The decomposition of N2O4 to NO2 is carried out at 280 K in chloroform. When equilibrium has been established, 0.2 mol of N2O4 and 2 × 10-3 mol of NO2 are present in 2 litre solution. The equilibrium constant for reaction N2O4 leftharpoons 2NO2 is

Question

The decomposition of N2O4 to NO2 is carried out at 280 K in chloroform. When equilibrium has been established, 0.2 mol of N2O4 and 2 × 10-3 mol of NO2 are present in 2 litre solution. The equilibrium constant for reaction N2O4 leftharpoons 2NO2 is (A) 1 × 10-2 (B) 2 × 10-3 (C) 1 × 10-5 (D) 2 × 10-5

Tardigrade Admin · Accepted Answer

K=([NO2]2/[N2O4])=([2×(10-3/2)]2/[(2)2])=(10-6/10-1)=10-5

$1 \times 1 0^{- 2}$

$2 \times 1 0^{- 3}$

$1 \times 1 0^{- 5}$

$2 \times 1 0^{- 5}$

$K = \frac{[ N O _{2} ] ^{2}}{[ N _{2} O _{4} ]} = \frac{[ 2 \times \frac{1 0 ^{- 3}}{2} ] ^{^{2}}}{[ \frac{2}{2} ]} = \frac{1 0 ^{- 6}}{1 0 ^{- 1}} = 1 0^{- 5}$

Q. The decomposition of N2​O4​ to NO2​ is carried out at 280 K in chloroform. When equilibrium has been established, 0.2mol of N2​O4​ and 2×10−3mol of NO2​ are present in 2litre solution. The equilibrium constant for reaction N2​O4​⇌2NO2​ is

1×10−2

2×10−3

1×10−5

2×10−5

K=[N2​O4​][NO2​]2​=[22​][2×210−3​]2​=10−110−6​=10−5

$1 \times 1 0^{- 2}$

$2 \times 1 0^{- 3}$

$1 \times 1 0^{- 5}$

$2 \times 1 0^{- 5}$

$K = \frac{[ N O _{2} ] ^{2}}{[ N _{2} O _{4} ]} = \frac{[ 2 \times \frac{1 0 ^{- 3}}{2} ] ^{^{2}}}{[ \frac{2}{2} ]} = \frac{1 0 ^{- 6}}{1 0 ^{- 1}} = 1 0^{- 5}$