Question

The equilibrium:
${NH}_{4}HS\left(s\right)\rightleftharpoons {NH}_3\left(g\right)+H_{2}S\left(g\right)$
is followed to set up at ${127}^{o}C$ in a closed vessel. The total pressure at equilibrium was $20atm$. The $K_c$ for the reaction is:

• A. $0.092M^2$
• B. $0.085M^2$
• C. $3.045M^2$
• D. None of these

Answer 1

Answer: (A)

$0.092M^2$

The equilibrium:

${NH}_{4}HS\left(s\right)\rightleftharpoons {NH}_3\left(g\right)+H_{2}S\left(g\right)$

is followed to set up at ${127}^{o}C$ in a closed vessel.

The total pressure at equilibrium was $20atm$.

$p_{{NH}_3}=p_{H_{2}S}=\frac{P_{total}}{2}=\frac{20}{2}atm=10atm$

$K_p=P_{NH3}\times P_{H2S}=10atm\times 10atm=100at{m}^2$

$K_p=K_c\times {\left(RT\right)}^{\Delta n}$

$100at{m}^2=K_c\times {(0.08206Latm/molK\times (127+273\left)K\right)}^{1+1-0}$

$100at{m}^2=K_c\times {1077.5\left(Latm/mol\right)}^2$

$K_c=0.092M^2$

The $K_c$ for the reaction is $0.092M^2$.