Question

The value of $K_p$ for the reaction.
$2H_{2}S\left(g\right)\iff 2H_2\left(g\right)+S_2\left(g\right)$ 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 $2H_{2}S\left(g\right)\iff 2H_2\left(g\right)+S_2\left(g\right)$.
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, $K_c<1.2\times {10}^{-2}$.