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Example &#8211; 02:</strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-32-1.png" alt="" class="wp-image-11109"/></figure></div>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Example &#8211; 03:</strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-33-1.png" alt="" class="wp-image-11111"/></figure></div>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Example &#8211; 04:&nbsp;</strong></p>



<p><strong>For the reaction N<sub>2(g)</sub> + 3H<sub>2(g)</sub> ⇌&nbsp; 2NH<sub>3(g)</sub>, the value of the equilibrium constant K<sub>p</sub> is 3.6 x 10<sup>-2</sup> at 500 K. Calculate the value of K<sub>c</sub> for the reaction at the same temperature. Take R = 0.0831 bar lit mol<sup>-1</sup> K<sup>-1</sup>.</strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-34-1.png" alt="" class="wp-image-11113"/></figure></div>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Example &#8211; 05:</strong></p>



<p><strong>For the reaction 2SO<sub>2(g)</sub>) + O<sub>2(g)</sub> ⇌ 2SO<sub>3(g)</sub>, the value of the equilibrium constant K<sub>p</sub> is 2.0 X 10<sup>10</sup> bar<sup>-1</sup> at 450 K. Calculate the value of K<sub>c</sub> for the reaction at the same temperature. Take R = 0.0831 bar lit&nbsp;mol<sup>-1</sup> K<sup>-1</sup>.</strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-35-1.png" alt="" class="wp-image-11115"/></figure></div>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Example &#8211; 06:</strong></p>



<p><strong>For the reaction 2NOCl<sub>(g)</sub> ⇌ 2NO<sub>(g)</sub>+ Cl<sub>2(g)</sub>, the value of the equilibrium constant K<sub>p</sub> is1.8 X 10<sup>-2</sup> at 500 K. Calculate the value of K<sub>c</sub> for the reaction at the same temperature. Take R = 0.0831 bar lit&nbsp;mol<sup>-1</sup> K<sup>-1</sup>.</strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-36-1.png" alt="" class="wp-image-11117"/></figure></div>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Example &#8211; 07:</strong></p>



<p><strong>For the reaction CaCO<sub>3(s)&nbsp;&nbsp;</sub>⇌ CaO<sub>(s)</sub>+ CO<sub>2(g)</sub>, the value of equilibrium constant K<sub>p</sub> is 167 at 1073 K. Calculate the value of K<sub>c</sub> for the reaction at the same temperature. Take R = 0.0831 bar lit&nbsp;mol<sup>-1</sup> K<sup>-1</sup>.</strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img loading="lazy" decoding="async" width="300" height="232" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-37.png" alt="" class="wp-image-11118"/></figure></div>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Example &#8211; 08:</strong></p>



<p><strong>Find the ratio of K<sub>p</sub>/K<sub>c</sub> for the reaction,&nbsp;CO(g) + 1/2O<sub>2(g)&nbsp;&nbsp;</sub>⇌&nbsp;&nbsp;2CO<sub>2(g)</sub>, at 500 K.&nbsp;Take R = 0.0831 bar lit&nbsp;mol<sup>-1</sup> K<sup>-1</sup>.</strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-38-1.png" alt="" class="wp-image-11120"/></figure></div>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Example &#8211; 09:</strong></p>



<p><strong>Find the ratio of&nbsp;K<sub>p</sub>/K<sub>c&nbsp;</sub> for the reaction,&nbsp;N<sub>2(g)</sub>&nbsp;+ O<sub>2(g)</sub>&nbsp;⇌&nbsp;&nbsp;&nbsp;2NO<sub>(g)</sub></strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-39-1.png" alt="" class="wp-image-11122"/></figure></div>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Example &#8211; 10:</strong></p>



