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Q.

At 300K and 1 atm pressure 40% of $N_{2} O_{4}$ is dissociated into $N O_{2}$ . The value of $K_{p}$ is

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a

$\frac{9}{49}$

b

$\frac{16}{21}$

c

$\frac{21}{16}$

d

$\frac{4}{7}$

answer is B.

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Detailed Solution

$N_{2} O_{4} ⇌ 2 {NO}_{2}$

$α = \sqrt{\frac{k_{p}}{k_{p} + 4 p}} 0.4 = \sqrt{\frac{k_{p}}{k_{p} + 4 (1)}} 16 \times 10^{- 2} = \frac{k_{p}}{k_{p} + 4} 64 \times 10^{- 2} + 16 \times 10^{- 2} k_{p} = k_{p} 64 \times 10^{- 2} = (1 - 16 \times 10^{- 2}) k_{p} k_{p} = \frac{16}{21}$

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