For the pair of reaction given below, (i) N _ 2 (g) + 3 H _ 2 ( g) ⇌ 2N H _ 3 ( g) (ii) (1)/(2) N _ 2 ( g) + (3)/(2) H _ 2 (g) ⇌ N H _ 3 (g) If at a particular temperature, K _ P _ 1 and K _ P _ 2 are the equilibrium

Correct answer: B. K _ P _ 1 = K _ P _ 2 ^ 2 K _ P _ 1 = p _ NH_ 3 ^ 2 p _ N _ 2 .p_ H _ 2 ^ 3 ...(i) K _ P _ 2 = P _ NH_ 3 p _ N _ 2 ^ (1)/(2) .p_ H _ 2 ^ (3)/(2) K _ P _ 2 ^ 2 = p _ NH_ 3 ^ 2 p _ N _ 2 .p_ H _ 2 ^ 3 ...(ii) Dividing equation (ii) by equation (i) K _ P _ 1 K _ P _ 2 ^ 2 =1 or K _ P _ 1 = K _ P _ 2 ^ 2

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

For the pair of reaction given below, (i) N _ 2 (g) + 3 H _ 2 ( g)...

For the pair of reaction given below,

(i)

{N}_{2} \left(g\right) + 3{H}_{2} \left(

g\right) ⇌

2N{H}_{3}\left(

g\right)

<br /> <br /> (ii) \dfrac{1}{2}{N}_{2}\left(

g\right) +

\dfrac{3}{2}{H}_{2}\left(g\right) ⇌ N{H}_{3}\left(g\right)

If at a particular temperature,

{K}_{{P}_{1}}

and

{K}_{{P}_{2}}

are the equilibrium constants for reaction (i) and (ii) respectively then,

Options

A ${K}_{{P}_{1}} \text{=}{\text{2K}_{{P}_{2}}$
B ${K}_{{P}_{1}} \text{=}{K}_{{P}_{2}}^{2}$
C ${\text{2K}_{{P}_{1}} \text{=}{K}_{{P}_{2}}$
D ${K}_{{P}_{1}}^{2} \text{=}{K}_{{P}_{2}}$

Correct answer

B. ${K}_{{P}_{1}} \text{=}{K}_{{P}_{2}}^{2}$

Step-by-step solution

{K}_{{P}_{1}} \text{=}\dfrac{{p}_{{\text{NH}_{3}}^{2}}{{p}_{{N}_{2}} {\text{.p}_{{H}_{2}}^{3}}

...(i)

{K}_{{P}_{2}} \text{=}\dfrac{{P}_{{\text{NH}_{3}}}{{p}_{{N}_{2}}^{\dfrac{1}{2}} {\text{.p}_{{H}_{2}}^{\dfrac{3}{2}}}

{K}_{{P}_{2}}^{2} \text{=}\dfrac{{p}_{{\text{NH}_{3}}^{2}}{{p}_{{N}_{2}} {\text{.p}_{{H}_{2}}^{3}}

...(ii)

Dividing equation (ii) by equation (i)

\dfrac{{K}_{{P}_{1}}}{{K}_{{P}_{2}^{2}}} \text{=1}

{K}_{{P}_{1}} \text{=}{K}_{{P}_{2}}^{2}

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