Question

${A_g}_ + 2{B_g} \rightleftharpoons 3{C_g} + {D_g}$.
The value of $Kp$ ​ = $0.05\;atm$ at $1000K$. The value of $Kc$ is represented by

$$A.{\text{ }}5 \times {10^{ - 4}}R \\\ B.{\text{ }}\dfrac{{5 \times {{10}^{ - 5}}}}{R} \\\ C.{\text{ }}5 \times {10^{ - 5}}{R^{ - 1}} \\\ D.{\text{ }}\dfrac{{5 \times {{10}^{ - 5}}}}{R} \\\$$

Answer 1

We must know that for any chemical reactions of gaseous molecules, the relation between $Kp$ and $Kc$ are given as follows,
$Kp = Kc{\left( {RT} \right)^{\Delta n}}$
Therefore, we can find the value of $Kc$ for the equilibrium reaction.

Complete step by step answer:
The equilibrium constant of any chemical reaction is given by the value of the reaction quotient when the reaction has reached equilibrium.
The equilibrium constant, describes the ratio of product and reactant concentrations at equilibrium in terms of partial pressures.
$Kp$ And $Kc$ are the equilibrium constant of any chemical reaction.
$Kc$ is an equilibrium constant that is a dependent value on concentrations of reactants and concentrations of products. Simply we can say that, $Kc$ is the ratio of product of product concentration raised to powers of their coefficient to product of reactant concentration raised to powers of their coefficient.
Equilibrium constants $Kc$ are expressed in molarity.
$Kp$ is an equilibrium constant that is dependent on the partial pressures applied by the gaseous components during the reaction. Simply we can say that, $Kp$ is the ratio of product of partial pressures of product raised to powers of their coefficient to product of reactant partial pressure raised to powers of their coefficient.
Equilibrium constant $Kp$ is are expressed in atmospheric pressure
Relations between $Kp$ and $Kc$ are given as.
$Kp = Kc{\left( {RT} \right)^{\Delta n}}$
Where,
$\Delta n$ represents the change in the number of moles of gas molecules.
So, from given in question,
$KP = 0.05\;atm,$
$T = 1000\;K$
∴ $\Delta n$=no of moles of product – no of moles reactant
∴ $\Delta n$=$4 - 3{\text{ }} = 1$
Substituting the values, we get
$Kp = Kc{\left( {RT} \right)^{\Delta n}}$
$0.05{\text{ }} = {\text{ }}Kc{\text{ }}{\left( {R{\text{ }} \times 1000} \right)^1}$
∴ $Kc{\text{ }} = \dfrac{{5 \times {{10}^{ - 5}}}}{R} = 5 \times {10^{ - 5}}{R^{ - 1}}$
Hence, the correct option is option D.

Note:
We must know that, when the change in the number of moles of gas molecules, i.e. $\Delta n$ is equal to zero, then $Kp$ becomes equal to $Kc$.