Question

The following reaction can be used to make ${\mathbf{H}}_{\mathbf{2}}$for the synthesis of ammonia from the greenhouse gases carbon dioxide and methane:

${\mathbf{CH}}_{\mathbf{4}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{CO}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{CO}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

(a) What is the percent yield of H2 when an equimolar mixture of ${\mathbf{CH}}_{\mathbf{4}}$and ${\mathbf{CO}}_{\mathbf{2}}$with a total pressure of 20.0 atm reaches equilibrium at 1200. K, at which ${\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{548}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$?

(b) What is the percent yield of ${\mathbf{H}}_{\mathbf{2}}$for this system at 1300. K, at which ${\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^7$?

(c) Use the van’t Hoff equation to find $\mathbf{∆}{\mathbf{H}}_{\mathbf{rxn}}^{\mathbf{0}}$ .

Short answer

(a) The percent yield of${\mathrm{H}}_2$ is 98.0%.

(b) The percent yield of${\mathrm{H}}_2$ is 99.2% at 1300 K.

(c) The standard heat Of reaction$∆{\mathrm{H}}_{\mathrm{rxn}}^0$ is $2.60\times {10}^5\mathrm{J}/\mathrm{mol}$.

Step-by-step solution

(a) Percent yield of H2

The equimolar mixture of${\mathrm{CH}}_4$and${\mathrm{CO}}_2$has a total pressure of 20.0 atm. It gives initial pressure of both reactants as${\mathrm{P}}_{{\mathrm{CH}}_4}=10.0\mathrm{atm}$and ${\mathrm{P}}_{{\mathrm{CO}}_2}=10.0\mathrm{atm}$.

Set up a reaction for the equilibrium:

Write the expression for the equilibrium constant ${\mathrm{K}}_{\mathrm{p}}$. Substitute ${\mathrm{K}}_{\mathrm{p}}=3.58\times {10}^6$and solve the expression to obtain the value of x:

localid="1654938155572" $$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{P}}_{\infty }^2{{\mathrm{P}}^2}_{{\mathrm{H}}_2}}{{\mathrm{P}}_{{\mathrm{CH}}_4}{\mathrm{P}}_{{\mathrm{CO}}_2}} \\ 3.58\times {10}^6=\frac{{\left(2\mathrm{x}\right)}^2{\left(2\mathrm{x}\right)}^2}{(10.0-\mathrm{x})(10.0-\mathrm{x})} \\ {(3.58\times {10}^6)}^{1/2}=\frac{{\left(2\mathrm{x}\right)}^2}{(10.0-\mathrm{x})} \\ 1.884\times {10}^3=\frac{{\left(2\mathrm{x}\right)}^2}{(10.0-\mathrm{x})} \end{gathered}$$

Solve the expression to obtain quadratic equation:

$4{\mathrm{x}}^2+1.884\times {10}^3\mathrm{x}-1.884\times {10}^4=0$

Solve the quadratic equation:

$$\begin{gathered} \mathrm{x}=\frac{-1.884\times {10}^3\sqrt[+]{\left(1.884\times {10}^3\right)2-4\times \left(4\right)\times (-1.884\times {10}^4)}}{2\times 4} \\ =9.796 \end{gathered}$$

Calculate ${\mathrm{PH}}_2$, the partial pressure of${\mathrm{H}}_2$:

$$\begin{gathered} {\mathrm{PH}}_2=2\mathrm{x} \\ =2\times 9.796 \\ =19.59\mathrm{atm} \end{gathered}$$

If the reaction proceeds to completion, the number of moles of ${\mathrm{H}}_2$formed would be double the number of moles of the ${\mathrm{CH}}_4$or ${\mathrm{CO}}_2$. Since pressure of a gas is directly proportional to its number of moles, the partial pressure of ${\mathrm{H}}_2$would be double the initial pressure of ${\mathrm{CH}}_4$or ${\mathrm{CO}}_2$as localid="1654941251931" ${\mathrm{PH}}_2=20.0\mathrm{atm}$. Determine the percent yield of ${\mathrm{H}}_2$

$$\begin{gathered} \mathrm{Percent}\mathrm{yield}=\frac{19.59\mathrm{atm}}{20.0\mathrm{atm}}\times 100 \\ =98.0\% \end{gathered}$$

Thus, the percent yield of is 98.0%.