<p><strong>The equilibrium constant K<sub>p</sub> for the reaction H<sub>2(g)</sub>&nbsp;&nbsp;+ I<sub>2(g)</sub>&nbsp; ⇌&nbsp; 2HI<sub>(g)</sub> is 130 at 510 K. Calculate K<sub>c</sub> for following reactions a)&nbsp;2HI<sub>(g)</sub> &nbsp;⇌ H<sub>2(g)</sub>&nbsp;&nbsp;+ I<sub>2(g)</sub>,&nbsp; b) HI<sub>(g)</sub> &nbsp;⇌ 1/2H<sub>2(g)</sub>&nbsp;&nbsp;+ 1/2I<sub>2(g)</sub>, c) 1/2H<sub>2(g)</sub>&nbsp;&nbsp;+ 1/2I<sub>2(g)</sub>&nbsp; ⇌&nbsp; HI<sub>(g)</sub>.</strong></p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-40-1.png" alt="" class="wp-image-11124"/></figure></div>



<p class="has-text-align-center">∴&nbsp; K<sub>c</sub> = 130</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-41-1.png" alt="" class="wp-image-11126" width="138" height="51"/></figure></div>



<p class="has-text-align-center">For the reaction 2HI<sub>(g)</sub> &nbsp;⇌ H<sub>2(g)</sub>&nbsp;&nbsp;+ I<sub>2(g)</sub>,</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-42-1.png" alt="" class="wp-image-11128" width="225" height="44"/></figure></div>



<p class="has-text-align-center">For the reaction, HI<sub>(g)</sub> &nbsp;⇌ 1/2H<sub>2(g)</sub>&nbsp;&nbsp;+ 1/2I<sub>2(g)</sub></p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-43-1.png" alt="" class="wp-image-11130" width="208" height="86"/></figure></div>



<p class="has-text-align-center">For reaction,&nbsp;1/2H<sub>2(g)</sub>&nbsp;&nbsp;+ 1/2I<sub>2(g)</sub>&nbsp; ⇌&nbsp; HI<sub>(g)</sub>.</p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-44-1.png" alt="" class="wp-image-11132" width="199" height="92"/></figure></div>



<h4 class="wp-block-heading"><strong>Science &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/" target="_blank">Chemistry</a> &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/chemical-equilibrium/" target="_blank">Chemical Equilibrium</a> &gt; Writing Expression for Equilibrium Constant</strong></h4>
<p>The post <a href="https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/expression-for-equilibrium-constant/11052/">Writing Expression for Equilibrium Constant</a> appeared first on <a href="https://thefactfactor.com">The Fact Factor</a>.</p>
]]></content:encoded>
					
					<wfw:commentRss>https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/expression-for-equilibrium-constant/11052/feed/</wfw:commentRss>
			<slash:comments>0</slash:comments>
		
		
			</item>
		<item>
		<title>Law of Mass Action</title>
		<link>https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/law-of-mass-action/11035/</link>
					<comments>https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/law-of-mass-action/11035/#respond</comments>
		
		<dc:creator><![CDATA[Hemant More]]></dc:creator>
		<pubDate>Thu, 02 Apr 2020 09:16:51 +0000</pubDate>
				<category><![CDATA[Physical Chemistry]]></category>
		<category><![CDATA[Active mass]]></category>
		<category><![CDATA[Chemical equilibrium]]></category>
		<category><![CDATA[Chemical reaction]]></category>
		<category><![CDATA[Chemistry]]></category>
		<category><![CDATA[Concentration quotient]]></category>
		<category><![CDATA[Dynamic equilibrium]]></category>
		<category><![CDATA[Endothermic reaction]]></category>
		<category><![CDATA[Equilibrium]]></category>
		<category><![CDATA[Equilibrium concentration]]></category>
		<category><![CDATA[Equilibrium constant.]]></category>
		<category><![CDATA[Equilibrium mixture]]></category>
		<category><![CDATA[Exothermic reaction]]></category>
		<category><![CDATA[Heterogeneous reaction]]></category>
		<category><![CDATA[Homogeneous reaction]]></category>
		<category><![CDATA[Irreversible reaction]]></category>
		<category><![CDATA[Law of chemical equilibrium]]></category>
		<category><![CDATA[Law of mass action]]></category>
		<category><![CDATA[Partial pressure]]></category>
		<category><![CDATA[Physical equilibrium]]></category>
		<category><![CDATA[Products]]></category>
		<category><![CDATA[Rate of chemical reaction]]></category>
		<category><![CDATA[Reactants]]></category>
		<category><![CDATA[Reversible reaction]]></category>
		<guid isPermaLink="false">https://thefactfactor.com/?p=11035</guid>

