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Public
", "licence": "Creative Commons Attribution-NonCommercial-NoDerivs 4.0 International"}, "statement": "", "advice": "The Gibbs free energy of a reacting mixture, $\\mathrm{\\Delta G}$, is given by the following equation:
\n$\\mathrm{\\Delta G=\\Delta G^\\circ + RTln(Q)}$,
\nwhere $\\mathrm{\\Delta G^\\circ}$ is the standard Gibbs free energy, in $\\mathrm{J \\space mol^{-1}}$; $\\mathrm{R=8.314\\space J \\space K^{-1}\\space mol^{-1}}$ is the universal gas constant; $\\mathrm{T}$ is the temperature of the system, in $\\mathrm{K}$; and $\\mathrm{Q}$ is the reaction quotient, given by:
\n$\\mathrm{Q=\\frac{[C]^c[D]^d}{[A]^a[B]^b}}$ for the reaction $\\mathrm{aA\\space +\\space bB\\space\\rightleftharpoons\\space cC\\space +\\space dD}$.
\n\n\nWe are asked to calculate $\\mathrm{\\Delta G}$ for the standard formation of ammonia. By considering the stoichiometries of hydrogen and nitrogen within ammonia, we can fairly easily deduce the chemical equation for this formation:
\n$\\mathrm{N_2\\space +\\space 3H_2\\space\\rightleftharpoons\\space 2NH_3}$ .
\n\nWe must, however, pay attention to the fact that we will eventually be using the $\\mathrm{\\Delta G^\\circ}$, for the standard formation of ammonia, i.e. the formation of 1 mole of ammonia under standard conditions. Thus, we half the stoichiometries of our equation:
\n$\\mathrm{0.5N_2\\space +\\space 1.5H_2\\space\\rightleftharpoons\\space NH_3}$ .
\n\nThis gives us a reaction quotient of:
\n$\\mathrm{Q=\\frac{[NH_3]}{[N_2]^{0.5}[H_2]^{1.5}}}$.
\nSince all three substances are gaseous at $\\mathrm{298\\space K}$, we swap out their 'concentrations' for their partial pressures:
\n$\\mathrm{Q=\\frac{p_{_{NH_3}}}{p_{_{N_2}}^{0.5}p_{_{H_2}}^{1.5}}}$ .
\n\nSubstituting this expression for $\\mathrm{Q}$ into our original equation, we now have:
\n$\\mathrm{\\Delta G=\\Delta G^\\circ + RTln(\\frac{p_{_{NH_3}}}{p_{_{N_2}}^{0.5}p_{_{H_2}}^{1.5}})}$ .
\n\n\nWe are given the temperature of the system, $\\mathrm{T=298\\space K}$; and the partial pressures of all three gases, $\\mathrm{p_{_{NH_3}}=\\var{dpformat(nh3,1)}\\space bar}$, $\\mathrm{p_{_{N_2}}=\\var{dpformat(n2,1)}\\space bar}$, and $\\mathrm{p_{_{H_2}}=\\var{dpformat(h2,1)}\\space bar}$ .
\nWe are also given the standard enthalpy and entropy for the formation of ammonia: $\\mathrm{\\Delta H^\\circ=\\var{enthalpy} \\space kJ \\space mol^{-1}}$, which we convert by multiplying by one thousand, to $\\mathrm{\\Delta H^\\circ =\\var{enth} \\space J \\space mol^{-1}}$; and $\\mathrm{\\Delta S^\\circ=\\var{sigformat(entropy,4)} \\space J \\space mol^{-1}}$. From these values, we can calculate the standard Gibbs free energy of formation, using the formula:
\n$\\mathrm{\\Delta G^\\circ =\\Delta H^\\circ - T\\Delta S^\\circ}$
\n$\\mathrm{\\Rightarrow\\space\\space\\space\\Delta G^\\circ =\\var{enth}-(298\\times \\var{entropy})=\\var{standard} \\space J \\space mol^{-1}}$
\n\nWe can thus calculate $\\mathrm{\\Delta G}$ for our reaction:
\n\n$\\mathrm{\\Delta G={standard} + (8.314\\times 298\\times ln[\\frac{\\var{dpformat(nh3,1)}}{{\\var{dpformat(n2,1)}}^{0.5}\\times{\\var{dpformat(h2,1)}}^{1.5}}])=\\var{sigformat(gibbs,3)}\\space J\\space mol^{-1}}$ .
\n\nTo quote our answer in the units asked for, we divide by one thousand to obtain:
\n$\\mathrm{\\Delta G=\\var{sigformat(gibbskj,3)}\\space kJ\\space mol^{-1}}$ .
\n\n\nThe Gibbs free energy we have calculated is negative, which tells us that the reaction will proceed spontaneously in the forward direction, i.e. ammonia will be formed. Thus, the quantity of ammonia will increase.
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\nWhat is the Gibbs free energy of formation of ammonia when the partial pressures of the $\\mathrm{N_2}$, $\\mathrm{H_2}$ and $\\mathrm{NH_3}$ (treated as perfect gases) are $\\mathrm{\\var{dpformat(n2,1)}\\space bar}$, $\\mathrm{\\var{dpformat(h2,1)}\\space bar}$, and $\\mathrm{\\var{dpformat(nh3,1)}\\space bar}$ respectively?
\n\n[[0]] $\\mathrm{kJ\\space mol^{-1}}$
\n\nWill the quantity of ammonia increase or decrease in this case?
\n\n[[1]]
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