Chemical Thermodynamics: Equilibrium Partial Pressures from Gibbs Free Energy of Formation

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Created this as a copy of Equilibrium Partial Pressures from Gibbs Free Energy of Formation.

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			Name	Type	Generated Value

i2	number	0.4
h2	number	0.7
temp	integer	713
gibbs	number	-11.57
k	number	49.5799478564
a	number	45.5799478564
b	number	-54.537942642
c	number	13.8823853998
xplus	number	0.8292451964
xminus	number	0.3672884814
finali2	number	0.0327115186
finalh2	number	0.3327115186
finalhi	number	0.7345769628
g2	number	-23140

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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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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 $\mathrm{\var{temp}\space K}$ . The standard Gibbs free energy of formation of $\mathrm{HI}$ at $\mathrm{\var{temp}\space K}$ is $\mathrm{\var{gibbs}\space kJ\space mol^{-1}}$ .

What is the equilibrium partial pressure of each of the three gases?

$\mathrm{I_2}$ : Gap 0. $\mathrm{bar}$

$\mathrm{H_2}$ : Gap 1. $\mathrm{bar}$

$\mathrm{HI}$ : Gap 2. $\mathrm{bar}$ .

Advice

The reaction of diatomic iodine and diatomic hydrogen to form hydrogen iodide is given by:

$\mathrm{I_2\space + \space H_2 \space \rightleftharpoons \space 2HI}$ .

As the reaction proceeds from the initial mixture, the partial pressures of the reactants will decrease and that of the product will increase. Let's denote the change in partial pressure of $\mathrm{I_2}$ (or $\mathrm{H_2}$ ) or as $x$ (with units of $\mathrm{bar}$ ), from the initial partial pressure to the final partial pressure at equilibrium. 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 two moles of $\mathrm{HI}$ are formed for every one mole of $\mathrm{I_2}$ used, the pressure increase for $\mathrm{HI}$ will be $2x$ .

Thus, we can tabulate the expressions for the initial partial pressure, final partial pressure, and partial pressure change for each species in the reaction:

$\mathrm{I_2}$ $\mathrm{H_2}$ $\mathrm{HI}$

$\mathrm{Initial \space Pressure\space (bar)}$ $\mathrm{\var{i2}}$ $\mathrm{\var{h2}}$ $\mathrm{0}$

$\mathrm{Pressure \space Change\space (bar)}$ $-x$ $-x$ $+2x$

$\mathrm{Final \space Pressure\space (bar)}$ $(\mathrm{\var{i2}}-x)$ $(\mathrm{\var{h2}}-x)$ $2x$

Now that we have an expression for each of our final partial pressures, we can substitute these into an expression for the equilibrium constant, $\mathrm{K}$ :

$\mathrm{K=\frac{p_{HI}^2}{p_{I_2}p_{H_2}}}$

$\Rightarrow\space\space\space \mathrm{K=}\frac{(2x)^2}{(\mathrm{\var{i2}}-x)(\mathrm{\var{h2}}-x)}=\frac{4\space x^2}{(\mathrm{\var{i2}}-x)(\mathrm{\var{h2}}-x)}$ .

Because there are $x^2$ and $x$ terms, we can rearrange this expression into a quadratic form; $ax^2+bx+c=0$

Note: 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.

This rearrangement is given as follows:

$\mathrm{K=}\frac{4\space x^2}{(\mathrm{\var{i2}}-x)(\mathrm{\var{h2}}-x)}$

$\Rightarrow\space\space\space \mathrm{K}(\mathrm{\var{i2}}-x)(\mathrm{\var{h2}}-x)=4x^2$

$\Rightarrow\space\space\space \mathrm{K}([\var{i2}\times\var{h2}]+[-\var{h2}x]+[-\var{i2}x]+x^2)=4x^2$

$\Rightarrow\space\space\space \mathrm{K}([\var{i2}\times\var{h2}]+[-\var{h2}-\var{i2}]x+x^2)=4x^2$

$\Rightarrow\space\space\space \mathrm{K}[\var{i2}\times\var{h2}]+\mathrm{K}[-\var{h2}-\var{i2}]x+\mathrm{K}x^2=4x^2$

$\Rightarrow\space\space\space \mathrm{K}[\var{i2}\times\var{h2}]+\mathrm{K}[-\var{h2}-\var{i2}]x+[\mathrm{K}-4]x^2=0$

$\Rightarrow\space\space\space [\mathrm{K}-4]x^2+\mathrm{K}[-\var{h2}-\var{i2}]x+\mathrm{K}[\var{i2}\times\var{h2}]=0$ .

