9.2 g N2O4 is heated in a 1L vessel till equilibrium state is established N2O4(g)2NO2(g) In equilibrium state 50 % N2O4 was dissociated, equilibrium constant will be (mol. wt. of N2O4 = 92)

Question

9.2 g N2O4 is heated in a 1L vessel till equilibrium state is established N2O4(g)<=>2NO2(g) In equilibrium state 50 % N2O4 was dissociated, equilibrium constant will be (mol. wt. of N2O4 = 92) (A) 0.3 (B) 0.2 (C) 0.1 (D) 0.4

Tardigrade Admin · Accepted Answer

Moles of N2O4 =(9.2/92.0) =0.1 mole beginmatrix&N2O4&<=>&2NO2(g)&& textInitiailly &0.1 textmoles&&0& textAt eqm.&0.05&&0.05 × 2 endmatrix K=([NO2]2/[N2O4]) =(0.1×0.1/0.05) =0.2

Q. $9.2 g N_{2} O_{4}$ is heated in a $1 L$ vessel till equilibrium state is established
$N_{2} O_{4 (g)} 2 N O_{2 (g)}$
In equilibrium state $50% N_{2} O_{4}$ was dissociated, equilibrium constant will be (mol. wt. of $N_{2} O_{4} = 92$ )

$0.3$

$0.2$

$0.1$

$0.4$

Moles of $N_{2} O_{4} = \frac{9.2}{92.0} = 0.1$ mole
$Initiailly At eqm. N_{2} O_{4} 0.1 moles 0.05 2 N O_{2} (g) 0 0.05 \times 2$
$K = \frac{[ N O _{2} ] ^{2}}{[ N _{2} O _{4} ]}$
$= \frac{0.1 \times 0.1}{0.05}$
$= 0.2$

Q. 9.2gN2​O4​ is heated in a 1L vessel till equilibrium state is established N2​O4(g)​​2NO2(g)​ In equilibrium state 50%N2​O4​ was dissociated, equilibrium constant will be (mol. wt. of N2​O4​=92)

0.3

0.2

0.1

0.4