Q102CP - Chemistry: Molecular Nature Of Matter And Change

Question

 Using  ${\mathbf{CH}}_{\mathbf{4}}$and steam as a source of ${\mathbf{H}}_{\mathbf{2}}$ for ${\mathbf{NH}}_{\mathbf{3}}$synthesis requires high temperatures. Rather than burning ${\mathbf{CH}}_{\mathbf{4}}$ separately to heat the mixture, it is more efficient to inject some ${\mathbf{O}}_{\mathbf{2}}$ into the reaction mixture. All of the ${\mathbf{H}}_{\mathbf{2}}$ is thus released for the synthesis, and the heat of reaction for the combustion of ${\mathbf{CH}}_{\mathbf{4}}$  helps maintain the required temperature. Imagine the reaction occurring in two steps:

$$\begin{gathered} \mathbf{2}{\mathbf{CH}}_{\mathbf{4}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{CO}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{34}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}\mathbf{at}\mathbf{1000}\mathbf{.}\mathbf{K} \\ {\mathbf{CO}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }{\mathbf{CO}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{374}\mathbf{ }\mathbf{at}\mathbf{ }\mathbf{1000}\mathbf{.}\mathbf{K} \end{gathered}$$

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

(b) What is Kp for the overall reaction? 

(c) What is Kc for the overall reaction? 

(d) A mixture of 2.0 mol of ${\mathbf{CH}}_{\mathbf{4}}$ ,1.0 mol of ${\mathbf{O}}_{\mathbf{2}}$, and 2.0 mol of steam with a total pressure of 30. atm reacts at 1000. K 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 for the reaction of methane, steam, and oxygen to form carbon dioxide and hydrogen as follows:

$2{\mathrm{CH}}_{4\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}+2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+6{\mathrm{H}}_{2\left(\mathrm{g}\right)}$

(b)  ${\mathrm{K}}_{\mathrm{p}}$for the overall reaction is $1.76\times {10}^{29}$

(c) The ${\mathrm{K}}_{\mathrm{c}}$for the overall reaction is $3.19\times {10}^{23}$.

(d) The final pressure is 48.0 atm.

Step 1: (a) Write the overall equation for the reaction of methane, steam, and oxygen to form carbon dioxide and hydrogen

(a) The overall equation for the reaction of methane, steam, and oxygen to form carbon dioxide and hydrogen is as follows:

$2{\mathrm{CH}}_{4\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}+2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+6{\mathrm{H}}_{2\left(\mathrm{g}\right)}$

The process you describe is known as steam reforming and usually involves a catalyst like nickel. But as Malcolm Sergeant says, it’s mostly about the heat; the temperatures of nearly 1000C are used.

The heat doesn’t cause the methane to combust because there is no free oxygen in the system for it to react with.

Step 2: (b)Kp for the overall reaction

${\mathrm{K}}_{\mathrm{p}}$is the equilibrium constant calculated from a reaction equation's partial pressures. It is used to express the relationship between reactants and product pressures.${\mathrm{K}}_{\mathrm{p}}$for the overall reaction is $1.76\times {10}^{29}$ 

Step 3: (c) What is Kc for the overall reaction?

${\mathrm{K}}_{\mathrm{c}}$  for the overall reaction:

We know,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}{\left(\mathrm{RT}\right)}^{∆\mathrm{n}} \\ {\mathrm{K}}_{\mathrm{c}}=\frac{{\mathrm{K}}_{\mathrm{p}}}{{\left(\mathrm{RT}\right)}^{∆\mathrm{n}}} \end{gathered}$$

   

Temperature, T= 1000 K

Gas constant, R= 0.08206 L.atm/mol.K

$$\begin{gathered} \mathrm{\Delta n}=\mathrm{number} \mathrm{of} \mathrm{mole} \mathrm{of} \mathrm{gaseous} \mathrm{product} - \mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{gaseous} \mathrm{reactant} \\ =[2+6 ]-[2+1+2 ] \\ =8-5 \\ =3 \end{gathered}$$

Substitute all these in Kc equation then we will get Kc.

Therefore

$$\begin{gathered} {\mathrm{K}}_{\mathrm{C}}=\frac{(1.76\times {10}^{29})}{(0.08206\times 1000{)}^3} \\ =3.19\times {10}^{23} \end{gathered}$$

Thus, the Kc for overall reaction is $3.19\times {10}^{23}$.

