Question

At T(K), consider the following gaseous reaction, which is in equilibrium: N$_2$O$_5 \rightleftharpoons 2$NO$_2 + \frac{1}{2}$O$_2$. What is the fraction of N$_2$O$_5$ decomposed at constant volume and temperature, if the initial pressure is 300 mm Hg and pressure at equilibrium is 480 mm Hg? (Assume all gases as ideal)

• A. 0.2
• B. 0.6
• C. 0.4
• D. 0.8

Hint:

Equilibrium pressure calculations: $P_{\text{total}} = \sum P_i$.

Answer 1

The Correct Option is C

Let the initial pressure of N$_2$O$_5$ be $P_0 = 300$ mm Hg. Let $x$ be the fraction of N$_2$O$_5$ decomposed. At equilibrium: Pressure of N$_2$O$_5$ = $P_0(1-x)$ Pressure of NO$_2$ = $2P_0x$ Pressure of O$_2$ = $\frac{1}{2}P_0x$ Total pressure at equilibrium is given as 480 mm Hg. Therefore, $$ P_{\text{total}} = P_0(1-x) + 2P_0x + \frac{1}{2}P_0x = P_0(1 + \frac{3}{2}x) $$ $$ 480 = 300(1 + \frac{3}{2}x) $$ $$ \frac{480}{300} = 1 + \frac{3}{2}x $$ $$ 1.6 = 1 + \frac{3}{2}x $$ $$ 0.6 = \frac{3}{2}x $$ $$ x = \frac{2 \times 0.6}{3} = 0.4 $$