Question

$1.50mol$ each of hydrogen and iodine were placed in a sealed $10L$ container maintained at $717K$. At equilibrium $1.25mol$ each of hydrogen and iodine were left behind. The equilibrium constant, $K_c$ for the reaction
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ at $717K$ is:

• A. $0.4$
• B. $0.16$
• C. $50$
• D. $25$

Answer 1

Answer: (B)

$\begin{array}{cc}\begin{array}{c}att=0\\ ateq.\end{array}\begin{array}{c}{H}_2\\ \begin{array}{c}1.50\\ 1.50-x\end{array}\end{array}+\begin{array}{c}{I}_2\\ \begin{array}{c}1.50\\ 1.50-x\end{array}\end{array}\begin{array}{c}2HI\\ \begin{array}{c}0\\ 2x\end{array}\end{array}& \begin{array}{c}1.50-x=1.25\\ \begin{array}{c}x=0.25mol.\\ \left[HI\right]=0.50mol.\end{array}\end{array}\\ K_e=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}=\frac{{\left(0.50\right)}^2}{\left(1.25\right)\left(1.25\right)}\Rightarrow \frac{0.50\times 0.50}{1.25\times 1.25}\Rightarrow \frac{4}{25}\Rightarrow 0.16& \end{array}$

Option B is correct.