// Numbas version: finer_feedback_settings
{"name": "Chemical Thermodynamics: Gibbs Free Energy of Formation 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Chemical Thermodynamics: Gibbs Free Energy of Formation 2", "tags": [], "metadata": {"description": "<p>Public</p>", "licence": "Creative Commons Attribution-NonCommercial-NoDerivs 4.0 International"}, "statement": "", "advice": "<p>The Gibbs free energy of a reacting mixture, $\\mathrm{\\Delta G}$, is given by the following equation:</p>\n<p>$\\mathrm{\\Delta G=\\Delta G^\\circ + RTln(Q)}$,</p>\n<p>where $\\mathrm{\\Delta G^\\circ}$ is the <strong>standard</strong> Gibbs free energy, in $\\mathrm{J \\space mol^{-1}}$;&nbsp; $\\mathrm{R=8.314\\space J \\space K^{-1}\\space mol^{-1}}$ is the universal gas constant;&nbsp; $\\mathrm{T}$ is the temperature of the system, in $\\mathrm{K}$; and $\\mathrm{Q}$ is the reaction quotient, given by:</p>\n<p>$\\mathrm{Q=\\frac{[C]^c[D]^d}{[A]^a[B]^b}}$&nbsp; &nbsp;for the reaction&nbsp; &nbsp; &nbsp; $\\mathrm{aA\\space +\\space bB\\space\\rightleftharpoons\\space cC\\space +\\space dD}$.</p>\n<p></p>\n<p></p>\n<p>We 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:</p>\n<p>$\\mathrm{N_2\\space +\\space 3H_2\\space\\rightleftharpoons\\space 2NH_3}$ .</p>\n<p></p>\n<p>We must, however, pay attention to the fact that we will eventually be using $\\mathrm{\\Delta G^\\circ}$, for the <strong>standard</strong> formation of ammonia, i.e. the formation of <strong>1 mole&nbsp;</strong>of ammonia under standard conditions. Thus, we half the stoichiometries of our equation:</p>\n<p>$\\mathrm{0.5N_2\\space +\\space 1.5H_2\\space\\rightleftharpoons\\space NH_3}$ .</p>\n<p></p>\n<p>This gives us a reaction quotient of:</p>\n<p>$\\mathrm{Q=\\frac{[NH_3]}{[N_2]^{0.5}[H_2]^{1.5}}}$.</p>\n<p>Since all three substances are gaseous at $\\mathrm{298\\space K}$, we swap out their 'concentrations' for their <strong>partial pressures</strong>:</p>\n<p>$\\mathrm{Q=\\frac{p_{_{NH_3}}}{p_{_{N_2}}^{0.5}p_{_{H_2}}^{1.5}}}$ .</p>\n<p></p>\n<p>Substituting this expression for $\\mathrm{Q}$ into our original equation, we now have:</p>\n<p>$\\mathrm{\\Delta G=\\Delta G^\\circ + RTln(\\frac{p_{_{NH_3}}}{p_{_{N_2}}^{0.5}p_{_{H_2}}^{1.5}})}$ .</p>\n<p></p>\n<p></p>\n<p>We 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 $\\mathrm{p_{_{H_2}}=\\var{dpformat(h2,1)}\\space bar}$ .</p>\n<p>We are also given the equilibrium constant for the standard formation of ammonia, $\\mathrm{k=\\var{scientificnumberlatex(siground(k,5))}}$. From this, we can calculate the standard Gibbs free energy of formation, using the formula:</p>\n<p>$\\mathrm{\\Delta G^\\circ =-RTln(k)}$</p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space\\Delta G^\\circ =-8.314\\times 298\\times ln(\\var{scientificnumberlatex(siground(k,5))})=\\var{standard} \\space J \\space mol^{-1}}$</p>\n<p></p>\n<p>We can thus calculate $\\mathrm{\\Delta G}$ for our reaction:</p>\n<p></p>\n<p>$\\mathrm{\\Delta G=\\var{standard} + (8.314\\times 298\\times ln[\\frac{\\var{dpformat(nh3,1)}}{{\\var{dpformat(n2,1)}}^{0.5}\\times{\\var{dpformat(h2,1)}}^{1.5}}])=\\var{sigformat(gibbs,3)}\\space J\\space mol^{-1}}$ .</p>\n<p></p>\n<p>To quote our answer in the units asked for, we divide by one thousand to obtain:</p>\n<p>$\\mathrm{\\Delta G=\\var{sigformat(gibbskj,3)}\\space kJ\\space mol^{-1}}$ .</p>\n<p></p>\n<p></p>\n<p>The Gibbs free energy we have calculated is <strong>negative</strong>, which tells us that the reaction will proceed spontaneously in the <strong>forward</strong> direction, i.e. ammonia will be formed. Thus, the quantity of ammonia will <strong>increase</strong>.</p>", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"n2": {"name": "n2", "group": "Ungrouped variables", "definition": "random(0.5 .. 4.5#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "h2": {"name": "h2", "group": "Ungrouped variables", "definition": "random(0.5 .. 4.5#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "nh3": {"name": "nh3", "group": "Ungrouped variables", "definition": "random(0.5 .. 4.5#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "q": {"name": "q", "group": "Ungrouped variables", "definition": "nh3/((n2^0.5)*(h2^1.5))", "description": "", "templateType": "anything", "can_override": false}, "standard_kj": {"name": "standard_kj", "group": "Ungrouped variables", "definition": "random(-17.5 .. -15.5#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "gibbs": {"name": "gibbs", "group": "Ungrouped variables", "definition": "(standard_kj*1000)+(298*8.314*ln(q))", "description": "", "templateType": "anything", "can_override": false}, "gibbskj": {"name": "gibbskj", "group": "Ungrouped variables", "definition": "gibbs/1000", "description": "", "templateType": "anything", "can_override": false}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "exp((-standard_kj*1000)/(8.314*298))", "description": "", "templateType": "anything", "can_override": false}, "standard": {"name": "standard", "group": "Ungrouped variables", "definition": "standard_kj*1000", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "k>1", "maxRuns": "1000"}, "ungrouped_variables": ["n2", "h2", "nh3", "q", "standard_kj", "gibbs", "gibbskj", "k", "standard"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "<p>The equilibrium constant for the standard formation of ammonia is found to be $\\mathrm{\\var{scientificnumberlatex(siground(k,5))}}$ at $\\mathrm{298\\space K}$.</p>\n<p>What 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?</p>\n<p></p>\n<p>[[0]] $\\mathrm{kJ\\space mol^{-1}}$</p>\n<p></p>\n<p>Will the quantity of ammonia increase or decrease in this case?</p>\n<p></p>\n<p>[[1]]</p>", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "3", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "gibbskj-(gibbskj/200)", "maxValue": "gibbskj+(gibbskj/200)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "sigfig", "precision": "3", "precisionPartialCredit": "50", "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": true, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "1_n_2", 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