Question

For the reaction: $C{O}_2\left(g\right)+H_2\left(g\right)\leftrightharpoons CO\left(g\right)+H_{2}O\left(g\right)$, K is found to be 16 at a given temperature. Originally equal numbers of moles of $H_2$ and $C{O}_2$ were placed in the flask. At equilibrium, the pressure of $H_2$ is 1.20 atm. What is the partial pressure of $CO$ and $H_{2}O$?

• A. 4.80 atm each
• B. 9.60 atm each
• C. 2.40 atm each
• D. 1.20 atm each

Answer 1

Answer: (A)

4.80 atm each

The given reaction is ;-

${CO}_2\left(g\right)+H_2\left(g\right)\rightleftharpoons CO\left(g\right)+H_{2}O\left(g\right)$

Initial moles ; $M$ $m$ $0$ $0$ (let)

At eqm : $M-m$ $M-m$ $m$ $m$ (let)

Now, given that equation pressure of $H_2=1.2atm$

total moles at equation $=M-m+M-m+2m=2M$

So, pressure of $H_2=\frac{M-m}{2M}\times P_{Total}$

$\Rightarrow 1.2=\frac{M-m}{2M}\times P_{Total}$

$\Rightarrow P_{Total}=\frac{2.4M}{(M-m)}=\frac{2.4}{(1-\frac{m}{M})}$

Now, $K_P=\frac{p_{CO}.p_{H_{2}O}}{P_{{CO}_2}.p_{H_2}}$

$\Rightarrow 16=\frac{m.m}{(M-m)(M-m)}$

$\Rightarrow \frac{m}{M-m}=4or\frac{m}{M-m}=-4$

$\Rightarrow \frac{M-m}{m}=\frac{1}{4}or\Rightarrow \frac{M-m}{m}=\frac{-1}{4}$

$\Rightarrow \frac{M}{m}=\frac{5}{4}or\Rightarrow \frac{M}{m}=\frac{3}{4}$

$\Rightarrow \frac{m}{M}=4/5or\Rightarrow \frac{m}{M}=\frac{4}{3}$ (not possible)

So, $P_{Total}=\frac{2.4}{(1-\frac{m}{M})}=\frac{2.4}{(1-\frac{4}{5})}=2.4\times 5=12$

$\therefore p_{CO}=\frac{m}{2M}\times P_{Total}=\frac{4}{5}\times 12=4.8atm$

$p_{H_{2}O}=p_{CO}=4.8atm$