Step 4: (d) What is the final pressure?

The equilibrium mixture reaction is

$2{\mathrm{CH}}_{4\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}+2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+6{\mathrm{H}}_{2\left(\mathrm{g}\right)}$

Initial mole of ${\mathrm{CH}}_{4\left(\mathrm{g}\right)}=2.00 \mathrm{mol}$

Initial mole of ${\mathrm{O}}_{2\left(\mathrm{g}\right)}=1.00 \mathrm{mol}$

Initial mole of ${\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}=2.00 \mathrm{mol}$

And the total pressure P= 30 atm 

The partial pressure of gases at initially are as follows:

Total moles of gases at initially,

$$\begin{gathered} \mathrm{n}=(2.00 \mathrm{mol}+1.00 \mathrm{mol}+2.00 \mathrm{mol}) \\ =5.00 \mathrm{moles} \end{gathered}$$

Mole fraction of, 

$$\begin{gathered} {\mathrm{CH}}_{4\left(\mathrm{g}\right)}=\frac{\mathrm{moles} \mathrm{of} {\mathrm{CH}}_{4\left(\mathrm{g}\right)}}{\mathrm{total} \mathrm{moles}} \\ =\frac{2.00 \mathrm{moles}}{5.00 \mathrm{moles}} \\ =0.40 \end{gathered}$$

 Mole fraction of,

$$\begin{gathered} {\mathrm{O}}_{2\left(\mathrm{g}\right)}=\frac{\mathrm{moles} \mathrm{of} {\mathrm{O}}_{2\left(\mathrm{g}\right)}}{\mathrm{total} \mathrm{moles}} \\ =\frac{1.00 \mathrm{moles}}{5.00 \mathrm{moles}} \\ =0.20 \end{gathered}$$

Mole fraction of,

$$\begin{gathered} {\mathrm{H}}_{2 }{\mathrm{O}}_{\left(\mathrm{g}\right)}=\frac{\mathrm{moles} \mathrm{of} {\mathrm{H}}_{2 }{\mathrm{O}}_{\left(\mathrm{g}\right)}}{\mathrm{total} \mathrm{moles}} \\ =\frac{2.00 \mathrm{moles}}{5.00 \mathrm{moles}} \\ =0.40 \end{gathered}$$

$$\begin{gathered} {\mathrm{P}}_{{\mathrm{CH}}_4}=\mathrm{Mole} \mathrm{fraction} \mathrm{of} {\mathrm{CH}}_{4\left(\mathrm{g}\right)}\times \mathrm{total} \mathrm{pressure} \\ =0.40\times 30 \mathrm{atm} \\ =12.0 \mathrm{atm} \\ {\mathrm{P}}_{{\mathrm{O}}_2}=\mathrm{Mole} \mathrm{fraction} \mathrm{of} {\mathrm{O}}_{2\left(\mathrm{g}\right)}\times \mathrm{total} \mathrm{pressure} \\ =0.20\times 30 \mathrm{atm} \\ =6.0 \mathrm{atm} \\ {\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}=\mathrm{Mole} \mathrm{fraction} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}\times \mathrm{total} \mathrm{pressure} \\ =0.40\times 30 \mathrm{atm} \\ =12.0 \mathrm{atm} \end{gathered}$$

If x atm of pressure decreased in the reactants then the equilibrium will be established.

We assumed the reaction had completed. Therefore all the reactants were consumed completely. So there is no pressure extracted by the reactants. Hence their pressure is equal to zero after the reaction is over. Therefore,

$$\begin{gathered} 6.0-\mathrm{x}=12.0-2\mathrm{x}=12.0-2\mathrm{x}=0 \\ 6.0-\mathrm{x}=0 \\ \mathrm{x}=6.0 \end{gathered}$$

Therefore the final pressure of the reaction is,

$$\begin{gathered} \mathrm{Final} \mathrm{pressure}={\mathrm{P}}_{{\mathrm{CO}}_2}+{\mathrm{P}}_{{\mathrm{H}}_2} \\ =\left(2\mathrm{x}\right)\mathrm{atm}+\left(6\mathrm{x}\right)\mathrm{atm} \\ =(2\times 6.0)\mathrm{atm}+36.0 \mathrm{atm} \\ =48.0 \mathrm{atm} \end{gathered}$$

The final pressure is 48.0 atm