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:

$a=\mathrm{K}-4$ ,

$b=\mathrm{K}[-\var{h2}-\var{i2}]$ ,

$c=\mathrm{K}[\var{i2}\times\var{h2}]$ .

This is an appropriate point at which to calculate the value of $\mathrm{K}$ . We do this using the following equation, which we rearrange:

$\mathrm{\Delta G^\circ=-RTln(K)}$

$\mathrm{\Rightarrow\space\space\space ln(K)=-\frac{{\Delta G^\circ}}{RT}}$

$\mathrm{\Rightarrow\space\space\space K=exp(-\frac{{\Delta G^\circ}}{RT})}$ .

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 standard Gibbs free energy of formation of $\mathrm{HI}$ , i.e. the Gibbs free energy change accompanying the formation of one mole of $\mathrm{HI}$ , but our reaction yields two moles of $\mathrm{HI}$ . Thus, we must double the given value, then of course convert it from $\mathrm{kJ}$ to $\mathrm{J}$ by multiplying by one thousand, to $\mathrm{\Delta G^\circ =\var{g2} \space J \space mol^{-1}}$ .

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:

$\mathrm{K=exp(-\frac{{\Delta G^\circ}}{RT})}$

$\mathrm{\Rightarrow\space\space\space K=exp(-\frac{{\var{g2}}}{8.314\times \var{temp}})=(\var{(sigformat(k,3))}\cdots)}$ .

Thus, we can calculate actual values for $a$ , $b$ , and $c$ :

$a=\mathrm{K}-4=(\var{(sigformat(k,3))}\cdots)-4=(\var{(sigformat(a,3))}\cdots)$ ,

$b=\mathrm{K}[-\var{h2}-\var{i2}]=(\var{(sigformat(k,3))}\cdots)\times[-\var{h2}-\var{i2}]=(\var{(sigformat(b,3))}\cdots)$ ,

$c=\mathrm{K}[\var{i2}\times\var{h2}]=(\var{(sigformat(k,3))}\cdots)\times\var{i2}\times\var{h2}=(\var{(sigformat(c,3))}\cdots)$ .

Now, we can substitute these values into the quadratic formula to solve for $x$ :

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\Rightarrow\space\space\space x=\frac{-(\var{(sigformat(b,3))}\cdots)\pm\sqrt{(\var{(sigformat(b,3))}\cdots)^2-[4\times (\var{(sigformat(a,3))}\cdots)\times (\var{(sigformat(c,3))}\cdots)]}}{2\times(\var{(sigformat(a,3))}\cdots)}$ .

As is standard when using the quadratic formula, we obtain two solutions:

$x_+=\var{(sigformat(xplus,3))}\space\mathrm{bar}$ , $x_-=\var{(sigformat(xminus,3))}\space\mathrm{bar}$ .

Bearing in mind that $x$ represents the decrease in partial pressure of each reactant, we can note that $x_+$ is greater than the initial partial pressures of our reactants, meaning that a decrease of $x_+$ would result in negative partial pressures, which are impossible; thus, $x_+$ must be a 'non-physical' solution and $x_-$ must be the solution we want - the actual decrease in the partial pressures of the iodine and hydrogen gases. We will hence equate $x$ to $x_-$ , so $x=\var{(sigformat(xminus,3))}\space\mathrm{bar}$ .

Finally, by referring back to the last row of our table, we can calculate the final partial pressure of each of the three species in the reaction:

$\mathrm{p_{f(I_2)}}=\var{i2}-x=\var{i2}-\var{(sigformat(xminus,3))}=\mathrm{\var{(sigformat(finali2,3))}\space bar}$ ,

$\mathrm{p_{f(H_2)}}=\var{h2}-x=\var{h2}-\var{(sigformat(xminus,3))}=\mathrm{\var{(sigformat(finalh2,3))}\space bar}$ ,

$\mathrm{p_{f(HI)}}=2x=2\times\var{(sigformat(xminus,3))}=\mathrm{\var{(sigformat(finalhi,3))}\space bar}$ .

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Chemical Thermodynamics: Equilibrium Partial Pressures from Gibbs Free Energy of Formation

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Chemical Thermodynamics: Equilibrium Partial Pressures from Gibbs Free Energy of Formation

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