SATHEE: Equilibrium Question 650

Equilibrium Question 650

Question: For the reaction equilibrium $N_{2} O_{4}$ ⇌ $2 N O_{2 (g)}$ , the concentrations of $N_{2} O_{4}$ and $N O_{2}$ at equilibrium are $4.8 \times 10^{- 2}$ and $1.2 \times 10^{- 2}, m o l, l i t r e^{- 1}$ respectively. The value of $K_{c}$ for the reaction is [AIEEE 2003]

Options:

A) $3.3 \times 10^{2}$ $m o l l i t r e^{- 1}$

B) $3 \times 10^{- 1}$ $m o l l i t r e^{- 1}$

C) $3 \times 10^{- 3}$ $m o l l i t r e^{- 1}$

D) $3 \times 10^{3}$ $m o l l i t r e^{- 1}$

Show Answer

Answer:

Correct Answer: C

Solution:

$K = \frac{{[N O_{2}]}^{2}}{[N_{2} O_{4}]}$

$= \frac{{[1.2 \times 10^{- 2}]}^{2}}{[4.8 \times 10^{- 2}]}$

$= 0.3 \times 10^{- 2}$

$= 3 \times 10^{- 3}$

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Equilibrium Question 650

Question: For the reaction equilibrium N2O4 ⇌ 2NO2(g) , the concentrations of N2O4 and NO2 at equilibrium are 4.8×10−2 and 1.2×10−2,mol,litre−1 respectively. The value of Kc for the reaction is [AIEEE 2003]

Options:

Answer:

Solution:

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Question: For the reaction equilibrium $N_{2} O_{4}$ ⇌ $2 N O_{2 (g)}$ , the concentrations of $N_{2} O_{4}$ and $N O_{2}$ at equilibrium are $4.8 \times 10^{- 2}$ and $1.2 \times 10^{- 2}, m o l, l i t r e^{- 1}$ respectively. The value of $K_{c}$ for the reaction is [AIEEE 2003]