9.2 grams of N2O4(g) is taken in a closed one litre vessel and heated till the following equilibrium is reached N2O4(g) leftharpoons 2NO2(g). At equilibrium, 50 % N2O4(g) is dissociated. What is the equilibrium constant (in mol litre-1) (Molecular weight of N2O4 = 92)?

Question

9.2 grams of N2O4(g) is taken in a closed one litre vessel and heated till the following equilibrium is reached N2O4(g) leftharpoons 2NO2(g). At equilibrium, 50 % N2O4(g) is dissociated. What is the equilibrium constant (in mol litre-1) (Molecular weight of N2O4 = 92)? (A) 0.1 (B) 0.4 (C) 0.2 (D) 2

Tardigrade Admin · Accepted Answer

Kc=([NO2]2/[N2O4])=(4×(0.05)2/0.05)=4 ×0.05 =0.2

$0.1$

$0.4$

$0.2$

$2$

$K_{c} = \frac{[ N O _{2} ] ^{2}}{[ N _{2} O _{4} ]} = \frac{4 \times ( 0.05 ) ^{2}}{0.05} = 4 \times 0.05 = 0.2$

Q. 9.2 grams of N2​O4(g)​ is taken in a closed one litre vessel and heated till the following equilibrium is reached N2​O4(g)​⇌2NO2(g)​. At equilibrium, 50%N2​O4(g)​ is dissociated. What is the equilibrium constant (in mol litre-1) (Molecular weight of N2​O4​=92)?

0.1

0.4

0.2

2

Kc​=[N2​O4​][NO2​]2​=0.054×(0.05)2​=4×0.05=0.2

$0.1$

$0.4$

$0.2$

$2$

$K_{c} = \frac{[ N O _{2} ] ^{2}}{[ N _{2} O _{4} ]} = \frac{4 \times ( 0.05 ) ^{2}}{0.05} = 4 \times 0.05 = 0.2$