// Numbas version: finer_feedback_settings
{"name": "Chemical Thermodynamics: Van't Hoff Equation 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Chemical Thermodynamics: Van't Hoff Equation 2", "tags": [], "metadata": {"description": "<p>Public</p>", "licence": "Creative Commons Attribution-NonCommercial-NoDerivs 4.0 International"}, "statement": "", "advice": "<p>The Van't Hoff equation is used to relate the equilibrium constant of a reaction at one temperature with that at another temperature:</p>\n<p>$\\mathrm{ln(K_2)-ln(K_1)=\\frac{\\Delta H^\\circ}{R}[(\\frac{1}{T_1})-(\\frac{1}{T_2})]}$ ,</p>\n<p>where $\\mathrm{T_1}$ and $\\mathrm{T_2}$ are the initial and final temperatures, in $\\mathrm{K}$ ;&nbsp; &nbsp;$\\mathrm{K_1}$ and $\\mathrm{K_2}$ are the corresponding initial and final equilibrium constants; $\\mathrm{\\Delta H^\\circ}$ is the standard enthalpy of the reaction; and $\\mathrm{R=8.314\\space J\\space K^{-1}\\space mol^{-1}}$ is the universal gas constant.</p>\n<p></p>\n<p>We are given an initial standard Gibbs free energy of formation of $\\mathrm{NO}$, $\\mathrm{{\\Delta G^\\circ}_f = +\\var{siground(g2,5)}\\space kJ\\space mol^{-1}}$ at a temperature of $\\mathrm{T_1=298 \\space K}$, and asked to find the new equilibrium constant, $\\mathrm{K_2}$, at temperature $\\mathrm{T_2=\\var{temp} \\space K}$.</p>\n<p>To use the Van't Hoff equation, we must obtain the initial equilibrium constant for the reaction, $\\mathrm{K_1}$, from $\\mathrm{{\\Delta G^\\circ}_f}$. We do this using the following equation, which we adapt to our specific situation and rearrange:</p>\n<p>$\\mathrm{\\Delta G^\\circ=-RTln(K)}$</p>\n<p></p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space \\Delta G^\\circ=-RT_1ln(K_1)}$</p>\n<p></p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space ln(K_1)=-\\frac{{\\Delta G^\\circ}}{RT_1}}$</p>\n<p></p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space K_1=exp(-\\frac{{\\Delta G^\\circ}}{RT_1})}$ .</p>\n<p></p>\n<p></p>\n<p>Before we substitute in our values, we must consider $\\mathrm{\\Delta G^\\circ}$ carefully - we have been given $\\mathrm{{\\Delta G^\\circ}_f=+\\var{siground(g2,5)}\\space kJ\\space mol^{-1}}$, the <strong>standard Gibbs free energy of formation</strong> of $\\mathrm{NO}$, i.e. the Gibbs free energy change accompanying the formation of <strong>one mole</strong> of $\\mathrm{NO}$, but our reaction yields <strong>two moles</strong> of $\\mathrm{NO}$. Thus, we must <strong>double</strong> the given value, then of course convert it from $\\mathrm{kJ}$ to $\\mathrm{J}$ by multiplying by one thousand, to $\\mathrm{\\Delta G^\\circ =+\\var{siground(gibbs,5)} \\space J \\space mol^{-1}}$. We can now substitute our values into our equation:</p>\n<p>$\\mathrm{K_1=exp(-\\frac{{\\Delta G^\\circ}}{RT_1})}$</p>\n<p></p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space K_1=exp(-\\frac{{\\var{siground(gibbs,5)}}}{8.314\\times 298})=\\var{scientificnumberlatex(k)}}$ .</p>\n<p></p>\n<p>We are also given the reaction's standard enthalpy, $\\mathrm{\\Delta H^\\circ=\\var{h}\\space kJ\\space mol^{-1}}$, which we convert by multiplying by one thousand, to $\\mathrm{\\Delta H^\\circ =\\var{enthalpy} \\space J \\space mol^{-1}}$. Unlike equilibrium constants and Gibbs free energy changes, reaction enthalpies are relatively independent of temperature, and so for these types of questions we can consider $\\mathrm{\\Delta H^\\circ}$ to be constant over the given temperature change.</p>\n<p><strong>Note</strong>: We do <strong>not</strong> need to change our reaction enthalpy to accommodate the fact that <strong>two moles</strong> of $\\mathrm{NO}$ are formed in the reaction shown - the value given to us is a standard enthalpy <strong>for the reaction shown</strong>;&nbsp;it is <strong>not the standard enthalpy of formation</strong> of $\\mathrm{NO}$, which would be half the value given to us, as the enthalpy of formation of $\\mathrm{NO}$ is explicitly defined as the enthalpy change for the reaction yielding <strong>one mole</strong> of $\\mathrm{NO}$.