The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500 K. N2=1.5×10-2M, H2=3.0×10-2M and [NH3] = 1.2 × 10 -2M. Calculate the equilibrium constant.

A
$3.55 \times 10^{2}$
B
$1.06 \times 10^{4}$
C
$5.53 \times 10^{3}$
D
$3.05 \times 10^{4}$

Solution:

$N_{2} (g) + 3 H_{2} (g) \overset{}{⇌} 2 {NH}_{3} (g)$

$K_{C} = \frac{{[{NH}_{3} (g)]}^{2}}{[N_{2} (g)] {[H_{2} (g)]}^{3}} = \frac{{(1.2 \times 10^{- 2})}^{2}}{(1.5 \times 10^{- 2}) {(3.0 \times 10^{- 2})}^{3}}$

= 3.55 $\times 10^{2}$

The following concentrations were obtained for the formation of ${NH}_{3}$ from $N_{2}$ and $H_{2}$ at equilibrium at 500 K. $[N_{2}] = 1.5 \times 10^{- 2} M, [H_{2}] = 3.0 \times 10^{- 2} M$ and $[{NH}_{3}] = 1.2 \times 10^{- 2} M . Calculate the equilibrium constant .$

Solution:

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