Question

Calculate the partial pressure of B at equilibrium in the following equilibrium.
$\text{A}\left( \text{s} \right)\rightleftharpoons \text{B}\left( \text{g} \right)+2\text{C}\left( \text{g} \right);\text{ }{{\text{K}}_{\text{p}}}=32\text{ at}{{\text{m}}^{3}}$

Answer 1

${{\text{K}}_{\text{p}}}$ is known as the equilibrium constant which is calculated by using the partial pressures of the reaction species. The partial pressures of the products are multiplied and the coefficients are raised to their respective power in order to obtain ${{\text{K}}_{\text{p}}}$ . So if we let the partial pressure of B as ‘p’ then partial pressure of C will be ‘2p’ as two moles of C are used and their coefficients will be raised to their powers.

Formula Used: ${{\text{K}}_{\text{p}}}={{\text{p}}_{\text{B}}}\times {{\text{p}}_{\text{C}}}$ where ${{\text{p}}_{\text{B}}}$ is partial pressure of B and ${{\text{p}}_{\text{C}}}$ is partial pressure of C.

Complete Step By Step Solution
For the given equilibrium reaction $\text{A}\left( \text{s} \right)\rightleftharpoons \text{B}\left( \text{g} \right)+2\text{C}\left( \text{g} \right);\text{ }{{\text{K}}_{\text{p}}}=32\text{ at}{{\text{m}}^{3}}$
Let us assume the partial pressure of B as ‘p’ and partial pressure of C as ‘2p’ because 2 moles of gas C are used. So, according to the equilibrium constant
We have,
${{\text{K}}_{\text{p}}}={{\text{p}}_{\text{B}}}\times {{\text{p}}_{\text{C}}}$
${{\text{K}}_{\text{p}}}={{\text{p}}^{1}}\times {{\left( 2\text{p} \right)}^{2}}$
${{\text{K}}_{\text{p}}}=\text{p}\times \text{4}{{\text{p}}^{2}}$
$32=4{{\text{p}}^{3}}$
${{\text{p}}^{3}}=\dfrac{32}{4}$
${{\text{p}}^{3}}=8$
$\Rightarrow \text{p}=2$
For gas B, p is raised to power 1 because coefficient of gas B is 1 and for gas C, 2p is raised to power 2 because coefficient for gas C is 2. The value of p comes out to be 2, so the partial pressure of gas B at equilibrium is 2 atm.

Additional Information
Partial pressure of any constituent gas is that pressure which it would have exerted if it alone occupied the entire volume at the same temperature conditions. The total pressure of an ideal gas is the sum of the partial pressures of its constituent gases. Partial pressure can be obtained by mole fraction of gas in the given mixture. The concept of partial pressure is important as it predicts the movement of gases. The partial pressure of a gas can be increased by increasing the number of moles of a gas. There is a law associated with the partial pressures of gases which is known as Dalton’s Law. According to this law the total pressure exerted by the mixture of gases is equal to the sum of the partial pressure exerted by its individual gases.

Note
The partial pressure of a gas can also be obtained by multiplying the total pressure with the mole fraction of that gas. The Dalton’s Law for partial pressure was developed on the basis because Dalton laid emphasis on the fact that due to low density and high compressibility, gases consist most of the empty space and when two or more gases occupy the same volume then they behave completely independently of each other.