Before equilibrium is set-up for the chemical reaction, N2O4 leftharpoons 2NO2, vapour density of the gaseous mixture was measured. If D is the theoretical value of vapour density, variation of x with D/d is shown by the graph. What is value of D/d at point Y?

Question

Before equilibrium is set-up for the chemical reaction, N2O4 leftharpoons 2NO2, vapour density of the gaseous mixture was measured. If D is the theoretical value of vapour density, variation of x with D/d is shown by the graph. What is value of D/d at point Y? (A) 0 (B) 1.5 (C) 1 (D) 0.5

Tardigrade Admin · Accepted Answer

N2O4(g) leftharpoons 2NO2(g) Degree of dissociation, x=(D-d/d) or x=(D/d)-1 At Y, x=0 ∴ (D/d)-1=0 or (D/d)=1

Q. Before equilibrium is set-up for the chemical reaction, $N_{2} O_{4} ⇌ 2 N O_{2}$ , vapour density of the gaseous mixture was measured. If $D$ is the theoretical value of vapour density, variation of $x$ with $D / d$ is shown by the graph.
What is value of $D / d$ at point $Y$ ?

$0$

$1.5$

$1$

$0.5$

$N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}$
Degree of dissociation, $x = \frac{D - d}{d}$ or $x = \frac{D}{d} - 1$
At $Y, x = 0$
$∴ \frac{D}{d} - 1 = 0$ or $\frac{D}{d} = 1$

Q. Before equilibrium is set-up for the chemical reaction, N2​O4​⇌2NO2​, vapour density of the gaseous mixture was measured. If D is the theoretical value of vapour density, variation of x with D/d is shown by the graph. What is value of D/d at point Y?

0

1.5

1

0.5

N2​O4(g)​⇌2NO2(g)​ Degree of dissociation, x=dD−d​ or x=dD​−1 At Y,x=0 ∴dD​−1=0 or dD​=1

Q. Before equilibrium is set-up for the chemical reaction, $N_{2} O_{4} ⇌ 2 N O_{2}$ , vapour density of the gaseous mixture was measured. If $D$ is the theoretical value of vapour density, variation of $x$ with $D / d$ is shown by the graph.
What is value of $D / d$ at point $Y$ ?

$0$

$1.5$

$1$

$0.5$

$N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}$
Degree of dissociation, $x = \frac{D - d}{d}$ or $x = \frac{D}{d} - 1$
At $Y, x = 0$
$∴ \frac{D}{d} - 1 = 0$ or $\frac{D}{d} = 1$