Question

The equilibrium constant $K_c$ for the reaction, $A\left(g\right)+2B\left(g\right)\rightleftharpoons 3C\left(g\right)$ is $2\times {10}^{-3}$. What would be the equilibrium partial pressure gas $C$ if initial pressure of gas $A$ & $B$ are $1$ & $2$ atm respectively.

• A. $0.0625$ atm
• B. $0.1875$ atm
• C. $0.20$ atm
• D. None of these

Answer 1

The correct option is C $0.20$ atm

The equilibrium reaction is $A\left(g\right)+2B\left(g\right)\rightleftharpoons 3C\left(g\right)$.
The value of $\Delta n$ is $3-(1+2)=0$.

The relationship between $K_p$ and $K_c$ is $K_p=K_c(RT{)}^{\Delta n}$.

When the value of $\Delta n$ is 0, the term $(RT{)}^{\Delta n}$ becomes equal to 1.
In such case, $K_p=K_c=2\times {10}^{-3}$.

The expression for $K_p$ is $K_p=\frac{P_C^3}{P_A\times P_B^2}$.
Substitute values in the above expression.

$2\times {10}^{-3}=\frac{P_C^3}{1\times 2^2}$

Thus, $P_C=0.2atm$.