101 The methane used to obtain ${\text{H}}_2$ for manufacture is impure and usually contains other hydrocarbons, such as propane,${\text{C}}_3{\text{H}}_8$ . Imagine the reaction of propane occurring in two steps:

$\begin{array}{l}{\text{C}}_3{\text{H}}_8\left(g\right)+3{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 3\text{CO}\left(g\right)+7{\text{H}}_2\left(g\right)\\ K_{\text{p}}=8.175\times {10}^{15}\text{at}1200\text{K}\\ \\ \text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\\ K_{\text{p}}=0.6944\text{at}1200\text{K}\end{array}$

(a) Write the overall equation for the reaction of propane and steam to produce carbon dioxide and hydrogen.

(b) Calculate$K_{\text{p}}$ for the overall process at 1200K

(c) When$1.00$ volume of ${\text{C}}_3{\text{H}}_8$and $4.00$volumes of${\text{H}}_2\text{O}$ , each at$1200.\text{K}$ and$5.0\text{atm}$ , are mixed in a container, what is the final pressure? Assume the total volume remains constant, that the reaction is essentially complete, and that the gases behave ideally.

(d) What percentage of the${\text{C}}_3{\text{H}}_8$ remains unreacted?

A hydrocarbon is a chemical molecule that is made up entirely of hydrogen and carbon atoms.

a)

The first step:

${\text{C}}_3{\text{H}}_8\left(g\right)+3{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 3\text{CO}\left(g\right)+7{\text{H}}_2\left(g\right)$

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 3 :

$3\text{CO}\left(g\right)+3{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 3{\text{CO}}_2\left(g\right)+3{\text{H}}_2\left(g\right)$

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

${\text{C}}_3{\text{H}}_8\left(g\right)+3{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 3\text{CO}\left(g\right)+7{\text{H}}_2\left(g\right)3\text{CO}\left(g\right)+3{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 3{\text{CO}}_2\left(g\right)+3{\text{H}}_2\left(g\right)$

So the overall equation will be:

Therefore, ${\text{C}}_3{\text{H}}_8\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons 3{\text{CO}}_2\left(g\right)+10{\text{H}}_2\left(g\right)$.

b)

$K_{\text{P}}=8.175\cdot {10}^{15}$- for the first step of the given reaction

$K_{\text{P}}=0.6944-$for the second step of the given reaction

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

localid="1662032357030" $\begin{array}{l}{K}_{\text{overall}}=K_{\text{P}}\cdot {\left(K_{\text{P}}\right)}^3\\ K_{\text{overall}}=8.175\cdot {10}^{15}\cdot {\left(0.6944\right)}^3\\ K_{\text{overall}}=8.175\cdot {10}^{15}\cdot 0.3348\\ K_{\text{overall}}=2.74\cdot {10}^{15}\end{array}$

Therefore, $K_{\text{overall}}=2.74\cdot {10}^{15}$.

c)

Firstly we will calculate the initial partial pressure of ${\text{C}}_3{\text{H}}_8$ as:

$\begin{array}{c}{P}_{C_3{H}_4}=\frac{\text{Volume}}{\text{Totalvolume}}\cdot \text{Totalpressure}{P}_{C_3{H}_4}\\ =\frac{1}{5}\cdot 5\text{atm}{P}_{C_3{H}_4}\\ =1\text{atm}\end{array}$

Now we will calculate the initial partial pressure of${\text{H}}_2\text{O}$as:

$\begin{array}{c}{P}_{{\text{H}}_2\text{O}}=\frac{\text{Volume}}{\text{Totalvolume}}\cdot \text{Totalpressure}\\ P_{{\text{H}}_2\text{O}}=\frac{4}{5}\cdot 5\text{atm}\\ P_{{\text{H}}_2\text{O}}=4\text{atm}\end{array}$

From the reaction stoichiometry, we can see that 6 moles of ${\text{H}}_2\text{O}$reacts with 1 mole of ${\text{C}}_3{\text{H}}_8$.

So, we will calculate the pressure of the reacted ${\text{C}}_3{\text{H}}_8$as:

$\begin{array}{l}{P}_{C_3{\text{H}}_4}=\frac{1\text{atm}}{1}\\ P_{{\text{C}}_3{\text{H}}_4}=1\text{atm}\end{array}$

Therefore, $P_{{\text{C}}_3{\text{H}}_4}=1\text{atm}$.

d)

To solve for the percentage of the remaining${\text{C}}_3{\text{H}}_8$. We must divide the remaining amount by its original amount.

$\%{\text{C}}_3{\text{H}}_8$Unreacted

$\begin{array}{l}=\frac{0.33{\text{atmC}}_3{\text{H}}_8}{1.00{\text{atmC}}_3{\text{H}}_8}\times 100\%\\ =33\%.\end{array}$

Hence, the required percentage of ${\text{C}}_3{\text{H}}_8$ is 33%.