</p>\n<p></p>\n<p>To calculate $\\mathrm{K_2}$, we must rearrange the Van't Hoff equation appropriately, then simply substitute in our given values:</p>\n<p>$\\mathrm{ln(K_2)-ln(K_1)=\\frac{\\Delta H^\\circ}{R}([\\frac{1}{T_1}]-[\\frac{1}{T_2}])}$</p>\n<p></p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space ln(K_2)=\\frac{\\Delta H^\\circ}{R}([\\frac{1}{T_1}]-[\\frac{1}{T_2}])+ln(K_1)}$</p>\n<p></p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space K_2=exp[\\frac{\\Delta H^\\circ}{R}([\\frac{1}{T_1}]-[\\frac{1}{T_2}])+ln(K_1)]}$</p>\n<p></p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space K_2=exp[\\frac{\\var{enthalpy}}{8.314}\\times([\\frac{1}{298}]-[\\frac{1}{\\var{temp}}])+ln(\\var{scientificnumberlatex(siground(k,3))})]=\\var{scientificnumberlatex(siground(k2,3))}}$ .</p>", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"k": {"name": "k", "group": "Ungrouped variables", "definition": "random(0.000001 .. 0.00009#1e - 7)", "description": "", "templateType": "randrange", "can_override": false}, "temp": {"name": "temp", "group": "Ungrouped variables", "definition": "random(1400 .. 1700#1)", "description": "", "templateType": "randrange", "can_override": false}, "k2": {"name": "k2", "group": "Ungrouped variables", "definition": "exp((enthalpy/8.314)*((1/298)-(1/temp))+ln(k))", "description": "", "templateType": "anything", "can_override": false}, "enthalpy": {"name": "enthalpy", "group": "Ungrouped variables", "definition": "random(115500 .. 130500#1000)", "description": "", "templateType": "randrange", "can_override": false}, "h": {"name": "h", "group": "Ungrouped variables", "definition": "enthalpy/1000", "description": "", "templateType": "anything", "can_override": false}, "g2": {"name": "g2", "group": "Ungrouped variables", "definition": "gibbs/2000", "description": "", "templateType": "anything", "can_override": false}, "gibbs": {"name": "gibbs", "group": "Ungrouped variables", "definition": "-8.314*298*ln(k)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "1000"}, "ungrouped_variables": ["k", "temp", "k2", "enthalpy", "h", "g2", "gibbs"], "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 standard Gibbs free energy of formation for nitric oxide is found to be $\\mathrm{{\\Delta G^\\circ}_f = +\\var{siground(g2,5)}\\space kJ\\space mol^{-1}}$ at $\\mathrm{298 K}$.</p>\n<p>The reaction&nbsp; &nbsp;$\\mathrm{N_2\\space +\\space O_2\\space \\rightleftharpoons\\space 2NO}$&nbsp; &nbsp;has standard enthalpy $\\mathrm{&Delta;H^\\circ = +\\var{h} \\space kJ \\space mol^{-1}}$. For this reaction, calculate $\\mathrm{K}$ at $\\mathrm{\\var{temp} K}$:</p>\n<p></p>\n<p><em>To enter your answer in scientific notation, use the following convention: a&times;10<sup>-b</sup></em><sup>&nbsp;</sup><em>should be input as \"ae-b\".</em></p>\n<p>[[0]]</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": "k2-(k2/200)", "maxValue": "k2+(k2/200)", "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", "scientific"], "correctAnswerStyle": "scientific"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "contributors": [{"name": "Frances Docherty", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4059/"}, {"name": "Tess Lynn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/16608/"}, {"name": "Michael McFadden", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18132/"}], "resources": []}]}], "contributors": [{"name": "Frances Docherty", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4059/"}, {"name": "Tess Lynn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/16608/"}, {"name": "Michael McFadden", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18132/"}]}