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Public
", "licence": "Creative Commons Attribution-NonCommercial-NoDerivs 4.0 International"}, "statement": "", "advice": "The chemical potential of a substance in a mixture, $\\mathrm{\\mu_x}$, is the contribution of that substance to the total Gibbs free energy, $\\mathrm{G}$, of the mixture. It is effectively the molar Gibbs free energy inherent to that substance under a particular set of conditions.
\nAs a substance is formed, more of its chemical potential is introduced - and as a substance is consumed, its chemical potential is removed - from the total $\\mathrm{G}$ value. Since $\\mathrm{\\mu_x}$ is molar (i.e. it results from one mole of the substance), the change in $\\mathrm{G}$ when a substance is formed or consumed is proportional to the number of moles of the substance formed or consumed which we denote $\\mathrm{\\Delta n}$:
\n$\\mathrm{\\Delta G=\\Delta n\\cdot \\mu_x}$
\n\nWe are told that $\\mathrm{\\Delta n}$ moles of $\\mathrm{O_2}$ are used up in the reaction:
\n$\\mathrm{2SO_2\\space +\\space O_2\\space \\rightleftharpoons\\space 2SO_3}$.
\n\nSince $\\mathrm{\\Delta n}$ moles are consumed, we will re-express the change in moles of $\\mathrm{O_2}$ as $\\mathrm{-\\Delta n}$.
\nFrom the reaction, we can see that twice as much $\\mathrm{SO_2}$ is consumed, so the change in moles of $\\mathrm{SO_2}$ is $\\mathrm{-2\\Delta n}$.
\nSimilarly, we can see that twice as much $\\mathrm{SO_3}$ is formed than $\\mathrm{O_2}$ consumed, but we must pay attention to the fact that that much $\\mathrm{SO_3}$ is being formed, so the change in moles of $\\mathrm{SO_3}$ is $\\mathrm{+2\\Delta n}$.
\n\nKnowing all this, we can express the net change in the Gibbs free energy of the mixture in terms of the chemical potentials of the three substances:
\n$\\mathrm{\\Delta G=-2\\Delta n\\mu_{_{SO_2}}-(1)\\Delta n\\mu_{_{O_2}}+2\\Delta n\\mu_{_{SO_3}}}$ .
\n\nAdditionally, $\\mathrm{\\Delta G_{reaction}}$ is defined as the molar Gibbs free energy change (i.e. the change per one mole of specified substance formed or consumed), which we simply divide our previous answer by $\\mathrm{\\Delta n}$ to obtain:
\n$\\mathrm{\\Delta G_{reaction}=\\frac{\\Delta G}{\\Delta n}=-2\\mu_{_{SO_2}}-(1)\\mu_{_{O_2}}+2\\mu_{_{SO_3}}}$ .
\n\n\n\nFinally, we are told that, for the certain set of conditions we are concerned with:
\n$\\mathrm{\\mu_{_{SO_3}}\\var{arrow[rand]}\\space \\mu_{_{SO_2}}+0.5\\mu_{_{O_2}}}$ .
\n\nWe can double this to compare it to our expression for $\\mathrm{\\Delta G_{reaction}}$:
\n$\\mathrm{2\\mu_{_{SO_3}}\\var{arrow[rand]}\\space 2\\mu_{_{SO_2}}+\\mu_{_{O_2}}}$ .
\n\nFrom this, we can deduce whether our $\\mathrm{\\Delta G_{reaction}}$ will be positive or negative:
\n$\\mathrm{\\Delta G_{reaction}=-2\\mu_{_{SO_2}}-\\mu_{_{O_2}}+2\\mu_{_{SO_3}}=2\\mu_{_{SO_3}}-(2\\mu_{_{SO_2}}+\\mu_{_{O_2}})}$ ,
\n\n$\\mathrm{2\\mu_{_{SO_3}}\\var{arrow[rand]}\\space 2\\mu_{_{SO_2}}+\\mu_{_{O_2}}}$
\n\n$\\mathrm{\\Rightarrow\\space\\space\\space 2\\mu_{_{SO_3}}-(2\\mu_{_{SO_2}}+\\mu_{_{O_2}})\\space \\var{arrow[rand]}\\space 0}$
\n\n$\\mathrm{\\Rightarrow\\space\\space\\space \\Delta G_{reaction}\\space \\var{arrow[rand]}\\space 0}$ .
\n\nThus, {outcome[rand]}
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\n$\\mathrm{2SO_2\\space +\\space O_2\\space \\rightleftharpoons\\space 2SO_3}$
\nwhen $\\mathrm{\\Delta n}$ moles of $\\mathrm{O_2}$ are used up?
\nExpress your answer in terms of chemical potentials, $\\mathrm{\\mu_{_X}}$ (your coefficients may be negative).
\n\n$\\mathrm{\\Delta G_{reaction}=}$ [[0]] $\\mathrm{\\mu_{_{SO_2}}} +$ [[1]] $\\mathrm{\\mu_{_{O_2}}} +$ [[2]] $\\mathrm{\\mu_{_{SO_3}}}$ .
\n\n\nGiven that $\\mathrm{\\mu_{_{SO_3}}\\var{arrow[rand]}\\space \\mu_{_{SO_2}}+0.5\\mu_{_{O_2}}}$ at a certain reaction coordinate, will the quantity of $\\mathrm{SO_3}$ increase or decrease?
\n[[3]]
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