Q22 / 150 · MHT CET 2026 (16 April Shift 2) · Chemistry
The heat of combustion of carbon to $\text{CO}_2$ is $-393.5\ \text{kJ/mol}$. The heat released upon formation of 35.2 g of $\text{CO}_2$ from carbon and oxygen gas is
- A$+315\ \text{kJ}$
- Correct answerB$-315\ \text{kJ}$
- C$-3.15\ \text{kJ}$
- D$-630\ \text{kJ}$
Solution
The combustion reaction of carbon is given by:
$\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$
The heat of combustion of carbon to $\text{CO}_2$ is $\Delta H = -393.5\ \text{kJ/mol}$.
Molar mass of $\text{CO}_2 = 12 + 2 \times 16 = 44\ \text{g/mol}$.
Number of moles in $35.2\ \text{g}$ of $\text{CO}_2$ is:
$n = \dfrac{35.2}{44} = 0.8\ \text{mol}$
The enthalpy change for the formation of $0.8\ \text{mol}$ of $\text{CO}_2$ is:
$\Delta H = 0.8\ \text{mol} \times (-393.5\ \text{kJ/mol}) = -314.8\ \text{kJ} \approx -315\ \text{kJ}$
The negative sign indicates that the heat is released during the process.
Answer: $-315\ \text{kJ}$
$\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$
The heat of combustion of carbon to $\text{CO}_2$ is $\Delta H = -393.5\ \text{kJ/mol}$.
Molar mass of $\text{CO}_2 = 12 + 2 \times 16 = 44\ \text{g/mol}$.
Number of moles in $35.2\ \text{g}$ of $\text{CO}_2$ is:
$n = \dfrac{35.2}{44} = 0.8\ \text{mol}$
The enthalpy change for the formation of $0.8\ \text{mol}$ of $\text{CO}_2$ is:
$\Delta H = 0.8\ \text{mol} \times (-393.5\ \text{kJ/mol}) = -314.8\ \text{kJ} \approx -315\ \text{kJ}$
The negative sign indicates that the heat is released during the process.
Answer: $-315\ \text{kJ}$
Hence, the correct answer is option: 'B'.