Question

The equilibrium constant ${K_p}$ for the following reaction at ${191^ \circ }C$ is $10.24$. What is ${K_c}$.
$B(s) + \dfrac{3}{2}{F_2}(g) \rightleftharpoons B{F_3}(g)$
(A) 6.7
(B) 0.61
(C) 8.30
(D) 7.6

Answer 1

Hint: At first we will note down the given data in the question and we will write the equation. We will write the relation between ${K_p}$ and ${K_c}$ . Then we will put the values in the reaction and then calculate the ${K_c}$ . We should check the units of the values we put in the equation. From there we can choose the correct answer.

Complete step-by-step solution:Step1. The equation given in the question is $B(s) + \dfrac{3}{2}{F_2}(g) \rightleftharpoons B{F_3}(g)$ .
There is one mole of gaseous product and one and half the gaseous reactant.
The ${K_p}$ = $10.24$ , ${K_p}$ is the equilibrium constant.
Temperature given is ${191^ \circ }C$.
The temperature we need must be in kelvin. The temperature in Kelvin is $191 + 273 = 463K$
Step2. The relation between ${K_c}$ and ${K_p}$ is :
${K_p} = {K_c}{(RT)^{\Delta n}}$
Here ${K_p}$ is equilibrium constant expressed in the pressure
${K_c}$ equilibrium constant expressed in the molarity.
R is the universal gas constant which is $0.082Latm/K/mol$
T is temperature
The $\Delta n = {n_{(gaseous{\text{ product)}}}} - {n_{(gaseous{\text{ reactant)}}}}$
Step3. When we substitute the values we get
$1.24 = {K_c}{(0.082 \times 463)^{ - \dfrac{1}{2}}}$
$\Rightarrow {K_c} = 1.24 \times \sqrt {0.082 \times 463} $
$\therefore {K_c} = 7.64$
So the value of the ${K_c}$ is $7.64$.

Hence the option (D) is the correct answer.

Note:The equilibrium state is the state of the reaction where there is enough time have passed and the reaction have come at halt. It loses its tendency to move forward. The equilibrium constant is the value of reaction quotient. The ${K_c}$ and ${K_p}$ represent the same thing but ${K_p}$is calculated by the partial pressure of the components of equation. While the ${K_c}$ is calculated by the molarity of the components in the equation.