// Numbas version: finer_feedback_settings
{"name": "Chemical Thermodynamics: Equilibrium Partial Pressures from Gibbs Free Energy of Formation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Chemical Thermodynamics: Equilibrium Partial Pressures from Gibbs Free Energy of Formation", "tags": [], "metadata": {"description": "<p>Public</p>", "licence": "Creative Commons Attribution-NonCommercial-NoDerivs 4.0 International"}, "statement": "", "advice": "<p>The reaction of diatomic iodine and diatomic hydrogen to form hydrogen iodide is given by:</p>\n<p>$\\mathrm{I_2\\space + \\space H_2 \\space \\rightleftharpoons \\space 2HI}$ .</p>\n<p></p>\n<p>As the reaction proceeds from the initial mixture, the partial pressures of the reactants will <strong>decrease</strong> and that of the product will <strong>increase</strong>. Let's denote the <strong>change</strong> in partial pressure of $\\mathrm{I_2}$ (or $\\mathrm{H_2}$) or as $x$ (with units of $\\mathrm{bar}$), from the <strong>initial</strong> partial pressure to the <strong>final</strong> partial pressure at <strong>equilibrium</strong>. Note that there is initially no $\\mathrm{HI}$ present, so it will have an initial partial pressure of $\\mathrm{0 \\space bar}$ in the mixture. As <strong>two</strong> moles of $\\mathrm{HI}$ are formed for every <strong>one</strong> mole of $\\mathrm{I_2}$ used, the pressure increase for $\\mathrm{HI}$ will be $2x$.</p>\n<p></p>\n<p></p>\n<p>Thus, we can tabulate the expressions for the initial partial pressure, final partial pressure, and partial pressure change for each species in the reaction:</p>\n<p></p>\n<p></p>\n<p></p>\n<p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $\\mathrm{I_2}$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $\\mathrm{H_2}$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $\\mathrm{HI}$</p>\n<p></p>\n<p>$\\mathrm{Initial \\space Pressure\\space (bar)}$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $\\mathrm{\\var{i2}}$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $\\mathrm{\\var{h2}}$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp;&nbsp; &nbsp; &nbsp; $\\mathrm{0}$</p>\n<p></p>\n<p>$\\mathrm{Pressure \\space Change\\space (bar)}$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;$-x$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;$-x$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp; &nbsp; &nbsp; $+2x$</p>\n<p></p>\n<p>$\\mathrm{Final \\space Pressure\\space (bar)}$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $(\\mathrm{\\var{i2}}-x)$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $(\\mathrm{\\var{h2}}-x)$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;$2x$</p>\n<p></p>\n<p></p>\n<p></p>\n<p>Now that we have an expression for each of our final partial pressures, we can substitute these into an expression for the <strong>equilibrium constant</strong>, $\\mathrm{K}$:</p>\n<p></p>\n<p>$\\mathrm{K=\\frac{p_{HI}^2}{p_{I_2}p_{H_2}}}$</p>\n<p></p>\n<p>$\\Rightarrow\\space\\space\\space \\mathrm{K=}\\frac{(2x)^2}{(\\mathrm{\\var{i2}}-x)(\\mathrm{\\var{h2}}-x)}=\\frac{4\\space x^2}{(\\mathrm{\\var{i2}}-x)(\\mathrm{\\var{h2}}-x)}$ .</p>\n<p></p>\n<p></p>\n<p>Because there are $x^2$ and $x$ terms, we can rearrange this expression into a <strong>quadratic form</strong>; $ax^2+bx+c=0$</p>\n<p></p>\n<p><strong>Note</strong>: It is best to keep $\\mathrm{K}$ as an algebraic term during rearranging, then substitue its value in afterwards; thus, we do not need to calculate its value quite yet.