At equilibrium, N 2 O 4(g) leftharpoons 2 NO 2(g) the observed molecular weight of N 2 O 4 is 80 g mol -1 at 350 K. The percentage dissociation of N 2 O 4(g) at 350 K is

Question

At equilibrium, N 2 O 4(g) leftharpoons 2 NO 2(g) the observed molecular weight of N 2 O 4 is 80 g mol -1 at 350 K. The percentage dissociation of N 2 O 4(g) at 350 K is (A) 10 % (B) 15 % (C) 20 % (D) 18 %

Tardigrade Admin · Accepted Answer

N 2 O 4(g) leftharpoons 2 NO 2(g) Degree of dissociation may be calculated as x=( N -m/(n-1) m) [∴ n=2 (number of moles produced by 1 mol )] =(92-80/(2-1) 80)=(12/80)=0.15 Percentage dissociation =0.15 × 100=15 %

Q. At equilibrium, $N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}$ the observed molecular weight of $N_{2} O_{4}$ is $80 g m o l^{- 1}$ at $350 K$ . The percentage dissociation of $N_{2} O_{4 (g)}$ at $350 K$ is

$10%$

$15%$

$20%$

$18%$

$N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}$

Degree of dissociation may be calculated as

$x = \frac{N - m}{( n - 1 ) m} [∴ n = 2$ (number of moles produced by $1 m o l)]$

$= \frac{92 - 80}{( 2 - 1 ) 80} = \frac{12}{80} = 0.15$

Percentage dissociation $= 0.15 \times 100 = 15%$

Q. At equilibrium, N2​O4(g)​⇌2NO2(g)​ the observed molecular weight of N2​O4​ is 80gmol−1 at 350K. The percentage dissociation of N2​O4(g)​ at 350K is

10%

15%

20%

18%