Question

For the reaction, $H _{2}+ I _{2} \rightleftharpoons 2 HI$, the equilibrium concentration of $H _{2}, I _{2}$ and $HI$ are $8.0, 3.0$ and $28.0 \,mol / L$ respectively. The equilibrium constant is

• A. 28.34
• B. 32.66
• C. 34.78
• D. 38.88

Answer 1

Answer: (B)

32.66

Answer 2

Given $\left[H_{2}\right]=8.0\, mol / L$
$\left[I_{2}\right]=3.0\, mol / L$
$[H I]=28\, mol / L$
$K=?$
$H_{2}+I_{2} 2 H I$
$\therefore K=\frac{[H I]}{\left[H_{2}\right]\left[I_{2}\right]}=\frac{(28)^{2}}{(8) \times(3)}$
$=\frac{28 \times 28}{24}$
$=32.66$