</p>\n<p></p>\n<p>This rearrangement is given as follows:</p>\n<p></p>\n<p>$\\mathrm{K=}\\frac{4\\space x^2}{(\\mathrm{\\var{i2}}-x)(\\mathrm{\\var{h2}}-x)}$</p>\n<p></p>\n<p>$\\Rightarrow\\space\\space\\space \\mathrm{K}(\\mathrm{\\var{i2}}-x)(\\mathrm{\\var{h2}}-x)=4x^2$</p>\n<p></p>\n<p>$\\Rightarrow\\space\\space\\space \\mathrm{K}([\\var{i2}\\times\\var{h2}]+[-\\var{h2}x]+[-\\var{i2}x]+x^2)=4x^2$</p>\n<p></p>\n<p>$\\Rightarrow\\space\\space\\space \\mathrm{K}([\\var{i2}\\times\\var{h2}]+[-\\var{h2}-\\var{i2}]x+x^2)=4x^2$</p>\n<p></p>\n<p>$\\Rightarrow\\space\\space\\space \\mathrm{K}[\\var{i2}\\times\\var{h2}]+\\mathrm{K}[-\\var{h2}-\\var{i2}]x+\\mathrm{K}x^2=4x^2$</p>\n<p></p>\n<p>$\\Rightarrow\\space\\space\\space \\mathrm{K}[\\var{i2}\\times\\var{h2}]+\\mathrm{K}[-\\var{h2}-\\var{i2}]x+[\\mathrm{K}-4]x^2=0$</p>\n<p></p>\n<p>$\\Rightarrow\\space\\space\\space [\\mathrm{K}-4]x^2+\\mathrm{K}[-\\var{h2}-\\var{i2}]x+\\mathrm{K}[\\var{i2}\\times\\var{h2}]=0$ .</p>\n<p></p>\n<p>We now have our equation in the quadratic form; thus, by comparing this equation with the general expression for the quadratic form $(ax^2+bx+c=0)$, we can deduce that, for our equation:</p>\n<p></p>\n<p>$a=\\mathrm{K}-4$ ,</p>\n<p></p>\n<p>$b=\\mathrm{K}[-\\var{h2}-\\var{i2}]$ ,</p>\n<p></p>\n<p>$c=\\mathrm{K}[\\var{i2}\\times\\var{h2}]$ .</p>\n<p></p>\n<p></p>\n<p>This is an appropriate point at which to calculate the value of $\\mathrm{K}$. We do this using the following equation, which we rearrange:</p>\n<p>$\\mathrm{\\Delta G^\\circ=-RTln(K)}$</p>\n<p></p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space ln(K)=-\\frac{{\\Delta G^\\circ}}{RT}}$</p>\n<p></p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space K=exp(-\\frac{{\\Delta G^\\circ}}{RT})}$ .</p>\n<p></p>\n<p>Before we substitute in our values, we must consider $\\mathrm{\\Delta G^\\circ}$ carefully - we have been given $\\mathrm{{\\Delta G^\\circ}_f=\\var{gibbs}\\space kJ\\space mol^{-1}}$, the <strong>standard Gibbs free energy of formation</strong> of $\\mathrm{HI}$, i.e. the Gibbs free energy change accompanying the formation of <strong>one mole</strong> of $\\mathrm{HI}$, but our reaction yields <strong>two moles</strong> of $\\mathrm{HI}$. 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 $\\mathrm{\\Delta G^\\circ =\\var{g2} \\space J \\space mol^{-1}}$.</p>\n<p>Noting that we are also given the temperature of the system, $\\mathrm{T=\\var{temp}\\space K}$, we can now substitute our values into our equation:</p>\n<p>$\\mathrm{K=exp(-\\frac{{\\Delta G^\\circ}}{RT})}$</p>\n<p></p>\n<p>$\\mathrm{\\Rightarrow\\space\\space\\space K=exp(-\\frac{{\\var{g2}}}{8.314\\times \\var{temp}})=(\\var{(sigformat(k,3))}\\cdots)}$ .</p>\n<p></p>\n<p></p>\n<p>Thus, we can calculate actual values for $a$, $b$, and $c$:</p>\n<p>$a=\\mathrm{K}-4=(\\var{(sigformat(k,3))}\\cdots)-4=(\\var{(sigformat(a,3))}\\cdots)$ ,</p>\n<p></p>\n<p>$b=\\mathrm{K}[-\\var{h2}-\\var{i2}]=(\\var{(sigformat(k,3))}\\cdots)\\times[-\\var{h2}-\\var{i2}]=(\\var{(sigformat(b,3))}\\cdots)$ ,</p>\n<p></p>\n<p>$c=\\mathrm{K}[\\var{i2}\\times\\var{h2}]=(\\var{(sigformat(k,3))}\\cdots)\\times\\var{i2}\\times\\var{h2}=(\\var{(sigformat(c,3))}\\cdots)$ .</p>\n<p></p>\n<p></p>\n<p>Now, we can substitute these values into the <strong>quadratic formula</strong> to solve for $x$:</p>\n<p></p>\n<p>$x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$</p>\n<p></p>\n<p>$\\Rightarrow\\space\\space\\space x=\\frac{-(\\var{(sigformat(b,3))}\\cdots)\\pm\\sqrt{(\\var{(sigformat(b,3))}\\cdots)^2-[4\\times (\\var{(sigformat(a,3))}\\cdots)\\times (\\var{(sigformat(c,3))}\\cdots)]}}{2\\times(\\var{(sigformat(a,3))}\\cdots)}$ .</p>\n<p></p>\n<p>As is standard when using the quadratic formula, we obtain <strong>two solutions</strong>:</p>\n<p></p>\n<p>$x_+=\\var{(sigformat(xplus,3))}\\space\\mathrm{bar}$ ,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;$x_-=\\var{(sigformat(xminus,3))}\\space\\mathrm{bar}$ .