For the reaction, 2 NO_2 arrow 2 NO+O_2 , the value of K_p is equal to K . What will the value of K_p for reaction, NO+(1)/(2) O_2 NO_2

Correct answer: C. (1)/(K) 2 NO_2 2 NO+O_2 K_p= P_ NO_2 ^2 · P_ O_2 P_ NO_2 =K (i) For reaction, NO+(1)/(2) O_2 leftharpoons NO_2 K_p^ = P_ NO_2 P_ NO · P_ O_2 ^ 1 / 2 (ii) From equation (i) and (ii) K_ p ^ = 1 ~K

JIPMER 2004

Exam: JIPMER
Year: 2004
Subject: Chemistry
Chapter: Chemical Equilibrium

JIPMER 2004 · Chemistry · Chemical Equilibrium

For the reaction, 2 NO_2 arrow 2 NO+O_2 , the value of K_p is equal...

For the reaction,

2 \mathrm{NO}_2 \rightarrow 2 \mathrm{NO}+\mathrm{O}_2

, the value of

K_p

is equal to

K

. What will the value of

K_p

for reaction,

\mathrm{NO}+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{NO}_2

Options

A $\sqrt{K}$
B $\frac{1}{K}$
C $\sqrt{\frac{1}{K}}$
D $\frac{1}{K^2}$

Correct answer

C. $\sqrt{\frac{1}{K}}$

Step-by-step solution

2 \mathrm{NO}_2 \quad 2 \mathrm{NO}+\mathrm{O}_2

K_p=\frac{\mathrm{P}_{\mathrm{NO}_2}^2 \cdot \mathrm{P}_{\mathrm{O}_2}}{\mathrm{P}_{\mathrm{NO}_2}}=K \quad \ldots(i)

For reaction,

\mathrm{NO}+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{NO}_2

K_p^{\prime}=\frac{\mathrm{P}_{\mathrm{NO}_2}}{\mathrm{P}_{\mathrm{NO}} \cdot \mathrm{P}_{\mathrm{O}_2}^{1 / 2}} \quad \ldots(ii)

From equation (i) and (ii)

\mathrm{K}_{\mathrm{p}}^{\prime}=\sqrt{\frac{1}{\mathrm{~K}}}

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