Question

The following concentrations were obtained for the formation of $NH_3 $ from $N_2 $ and $H_2 $ at equilibrium at $500 K. [N_2] = 1.5 x 10^{-2} M. [H_2] = 3.0 x 10^{-2} M$ and $[NH+3] = 1.2 x 10^{-2} M. $ The equilibrium constant is

• A. $4.06 \times 10^{-2} $
• B. $3.55 \times 10^2 $
• C. $3 \times 10^4 $
• D. $3.5 \times 10^{-7} $

Answer 1

Answer: (B)

The equilibrium constant for the reaction
$N_2{g} + 3H _2{g} \rightleftharpoons 2NH_3{g } $ can be written as,
$ k_{c }=\frac{\left[NH_{3h}\right]^{2}}{\left[N_{2g}\right]\left[H_{2g}\right]^{3} } $
$= \frac{\left(1.2\times10^{-2}\right)^{2}}{\left(1.5\times10^{-2}\right)\left(3.0\times10^{-2}\right)^{3}}=3.55\times10^{2}$