Question

Ammonium hydrogen sulphide dissociates as follows
$N{H}_{4}HS\left(s\right)\rightleftharpoons N{H}_3\left(g\right)+H_{2}S\left(g\right)$
If solid $N{H}_{4}HS$ is placed in an evacuated flask at certain temperature it will dissociate until the total pressure is $600$ torr.
a) Calculate the value of equillibrium constant for the dissociation reaction
b) Addional $N{H}_3$ is introduced into the equillibrium mixture without changing the temperature until partial pressure of $N{H}_3$ is $750$ torr, what is the partial pressure of $H_{2}S$ under these conditions? What is the total pressure in the flask?

Answer 1

$N{H}_{4}H{S}_{\left(S\right)}\rightleftharpoons N{H}_{3\left(g\right)}+H_2{S}_{\left(g\right)}Equilibrium:PP$

$a)$ Given $P+P=600⟹P=300torr$

$\therefore K_P=P\times P=(300{)}^2=9\times {10}^4$

$b)$ Given $P_{N{H}_3}=750torr,$ Let $P_2$ be partial pressure of $H_{2}S$

$\therefore K_P=9\times {10}^4=P_{N{H}_3}\times P_2$

i.e. $9\times {10}^4=750\times P_2⟹P_2=120torr$

$\therefore P_{total}=P_{N{H}_3}+P_2=750+120=870torr$