Question

A reaction system in equilibrium according to the equation $2S{O}_2+O_2\rightleftharpoons 2S{O}_3$ in 1 litre reaction vessel at a given temperature was found to contain 0.11 mol of $S{O}_2$, 0.12 of $S{O}_3$ and 0.05 mol of $O_{2.}$ Another 1 litre reaction vessel contains 64 g of $S{O}_2$ at the same temperature. What mass of $O_2$ must be added to this vessel in order that at equilibrium half of $S{O}_2$is oxidised to $S{O}_3$?

• A. 9.34 g
• B. 10.87 g
• C. 21.3 g
• D. 32.4 g

Answer 1

The correct option is A 9.34 g

$2S{O}_2+O_2\to 2S{O}_3$
$K_c=0.12\times 0.12/0.11\times 0.11\times 0.05=23.8$
Now,
$2S{O}_2+O_2\to 2S{O}_3$

$2S{O}_2$	$O_2$$	$2S{O}_3$
$\frac{64}{64}=1-2x$	$a-x$	$2x=0.5$

So mass of oxygen added ${0.5}^2/(0.5\times 0.5\times (a-0.25\left)\right)=23.8$
$a-x=0.04$

$a=0.29$
Mass of oxygen added $=0.29\times 32=\mathrm{9.34.}$