					<description><![CDATA[<p>Science &#62; Chemistry &#62; Chemical Equilibrium &#62; Law of Mass Action In this article, we shall study the law of mass action and its application to chemical equlibrium. Rate of Chemical Reaction: The change in the concentration of the reactants (or products) per unit time is called the rate of reaction Mathematically it can be [&#8230;]</p>
<p>The post <a href="https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/law-of-mass-action/11035/">Law of Mass Action</a> appeared first on <a href="https://thefactfactor.com">The Fact Factor</a>.</p>
]]></description>
										<content:encoded><![CDATA[
<h4 class="wp-block-heading"><strong>Science &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/" target="_blank">Chemistry</a> &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/chemical-equilibrium/" target="_blank">Chemical Equilibrium</a> &gt; Law of Mass Action</strong></h4>



<p>In this article, we shall study the law of mass action and its application to chemical equlibrium.</p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Rate of Chemical Reaction:</strong> </p>



<p>The change in the concentration of the reactants (or products) per unit time is called the rate of reaction Mathematically it can be expressed as, The rate of Reaction = Change in Concentration of Products or Reactants/time in which change is taking place Hence the rate of chemical reaction is the change in concentration of the reactants in unit time. It&#8217;s S.I. unit is mol dm<sup>-3&nbsp;</sup>s<sup>-1</sup>. </p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Active Mass:</strong> </p>



<p>Active mass is the molar concentration per unit volume of that substance. It is denoted by enclosing the symbol or formula of that substance in the square bracket. For solutions, it is expressed in “ moles dm<sup>-3</sup>”. For gases, it is expressed in “ mol dm<sup>-3</sup>” or pressure in the atmosphere (atm) or bar or pascal (Pa). </p>



<p class="has-text-color has-background has-medium-font-size has-luminous-vivid-orange-color has-very-light-gray-background-color"><strong>Factors Affecting the Rate of Chemical Reaction:</strong> </p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Concentration:</strong> </p>



<p>If the concentration of reactants increases then the rate of reaction increases. </p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Pressure:</strong> </p>



<p>Change in pressure plays an important role in gaseous reactions. There can be three types of gaseous reactions: </p>



<h4 class="wp-block-heading"><strong>Reaction with the increase&nbsp; in volume:</strong> </h4>



<p>The reduction in pressure for a gaseous reaction accompanied by an increase in volume increases the rate of reaction. </p>



<p class="has-text-align-center">e.g.&nbsp; PCI<sub>5(g) &nbsp;</sub>&nbsp; ⇌ &nbsp; &nbsp;PCl<sub>3(g)</sub> &nbsp; + &nbsp; &nbsp; Cl<sub>2(g)</sub> </p>



<h4 class="wp-block-heading"><strong>Reaction with the decrease in volume:</strong> </h4>



<p>The increase in pressure for a gaseous reaction accompanied by a decrease in volume increases the rate of reaction. </p>



<p class="has-text-align-center">e.g. N<sub>2(g)</sub> + &nbsp;3 H<sub>2(g)</sub> &nbsp; ⇌ &nbsp; &nbsp;2NH<sub>3(g)</sub> </p>



<h4 class="wp-block-heading"><strong>Reaction with no change in volume:</strong> </h4>



<p>The gaseous reaction of equilibrium, involving no change in volume is independent of pressure. </p>



<p class="has-text-align-center">e.g.&nbsp; H<sub>2(g)</sub> + Cl<sub>2(g)</sub>&nbsp; ⇌ &nbsp;2HCl<sub>(g)</sub> </p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Temperature:</strong></p>



<p>In general rate of reaction increases with increase in temperature. </p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Catalyst:</strong> </p>



<p>In a chemical reaction, catalyst increases the rate of reaction. </p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Light:</strong> </p>



<p>The reactions which take place in the presence of light are called photochemical reactions. Such reactions are influenced by light. </p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Size of particles:</strong></p>



<p>If solid reactants are used in powdered form instead of granular form, the rate of reaction increases due to an increase in the surface area available for reaction</p>