(b) Percent yield of H2 for this system at 1300 K

Substitute${\mathrm{K}}_{\mathrm{p}}=2.626\times {10}^7$and solve the expression to obtain the value of x:

role="math" localid="1654939776209" $$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{P}}_{\infty }^2{\mathrm{P}}_{{\mathrm{H}}_2}^2}{{\mathrm{P}}_{{\mathrm{CH}}_4}{\mathrm{P}}_{{\mathrm{CO}}_2}} \\ 2.626\times {10}^7=\frac{{\left(2\mathrm{x}\right)}^2{\left(2\mathrm{x}\right)}^2}{(10.0-\mathrm{x})(10.0-\mathrm{x})} \\ {(2.626\times {10}^7)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{{\left(2\mathrm{x}\right)}^2}{(10.0-\mathrm{x})} \\ 5.124\times {10}^3=\frac{{\left(2\mathrm{x}\right)}^2}{(10.0-\mathrm{x})} \end{gathered}$$

Solve the expression to obtain quadratic equation:

$4{\mathrm{x}}^2+5.124\times {10}^3\mathrm{x}-5.124\times {10}^4=0$

Solve the quadratic equation:

$$\begin{gathered} \mathrm{x}=\frac{-5.124\times {10}^3\sqrt[+]{(5.124\times {10}^3)2-4\times \left(4\right)\times (-5.124\times {10}^4)}}{2\times 4} \\ =9.992 \end{gathered}$$

Calculate ${\mathrm{PH}}_2$,the partial pressure of ${\mathrm{H}}_2$

$$\begin{gathered} {\mathrm{PH}}_2=2\mathrm{x} \\ =2\times 9.92 \\ =19.84 \end{gathered}$$

If the reaction proceeds to completion, the number of moles of ${\mathrm{H}}_2$formed would be double the number of moles of the ${\mathrm{CH}}_4$or${\mathrm{CO}}_2$ . Since pressure of a gas is directly proportional to its number of moles, the partial pressure of ${\mathrm{H}}_2$would be double the initial pressure of ${\mathrm{CH}}_4$or ${\mathrm{CO}}_2$as${\mathrm{PH}}_2=20.0\mathrm{atm}$ . Determine the percent yield of${\mathrm{H}}_2$.

$$\begin{gathered} \mathrm{Percent}\mathrm{yield}=\frac{19.84\mathrm{atm}}{20.0\mathrm{atm}}\times 100 \\ =99.2\% \end{gathered}$$

Thus, the percent yield ${\mathrm{H}}_2$of is 99.2%.

Step 3: (c) Use the Van’t Hoff equation to find ∆Hrxn0

Write Van’t Hoff equation for the equilibrium constant ${\mathrm{K}}_2$at temperature ${\mathrm{T}}_2$and equilibrium constant ${\mathrm{K}}_1$at temperature${\mathrm{T}}_1$:

role="math" localid="1654940789466" $\mathrm{In}\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{∆{\mathrm{H}}_{\mathrm{rxn}}^0}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$

Where R is the gas constant and $∆{\mathrm{H}}_{\mathrm{rxn}}^0$ is the standard heat of reaction.

Substitute ${\mathrm{K}}_1=2.626\times {10}^7,{\mathrm{K}}_2=3.548\times {10}^6,{\mathrm{T}}_1=1300\mathrm{K},{\mathrm{T}}_2=1200\mathrm{K}\mathrm{and}\mathrm{R}=0.0821\mathrm{L}.\mathrm{atm}/\mathrm{mol}.\mathrm{K}.$Determine the standard heat of reaction $∆{\mathrm{H}}_{\mathrm{rxn}}^0$:

role="math" localid="1654941146247" $$\begin{gathered} \mathrm{In}\left(\frac{3.548\times {10}^6}{2.626\times {10}^7}\right)=-\frac{∆{\mathrm{H}}_{\mathrm{rxn}}^0}{8.314\mathrm{J}/\mathrm{mol}.\mathrm{K}}\left(\frac{1}{1200\mathrm{K}}-\frac{1}{1300\mathrm{K}}\right) \\ ∆{\mathrm{H}}_{\mathrm{rxn}}^0=2.60\times {10}^5\mathrm{J}/\mathrm{mol} \end{gathered}$$

Thus, the standard heat Of reaction $∆{\mathrm{H}}_{\mathrm{rxn}}^0$is $2.60\times {10}^5\mathrm{J}/\mathrm{mol}$.