</p>\n<p></p>\n<p></p>\n<p>Bearing in mind that $x$ represents the <strong>decrease</strong> in partial pressure of each reactant, we can note that $x_+$ is <strong>greater</strong> than the initial partial pressures of our reactants, meaning that a decrease of $x_+$ would result in negative partial pressures, which are <strong>impossible</strong>; thus, $x_+$ must be a '<strong>non-physical</strong>' solution and $x_-$ must be the solution we <strong>want</strong> - the actual decrease in the partial pressures of the iodine and hydrogen gases. We will hence equate $x$ to $x_-$ , so $x=\\var{(sigformat(xminus,3))}\\space\\mathrm{bar}$ .</p>\n<p></p>\n<p>Finally, by referring back to the last row of our table, we can calculate the <strong>final partial pressure</strong> of each of the three species in the reaction:</p>\n<p></p>\n<p>$\\mathrm{p_{f(I_2)}}=\\var{i2}-x=\\var{i2}-\\var{(sigformat(xminus,3))}=\\mathrm{\\var{(sigformat(finali2,3))}\\space bar}$ ,</p>\n<p></p>\n<p>$\\mathrm{p_{f(H_2)}}=\\var{h2}-x=\\var{h2}-\\var{(sigformat(xminus,3))}=\\mathrm{\\var{(sigformat(finalh2,3))}\\space bar}$ ,</p>\n<p></p>\n<p>$\\mathrm{p_{f(HI)}}=2x=2\\times\\var{(sigformat(xminus,3))}=\\mathrm{\\var{(sigformat(finalhi,3))}\\space bar}$ .</p>", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"i2": {"name": "i2", "group": "Ungrouped variables", "definition": "random(0.1 .. 0.9#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "h2": {"name": "h2", "group": "Ungrouped variables", "definition": "random(0.1 .. 0.9#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "temp": {"name": "temp", "group": "Ungrouped variables", "definition": "random(650 .. 750#1)", "description": "", "templateType": "randrange", "can_override": false}, "gibbs": {"name": "gibbs", "group": "Ungrouped variables", "definition": "-11.57", "description": "", "templateType": "number", "can_override": false}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "exp(-(g2)/(8.314*temp))", "description": "", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "k-4", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "-k*(i2+h2)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "k*i2*h2", "description": "", "templateType": "anything", "can_override": false}, "xplus": {"name": "xplus", "group": "Ungrouped variables", "definition": "(-b+sqrt((b^2)-(4*a*c)))/(2*a)", "description": "", "templateType": "anything", "can_override": false}, "xminus": {"name": "xminus", "group": "Ungrouped variables", "definition": "(-b-sqrt((b^2)-(4*a*c)))/(2*a)", "description": "", "templateType": "anything", "can_override": false}, "finali2": {"name": "finali2", "group": "Ungrouped variables", "definition": "i2-xminus", "description": "", "templateType": "anything", "can_override": false}, "finalh2": {"name": "finalh2", "group": "Ungrouped variables", "definition": "h2-xminus", "description": "", "templateType": "anything", "can_override": false}, "finalhi": {"name": "finalhi", "group": "Ungrouped variables", "definition": "2*xminus", "description": "", "templateType": "anything", "can_override": false}, "g2": {"name": "g2", "group": "Ungrouped variables", "definition": "gibbs*2000", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "1000"}, "ungrouped_variables": ["i2", "h2", "temp", "gibbs", "k", "a", "b", "c", "xplus", "xminus", "finali2", "finalh2", "finalhi", "g2"], "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>In an experiment, an initial mixture of $\\mathrm{\\var{i2}\\space bar}$ $\\mathrm{I_2}$ and $\\mathrm{\\var{h2}\\space bar}$ $\\mathrm{H_2}$ reacts to form an equilibrium mixture of the two reactant gases and their product, $\\mathrm{HI}$, at $\\mathrm{\\var{temp}\\space K}$. 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