Question

The equilibrium $N_2+O_2\rightleftharpoons 2NO$ is established in a vessel with a capacity of 2.5 L. The initial amount of $N_2$ is 2 mol and that of $O_2$ is 4 mol. Half a mole of $N_2$ is used up at the equilibrium. The molar concentration of NO is

• A. 0.2
• B. 0.4
• C. 0.6
• D. 0.1

Answer 1

Answer: (B)

0.4

$N_2+O_2⟶2No$

At $t=0$ 2 moles 4 moles

At $t=t_{eq}$ $\frac{1}{2}$ moles of $N_2$ is used up.

let at $t-t_{eq}O$ x moles is used up.

$N_2+O_2⟶2No$

$t=t_{eq}=$$2-x$ $4-x$ $2x$

where $x=0.5$

$\therefore t=t_{eq}2-0.54-0.52\times 0.5$

$=1.53.51$

$\therefore$ the molar concentration of $NO$ at $t=t_{eq}$

$=\frac{molesofNo}{Volume}$

$=\frac{1}{2.5}=0.4$