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Standard XII

Chemistry

Characteristics of Equilibrium Constant

Ammonium hydr...

Question

Ammonium hydrogen sulphide dissociates according to the equations:
${N H}_{4} H S (s) ⇌ {N H}_{3} (g) + H_{2} S (g)$
If the observed pressure at equilibrium is $1.12 a t m$ at $380 K$ , what is the equilibrium constant $K_{p}$ of the reaction?

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Solution

${N H}_{4} H S (s) ⇌ {N H}_{3} (g) + H_{2} S (g)$
Total pressure is 1.12 atm. One half of this is equal to partial pressure of ammonia. It is also equal to partial pressure of hydrogen sulphide.
$p_{{N H}_{3}} = p_{H_{2} S} = \frac{1}{2} \times 1.12 = 0.560 a t m$
$K_{p} = p_{{N H}_{3}} \times p_{H_{2} S} = 0.560 \times 0.560 = 0.314$
The equilibrium constant is $0.314$ .

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Ammonium hydrogen sulphide dissociates according to the equations:NH4HS(s)⇌NH3(g)+H2S(g)If the observed pressure at equilibrium is 1.12atm at 380K, what is the equilibrium constant Kp of the reaction?

NH4HS(s)⇌NH3(g)+H2S(g)Total pressure is 1.12 atm. One half of this is equal to partial pressure of ammonia. It is also equal to partial pressure of hydrogen sulphide.pNH3=pH2S=12×1.12=0.560atmKp=pNH3×pH2S=0.560×0.560=0.314The equilibrium constant is 0.314.

Ammonium hydrogen sulphide dissociates according to the equations:
${N H}_{4} H S (s) ⇌ {N H}_{3} (g) + H_{2} S (g)$
If the observed pressure at equilibrium is $1.12 a t m$ at $380 K$ , what is the equilibrium constant $K_{p}$ of the reaction?