Q17.101P - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

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

b)  .$K_{\text{overall }}=2.74\cdot {10}^{15}$

c)  .$P_{{\text{C}}_3{\text{H}}_4}=1 \text{atm}$

d) The percentage of the remaining ${\text{C}}_3{\text{H}}_8$  is 33%.

Step 1: Definition of hydrocarbon

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

Step 2: Write the overall equation

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

Step 3: Calculate K p

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:

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

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

Step 4: Calculate the final pressure

c)

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

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

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

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

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

Step 5: Calculate the percentage of the C 3 H 8

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

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

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