Question

Two moles of ammonia is introduced in a evacuated $500mL$ vessel at high temperature. The decomposition reaction is $2N{H}_3\left(g\right)\leftrightharpoons N_2\left(g\right)+3H_2\left(g\right)$ At the equilibrium $N{H}_3$, becomes $1$ mole then the K would be

• A. $0.42$
• B. $6.75$
• C. $1.7$
• D. $1.5$

Answer 1

Answer: (B)

$6.75$

$2N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$

$1$ $0.5$ $1.5$

Now $Kc=\frac{\left[N_2\right][H_2{]}^3}{[N{H}_3{]}^2}$

$=\frac{\left[0.5/0.5\right][1.5/0.5{]}^3}{[1.0/0.5{]}^2}$

$Kc=6.75$