Question

At 273K and 1 atm $a$ L of $N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$. To what extent has the decomposition proceeded when the original volume is 25% less than that of existing volume?

• A. $\alpha$ = 40%
• B. $\alpha$ = 10%
• C. $\alpha$ = 33%
• D. $\alpha$ = 67%

Answer 1

The correct option is C $\alpha$ = 33%

$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$
a 0 Moles before equilibrium
(a-x) 2x Moles at equilibrium
Given that original volume$=\frac{75}{100}\times$ existing volume at equilibrium
Since, $Moles\propto volume\left(atconstantPandT\right)$
$\therefore InitialMoles=\frac{75}{100}\times molesatequilibrium$
$a=\frac{75}{100}(a+x)$
$\therefore x=\frac{25}{75}a=0.33a$
Now,%decomposition
$\left(\alpha \right)$ =$\frac{x}{a}=\frac{0.33a}{a}=0.33or33$ %