<p class="has-text-color has-background has-medium-font-size has-luminous-vivid-orange-color has-very-light-gray-background-color"><strong>Law of Mass Action:</strong></p>



<p>In 1864, Norwegian chemists Cato Guldberg and Peter Wage put forward the law of mass action</p>



<p>Law of mass action states that &nbsp;“The rate of a chemical reaction is directly proportional to the product of active masses of reactants, at given temperature at that instant”.</p>



<p>Where active mass is molar concentration per unit volume of that substance. &nbsp;It is denoted by enclosing the symbol or the formula of that substance in the square bracket and expressed in mol dm<sup>-3</sup>.</p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Explanation&nbsp;of Law of Mass Action:</strong></p>



<p class="has-text-align-center">Consider a hypothetical reaction </p>



<p class="has-text-align-center">A + B →&nbsp;&nbsp;Products.</p>



<p class="has-text-align-center">According to the law of mass action the rate of reaction </p>



<p class="has-text-align-center">R ∝ [A] [B].</p>



<p class="has-text-align-center">For any general reaction </p>



<p class="has-text-align-center">aA + bB&nbsp;→ Products.</p>



<p class="has-text-align-center">According to the law of mass action the rate &nbsp;of reaction </p>



<p class="has-text-align-center">R ∝ [A]<sup>a</sup> [B]<sup>b</sup></p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Equilibrium mixture:</strong></p>



<p>The mixture of reactants and products formed at the chemical equilibrium state is called equilibrium mixture.</p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Equilibrium Concentrations:</strong></p>



<p>The concentrations of the reactants and products in a chemical equilibrium state for a reversible reaction are called equilibrium concentrations.</p>



<p class="has-text-color has-background has-medium-font-size has-luminous-vivid-orange-color has-very-light-gray-background-color"><strong>The Expression for Equilibrium Constant in Terms of Concentrations:</strong></p>



<p class="has-text-align-center">Consider a hypothetical reversible reaction </p>



<p class="has-text-align-center">A + B &nbsp;⇌ &nbsp;C + D</p>



<p class="has-text-align-center">According to the law of mass action the rate of the&nbsp;forward reaction</p>



<p class="has-text-align-center">R<sub>f</sub> ∝ [A] [B]</p>



<p class="has-text-align-center">∴ &nbsp;R<sub>f&nbsp;</sub> = &nbsp;K<sub>f</sub> [A] [B]</p>



<p class="has-text-align-center">Where K<sub>f &nbsp;</sub>= rate constant for the&nbsp;forward reaction</p>



<p class="has-text-align-center">According to the law of mass action the rate of the&nbsp;backward reaction</p>



<p class="has-text-align-center">R<sub>b</sub> ∝ [C] [D]</p>



<p class="has-text-align-center">∴ &nbsp;R<sub>b</sub> = &nbsp;K<sub>b</sub> [C] [D]</p>



<p class="has-text-align-center">Where K<sub>b</sub> = rate constant for the&nbsp;backward reaction</p>



<p class="has-text-align-center">Now, for a chemical equilibrium,</p>



<p class="has-text-align-center">R<sub>f&nbsp;</sub>= R<sub>b</sub></p>



<p class="has-text-align-center">K<sub>f&nbsp;</sub>[A] [B] = K<sub>b</sub> [C] [D]</p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img loading="lazy" decoding="async" width="116" height="98" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-04.png" alt="" class="wp-image-11040"/></figure></div>



<p class="has-text-align-center">Where K<sub>C</sub> is known as equilibrium constant and the equation is called “mass law equation”.</p>



<p class="has-text-align-center">Consider a general reversible reaction</p>



<p class="has-text-align-center">aA + bB + cC + … <strong>&nbsp;&nbsp;⇌&nbsp;&nbsp;</strong>lL&nbsp; + mM&nbsp; + nN + &#8230;.,&nbsp; &nbsp; Then,</p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img loading="lazy" decoding="async" width="159" height="51" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-05.png" alt="Law of Mass Action" class="wp-image-11043"/></figure></div>



