Question

Ammonium Hydrogen sulphide dissociates according to the equation,

$N{H}_{4}HS$(s) $\leftrightharpoons$ $N{H}_3$(g) + $H_{2}S$(g) if the observed pressure of the mixture is 1.12 atm at ${106}^{\circ }$C. What is the KP of the reaction -

• A. 0.2136
• B. 0.3136
• C. 0.4136
• D. 0.5126

Answer 1

The correct option is B

0.3136

The reaction is $N{H}_{4}HS$ (s) $\leftrightharpoons$ $N{H}_3$ (g) + $H_{2}S$ (g)

at equilibrium 1 - $\alpha$

Total moles of $N{H}_3$ + $H_{2}S$ = 2 $\alpha$ ($N{H}_{4}HSinsolidphase$)

Partial pressure = $\frac{molesofsubstance}{totalmoles}$ $\times$ total pressure

$P_{N{H}_3}$ = $\frac{\alpha }{2\alpha }$. p = 0.5 p

$P_{H_{2}S}$ = $\frac{\alpha }{2\alpha }$.p = 0.5p

$K_p$ = $P_{N{H}_3}$. $P_{H_{2}S}$ = ( p $\times$ 0.5) $\times$ (0.5 $\times$ p) = 0.25 $p^2$

Substituting value of p = 1.12 atm

$K_p$ = 0.25 $\times$ 1.12 $\times$ 1.12 = 0.3136