The thermal decomposition of lithium peroxide (Li2O2) is given by the equation shown - HSC - SSCE Chemistry - Question 15 - 2024 - Paper 1

To determine the ratio of the amount of $O_2(g)$ in containers P and Q, we begin by analyzing the equilibrium equation given:

$2Li_2O_2(s) \rightleftharpoons 2Li_2O(s) + O_2(g)$

From this equation, it can be seen that for every 2 moles of lithium peroxide decomposed, 1 mole of oxygen gas is produced.

Let’s denote the amount of Li2O2 in container Q as $x$; therefore, in container P, it will be $2x$. As both containers reach equilibrium, we must consider the relationships:

• For container Q:

  • The decomposition of $x$ moles of Li2O2 will produce rac{x}{2} moles of $O_2(g)$.

• For container P:

  • The decomposition of $2x$ moles of Li2O2 will produce rac{2x}{2} = x moles of $O_2(g)$.

Now, we can define the ratio of $[O_2(g)]$ in P to that in Q:

$\text{Ratio} = \frac{[O_2(g)]_P}{[O_2(g)]_Q} = \frac{x}{\frac{x}{2}} = 2:1$

Thus, the final ratio of $[O_2(g)]$ in container P to $[O_2(g)]$ in container Q is 2:1.