A mixture of and in 2:1 mole ratio is used to prepare water vapour by the reaction: . The total pres — States of Matter and Gaseous State Chemistry Question
Question
A mixture of $H_2$ and $O_2$ in 2:1 mole ratio is used to prepare water vapour by the reaction: $2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$. The total pressure of gases in the container is 4.5 atm at 57°C, before the reaction. The final total pressure of gases (in atm) at 127°C after reaction assuming 80% yield of water vapour is
💡 Solution & Explanation
Initially $P_{total} = 4.5\text{ atm}$. Due to 2:1 ratio, $P_{H_2} = 3.0\text{ atm}$ and $P_{O_2} = 1.5\text{ atm}$. At 80% yield, $0.8 \times 3.0 = 2.4\text{ atm}$ of $H_2$ reacts with $1.2\text{ atm}$ of $O_2$ to form $2.4\text{ atm}$ of $H_2O$. Remaining: $H_2 = 0.6\text{ atm}$, $O_2 = 0.3\text{ atm}$. Total pressure at 57°C is $0.6 + 0.3 + 2.4 = 3.3\text{ atm}$. When heated to 127°C (400 K) from 57°C (330 K), $P_{final} = 3.3 \times \frac{400}{330} = 4.0\text{ atm}$. Therefore, correct answer is 4.