<p>The equilibrium constant ‘ K<sub>C</sub>‘ is defined as the ratio of the product of the equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric coefficients of the substance in the balanced chemical equation.</p>



<p>Concentration is expressed in terms of mol dm<sup>-3</sup>&nbsp;or mol L<sup>-1</sup> or M. The equilibrium constant is denoted by Kc.</p>



<p>More is the numerical value of ‘K’ then more is the concentration of products in comparison with reactants &amp; vis-a-vis.</p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Concentration Quotient of a Chemical Reaction:</strong></p>



<p>At a given temperature, the ratio of the product of the equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric coefficients of the substance in the balanced chemical equation is called concentration ratio or concentration quotient. It is denoted by Q<sub>C</sub>.</p>



<p>At equilibrium, the concentration ratio is equal to the equilibrium constant Kc.</p>



<p>Its significance is that it helps in the prediction of the direction in which the net reaction is proceeding at given concentrations or partial pressures of reactants and products.</p>



<p>The generalization is</p>



<ul class="wp-block-list"><li>If Q<sub>C</sub> &gt; K<sub>C</sub>: The reaction is taking place in a backward direction i.e. in the direction of reactants.</li><li>If Q<sub>C</sub> &lt; K<sub>C</sub>: The reaction is taking place in a forward direction i.e. in the direction of products.</li><li>If Q<sub>C</sub> = K<sub>C</sub>: The reaction is in an equilibrium state and hence no net reaction is taking place.</li></ul>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Law of Chemical Equilibrium:</strong></p>



<p>At a given temperature, the ratio of the product of the equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric coefficients of the substance in the balanced chemical equation has a constant value.</p>



<p class="has-text-color has-background has-medium-font-size has-luminous-vivid-orange-color has-very-light-gray-background-color"><strong>The Expression for Equilibrium Constant in Terms of Partial Pressure:</strong></p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>Partial Pressure:</strong></p>



<p>It is the pressure exerted by a gas in a mixture of gases if it alone occupies the entire volume of the mixture of gases. The partial pressure of a gaseous component is proportional to the mole fraction. The partial pressure of a gas is calculated using the following formula</p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img loading="lazy" decoding="async" width="235" height="47" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-06.png" alt="" class="wp-image-11046"/></figure></div>



<p class="has-text-align-center">Where, n = Number of moles of gaseous component</p>



<p class="has-text-align-center">N = Total moles of a gaseous system</p>



<p class="has-text-align-center">P = Total pressure of the gaseous system</p>



<p class="has-text-color has-medium-font-size has-vivid-red-color"><strong>The Expression for Equilibrium Constant:</strong></p>



<p>In the gaseous system, the concentrations in concentration quotients are replaced by partial pressure, because for given temperature the partial pressure of the gas is directly proportional to its concentration.</p>



<p>Consider a hypothetical reversible reaction</p>



<p class="has-text-align-center">aA<sub>(g)</sub> + bB<sub>(g)</sub> ⇌&nbsp; cC(g) + dD<sub>(g)</sub></p>



<p class="has-text-align-center">Then the equilibrium constant in terms of partial pressures is given by</p>



<div class="wp-block-image"><figure class="aligncenter size-large"><img loading="lazy" decoding="async" width="134" height="68" src="https://thefactfactor.com/wp-content/uploads/2020/04/Chemical-Equilibrium-07.png" alt="" class="wp-image-11048"/></figure></div>



<h4 class="wp-block-heading"><strong>Science &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/" target="_blank">Chemistry</a> &gt; <a rel="noreferrer noopener" href="https://thefactfactor.com/chemistry/chemical-equilibrium/" target="_blank">Chemical Equilibrium</a> &gt; Law of Mass Action</strong></h4>
<p>The post <a href="https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/law-of-mass-action/11035/">Law of Mass Action</a> appeared first on <a href="https://thefactfactor.com">The Fact Factor</a>.</p>
]]></content:encoded>
					
					<wfw:commentRss>https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/law-of-mass-action/11035/feed/</wfw:commentRss>
			<slash:comments>0</slash:comments>
		
		
			</item>
	</channel>
</rss>
