Question

For the reaction: $C{O_2}\left( g \right) + {H_2}\left( g \right) \rightleftharpoons CO\left( g \right) + {H_2}O\left( g \right)$, K is found to be 16 at a given temperature. Originally equal numbers moles of ${H_2}$ and $C{O_2}$ were placed in the flask. At equilibrium, the pressure of ${H_2}$ is $1.20atm$. What is the partial pressure of $CO$ and ${H_2}O$?
A.$\text{4.80atm each}$
B.$\text{9.60atm each}$
C.$\text{2.40atm each}$
D.$\text{1.20atm each}$

Answer 1

We can calculate the partial pressures of $CO$ and ${H_2}O$ using the moles of $CO$, total moles and the total partial pressure. The moles of $CO$ and total moles are divided. The obtained product is then multiplied by the total partial pressure to get the partial pressures of $CO$ and ${H_2}O$.
Formula used: We can calculate the partial pressure of $CO$ using the formula,
${P_{CO}} = \dfrac{{{\text{Moles of CO}}}}{{{\text{Total moles}}}} \times {P_{total}}$

Complete step by step answer:
Given data contains,
The equilibrium pressure is 16.
The pressure of ${H_2}$ is $1.20atm$.
We can write the given reaction as,
$C{O_2}\left( g \right) + {H_2}\left( g \right) \rightleftharpoons CO\left( g \right) + {H_2}O\left( g \right)$
The initial change equilibrium table is written as,
$C{O_2}\left( g \right) + {H_2}\left( g \right) \rightleftharpoons CO\left( g \right) + {H_2}O\left( g \right)$

${\text{Initial moles}}$	$M$	$m$	$0$	$0$
${\text{Format files}}$	$M - m$	$M - m$	$m$	$m$

Now, let us calculate the total moles present at equation
Total moles present at the equation=$M - m + M - m + 2m = 2M$
The total moles present at the equation is $2M$.
The total vapour pressure is calculated as using the formula of pressure of ${H_2}$,
Pressure of ${H_2}$=$\dfrac{{M - m}}{{2M}} \times {P_{total}}$
Let us substitute the value of pressure of ${H_2}$ in the above expression.
1.2atm=$\dfrac{{M - m}}{{2M}} \times {P_{total}}$
${P_{total}} = \dfrac{{2.4M}}{{\left( {M - m} \right)}}$
${P_{total}} = \dfrac{{2.4M}}{{\left( {1 - \dfrac{m}{M}} \right)}}$
We can write the ${K_P}$ expression using the partial pressure of carbon dioxide, partial pressure of carbon monoxide, partial pressure of water and partial pressure of hydrogen.
${K_P} = \dfrac{{{P_{CO}} \cdot {P_{{H_2}O}}}}{{{P_{C{O_2}}} \cdot {P_{{H_2}}}}}$
Let us substitute the values now.
$16 = \dfrac{{m \cdot m}}{{\left( {M - m} \right) \cdot \left( {M - m} \right)}}$
Solving the equation,
$\dfrac{m}{{M - m}} = - 4$ or $\dfrac{m}{{M - m}} = 4$
$\dfrac{{M - m}}{m} = \dfrac{{ - 1}}{4}$ or$\dfrac{{M - m}}{m} = \dfrac{1}{4}$
We get,
$\Rightarrow \dfrac{M}{m} = \dfrac{5}{4}$ or $\dfrac{M}{m} = \dfrac{3}{4}$
Reciprocating the equation, we get
$\Rightarrow \dfrac{m}{M} = \dfrac{4}{5}\,$or $\dfrac{m}{M} = \dfrac{4}{3}$
The value $\dfrac{m}{M} = \dfrac{4}{3}$ is not possible.
Let us now calculate the total partial pressure by substituting the value of $\dfrac{m}{M}$.
${P_{total}} = \dfrac{{2.4M}}{{\left( {1 - \dfrac{m}{M}} \right)}}$
Substituting the values we get,
$\Rightarrow {P_{total}} = \dfrac{{2.4M}}{{\left( {1 - \dfrac{4}{5}} \right)}}$
On simplifying we get,
$\Rightarrow {P_{total}} = 2.4 \times 5$
$\Rightarrow {P_{total}} = 12$
We have calculated the total partial pressure as 12atm.
Let us now calculate the partial pressures of $CO$.
${P_{CO}} = \dfrac{{{\text{Moles of CO}}}}{{{\text{Total mass}}}} \times {P_{total}}$
${P_{CO}} = \dfrac{m}{{2M}} \times {P_{total}}$
Substituting the values we get,
$\Rightarrow {P_{CO}} = \dfrac{1}{2} \times \dfrac{4}{5} \times 12$
$\Rightarrow {P_{CO}} = 4.8atm$
The partial pressure of water will be equal to partial pressure of carbon monoxide.
${P_{{H_2}O}} = {P_{CO}} = 4.8atm$
The partial pressure of water is $4.8atm$.
The partial pressure of carbon monoxide is $4.8atm$.

Therefore, the option (A) is correct.

Note: We must remember that there is no change in solubility of liquids and solids with pressure changes, whereas in gases, the solubility increases with increase in pressure. An example of solubility of gases with changes in pressure is carbonated beverages. All carbonated beverages are kept under pressure to raise the amount of carbon dioxide dissolved in the solution. When we open the bottle, the pressure present above the solution reduces and it leads to effervescences of the solution, and few amounts of carbon dioxide bubbles.