Question

The value of $k_p$ for the reaction
$2AB\left(g\right)\iff 2A\left(g\right)+B_2\left(g\right)$ is $1.2\times {10}^{-2}$ at ${1065}^{\circ }C$. The value of $k_c$ for this reaction is:

• A. $1.2\times {10}^{-2}$
• B. $<1.2\times {10}^{-2}$
• C. 83
• D. $>1.2\times {10}^{-2}$

Answer 1

The correct option is B $<1.2\times {10}^{-2}$

The equilibrium reaction is $2AB\left(g\right)\iff 2A\left(g\right)+B_2\left(g\right)$.
The relationship between $K_p$ and $K_c$ is $K_p=K_c(RT{)}^{\Delta n}$.
For the equilibrium reaction, $\Delta n=2+1-2=1>0$.
Hence, $K_p>K_c$.
But $K_p=1.2\times {10}^{-2}$.
Hence, $K_c<1.2\times {10}^{-2}$.