Question

For reaction $PC{l_3}(g) + C{l_2}(g) \rightleftharpoons PC{l_5}(g)$, the value of ${K_c}$ at $250^\circ C$ is 26. The value of ${K_p}$ at this temperature will be:
(a) 0.61
(b) 0.83
(c) 0.57
(d) 0.46

Answer 1

Hint: ${K_c}$ and ${K_p}$ are the equilibrium constant used for ideal gas mixture and used when the equilibrium concentration is expressed as atmospheric pressure and molarity. For the general reaction $2A(g) + B(g) \to 2C(g)$
The ${K_p}$ is given as shown below.
${K_p} = \dfrac{{{P^2}c}}{{P_A^2{P_B}}}$
To derive the relation between ${K_p}$ and ${K_c}$, use the ideal gas equation and substitute the value of pressure in the above equation.

Complete step by step answer:
The reaction is shown below.
$PC{l_3}(g) + C{l_2}(g) \rightleftharpoons PC{l_5}(g)$
In this one mol of phosphorus trichloride reacts with one mole of chlorine to give one mol of phosphorus pentachloride.
The formula for calculating the ${K_P}$ is given below.
${K_p} = {K_c} \times {(RT)^{\Delta n}}$
Where,
${K_P}$ is the equilibrium constant when concentration is expressed in atmospheric pressure.
${K_c}$ is the equilibrium constant when concentration is expressed in molarity.
R is the universal gas constant
T is the temperature
$\Delta n$ is the change in moles.
R = 0.0821
T = 250 + 273.15 =523.15
${K_c}$is 26.
The change in a mole is calculated as the number of moles of gaseous product minus the number of moles of gaseous reactant
$\Delta n = 1 - 2$
$\Rightarrow n = - 1$
To calculate the value of ${K_p}$, substitute the values in the given equation as shown below.
${K_p} = \dfrac{{26}}{{0.0821 \times 523.15}}$
$\Rightarrow {K_p} = 0.61$
Thus, the value of ${K_p}$ at this 250 degree Celsius temperature is 0.61.

Thus, the correct option is option (a).

Note:
During the reaction, when the change in the number of moles of gaseous reactant and product is zero then both the equilibrium constant ${K_p}$ and ${K_c}$ will be equal. This equation is only applied for gaseous molecules, not for solid molecules or liquid molecules.