The decomposition of N2O4 to NO2 is carried out at 280oC in chloroform- When equilibrium is reached- 0-2 mol of N2O4 and 2- 10-3 mol of NO2 are present in a 2L solution- The equilibrium constant for the reaction N2O4- 2NO2 is-

The correct option is D 1× 10^ 5 N_2O_4⇌ 2NO_2 Initial l nbsp; nbsp; 0Final 1 x nbsp; 2x 1 x=0.2 mol nbsp; nbsp;2x=2× 10^ 3mol ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

The decomposi...

Question

The decomposition of $N_{2} O_{4}$ to $N O_{2}$ is carried out at $280^{o} C$ in chloroform. When equilibrium is reached, 0.2 mol of $N_{2} O_{4}$ and $2 \times 10^{- 3}$ mol of $N O_{2}$ are present in a 2L solution. The equilibrium constant for the reaction $N_{2} O_{4} ⇌ 2 N O_{2}$ is:

1 \times 10^{- 2}

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2 \times 10^{- 3}

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1 \times 10^{- 5}

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2 \times 10^{- 5}

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Solution

The correct option is D $1 \times 10^{- 5}$
$N_{2} O_{4} ⇌ 2 N O_{2}$
Initial l 0
Final 1-x 2x
$1 - x = 0.2 m o l 2 x = 2 \times 10^{- 3} m o l$
$[N_{2} O_{4}] = \frac{0.2}{2} = 0.1 M [N_{2}] = \frac{2 \times 10^{- 3}}{2} = 10^{- 3} M$
$K = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]} = \frac{(10^{- 3})^{2}}{0.1} = 10^{- 5}$

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Similar questions

Q. A vessel contains

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The decomposition of N2O4 to NO2 is carried out at 280oC in chloroform. When equilibrium is reached, 0.2 mol of N2O4 and 2×10−3 mol of NO2 are present in a 2L solution. The equilibrium constant for the reaction N2O4⇌2NO2 is:

The correct option is D 1×10−5N2O4⇌2NO2Initial l 0Final 1-x 2x1−x=0.2mol2x=2×10−3mol[N2O4]=0.22=0.1M[N2]=2×10−32=10−3MK=[NO2]2[N2O4]=(10−3)20.1=10−5

The decomposition of $N_{2} O_{4}$ to $N O_{2}$ is carried out at $280^{o} C$ in chloroform. When equilibrium is reached, 0.2 mol of $N_{2} O_{4}$ and $2 \times 10^{- 3}$ mol of $N O_{2}$ are present in a 2L solution. The equilibrium constant for the reaction $N_{2} O_{4} ⇌ 2 N O_{2}$ is: