Q17.102P - Chemistry: Molecular Nature Of Matter And Change

Question

Using ${\text{CH}}_4$ and steam as a source of ${\text{H}}_2$for ${\text{NH}}_3$ synthesis requires high temperatures. Rather than burning ${\text{CH}}_4$separately to heat the mixture, it is more efficient to inject some ${\text{O}}_2$ into the reaction mixture. All of the ${\text{H}}_2$ is thus released for the synthesis, and the heat of reaction for the combustion of ${\text{CH}}_4$ helps maintain the required temperature. Imagine the reaction occurring in two steps:

$\begin{array}{l}2{\text{CH}}_4\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2\text{CO}\left(g\right)+4{\text{H}}_2\left(g\right)\\ K_{\text{p}}=9.34\times {10}^{28}\text{ at }1000\text{K}\\ \\ \text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\\ K_{\text{p}}=1.374\text{ at }1000\text{K}\end{array}$

(a) Write the overall equation for the reaction of methane, steam, and oxygen to form carbon dioxide and hydrogen.

(b) What is $K_{\text{p}}$  for the overall reaction?

(c) What is $K_{\text{c}}$  for the overall reaction?

(d) A mixture of $2.0 \text{mol}$ of ${\text{CH}}_4,1.0 \text{mol}$ of ,${\text{O}}_2$ and $2.0 \text{mol}$ of steam with a total pressure of 30atm reacts at 1000K at constant volume. Assuming that the reaction is complete and the ideal gas law is a valid approximation, what is the final pressure?

Step-by-Step Solution

Answer

(a) The overall equation is .$2{\text{CH}}_4\left(g\right)+{\text{O}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 2{\text{CO}}_2\left(g\right)+6{\text{H}}_2\left(g\right)$

(b) .$K_{\text{overall }}=1.76\cdot {10}^{29}$

(c) .$K_{\text{c}}=3.19\cdot {10}^{23}$

(d).$P_{\text{final }}=48 \text{atm}$

Step 1: Definition of pressure

Pressure is defined as a measure of the force applied over a unit area.

Step 2: Write the overall equation

a)

The first step:  $2{\text{CH}}_4\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2\text{CO}\left(g\right)+4{\text{H}}_2\left(g\right)$

The second step:

$\text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)$

Multiply the second step reaction by 2 :

$2\text{CO}\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 2{\text{CO}}_2\left(g\right)+2{\text{H}}_2\left(g\right)$

Now add the equation for the first step and the multiplied equation for the second step:

$2{\text{CH}}_4\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2\text{CO}\left(g\right)+4{\text{H}}_2\left(g\right)2\text{CO}\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 2{\text{CO}}_2\left(g\right)+2{\text{H}}_2\left(g\right)$

So, the overall equation will be:

.$2{\text{CH}}_4\left(g\right)+{\text{O}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 2{\text{CO}}_2\left(g\right)+6{\text{H}}_2\left(g\right)$

Step 3: Calculate K p

b)

The values are:

$K_{\text{P}}=9.34\cdot {10}^{28}-$ for the first step of the given reaction

$K_{\text{P}}=1.374$- for the second step of the given reaction

$K_{\text{P}}$ for the overall process will be calculated as:

$K_{\text{overall }}=K_{\text{P}}\cdot {\left(K_{\text{P}}\right)}^2$

${}_{\text{overall }}=9.34\cdot {10}^{28}\cdot {\left(1.374\right)}^2$

$K_{\text{overall }}=9.34\cdot {10}^{28}\cdot 1.8878$

$K_{\text{overall }}=1.76\cdot {10}^{29}$

Therefore, $K_{\text{overall }}=1.76\cdot {10}^{29}$.

Step 4: Calculate K c

c)

$\begin{array}{c}{K}_{\text{p}}=1.76\cdot {10}^{29}\\ \text{R}=0.0821 \text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}\\ \text{T}=1000 \text{K}\end{array}$

It is expressed as:

$K_{\text{c}}=\frac{K_{\text{p}}}{{\left(RT\right)}^{\text{singlas}{ }_{\text{gl}}}}$

Firstly we need to calculate $\Delta n_{\text{gas }}$ as:

$\Delta n_{\text{gas }}=n_{\text{products }}-n_{\text{reactants }}$

$\Delta n_{\text{gas }}=8-5$

$\Delta n_{\text{gas }}=3$

So now as we have all the needed values, we can calculate $K_c$:

$K_{\text{c}}=\frac{K_{\text{p}}}{{\left(\text{RT}\right)}^{\Delta n_{\text{gas }}}}$

$K_{\text{c}}=\frac{1.76\cdot {10}^{29}}{{(0.0821 \text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}\cdot 1000 \text{K})}^3}$

$K_{\text{c}}=3.19\cdot {10}^{23}$

Therefore, $K_{\text{c}}=3.19\cdot {10}^{23}$.

Step 5: Calculate the final pressure

d)

We will calculate the final pressure from the equation:

.$P_{\text{initial }}=30 \text{atm}$

We will calculate the final pressure from the equation:

$\frac{n_{\text{initial }}}{P_{\text{intrial }}}=\frac{n_{\text{(nimal }}}{P_{\text{final }}}$

But firstly we have to calculate the initial moles of the gas and the final moles of the gas:

$n_{\text{initial }}=n_{{\text{CH}}_4}+n_{{\text{O}}_2}+n_{{\text{H}}_2\text{O}}$

$n_{\text{initial }}=2 \text{mol}+1 \text{mol}+2 \text{mol}$

$n_{\text{initial }}=5 \text{mol}$

$n_{\text{final }}=n_{{\text{CO}}_2}+n_{{\text{H}}_2}$

$n_{\text{final }}=2 \text{mol}+6 \text{mol}$

.$n_{\text{final }}=8 \text{mol}$

So now we can express and calculate the final pressure:

$\frac{n_{\text{initial }}}{P_{\text{initial }}}=\frac{n_{\text{final }}}{P_{\text{final }}}$

$\frac{5 \text{mol}}{30 \text{atm}}=\frac{8 \text{mol}}{P_{\text{final }}}$

$P_{\text{final }}=8 \text{mol}\cdot \frac{30 \text{atm}}{5 \text{mol}}$

$P_{\text{final }}=48 \text{atm}$

.

Therefore the final pressure is .$P_{\text{final }}=48 \text{atm}$