Question

A mixture of $S{O}_3,S{O}_2$ and $O_2$ gases is maintained at equilibrium in 10 litre. flask at a temperature at which Kc for the reaction, $2S{O}_2\left(g\right)+O_2\left(g\right)\rightharpoonup \leftharpoondown 2S{O}_3\left(g\right)$ is $100mo{l}^{-1}$ litre. At equilibrium.

• A. If no. of mole of $S{O}_3,S{O}_2$ in the flask are same, how many moles of $O_2$ are present 0.1
• B. If no. of mole of $S{O}_3$ in the flask are twine the no.of mole of $S{O}_2$, how many moles of $O_2$ are present 0.4
• C. Both A and B
• D. None

Answer 1

Answer: (C)

Both A and B

$2S{O}_2+O_2=2S{O}_3$

$k_c=\frac{[S{O}_3{]}^2}{\left[S{O}_2{]}^2\right[O_2]}$

(1) From given information, moles of sulfur trioxide and sulfur dioxide are same. Let's say at equilibrium, we have n moles of sulfur trioxide and m moles of oxygen. Moles of sulfur dioxide will also be n since sulfur trioxide and sulfur dioxide has same moles.

Kc is given as 100. Divide the moles of the liters to get the concentrations and plug in the values in the equilibrium expression.

$\left[S{O}_3\right]=n/10$

$\left[S{O}_2\right]=n/10$

$\left[O_2\right]=m/10$

$100=\frac{[n/10{]}^2}{\left[n/10{]}^2\right[m/10]}$

$100=10/m$

$m=0.1$

So, there are 0.1 moles of at equilibrium.

(2) Let's say there are n moles of sulfur dioxide and m moles of oxygen. Then moles of sulfur trioxide will be 2n as the moles of it are twice of sulfur dioxide.

$\left[S{O}_3\right]=2n/10$

$\left[S{O}_2\right]=n/10$

$\left[O_2\right]=m/10$

$100=\frac{[2n/10{]}^2}{\left[n/10{]}^2\right[m/10]}$

$100=40/m$

$m=0.4$

So, there are 0.4 moles of at equilibrium.