Q52 / 180 · COMEDK 2026 (9 May Shift 1) · Chemistry
The heat of combustion of carbon to $\text{CO}_2$ is -393.5 kJ $\text{mol}^{-1}$.
The heat released on the formation of 35.2 g of $\text{CO}_2$ by combustion of C is:
The heat released on the formation of 35.2 g of $\text{CO}_2$ by combustion of C is:
- A$+620\ \text{kJ}$
- Correct answerB$-315\ \text{kJ}$
- C$-325\ \text{kJ}$
- D$+215\ \text{kJ}$
Solution
The combustion reaction of carbon is given by:
$\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \quad \Delta H = -393.5\ \text{kJ mol}^{-1}$
Molar mass of $\text{CO}_2 = 12 + 2 \times 16 = 44\ \text{g mol}^{-1}$
Number of moles of $\text{CO}_2$ formed $= \dfrac{35.2}{44} = 0.8\ \text{mol}$
Enthalpy change for the formation of $0.8\ \text{mol}$ of $\text{CO}_2 = 0.8 \times (-393.5\ \text{kJ}) = -314.8\ \text{kJ}$
Rounding to the nearest integer, the heat released (enthalpy change) is approximately $-315\ \text{kJ}$.
Answer: $-315\ \text{kJ}$
$\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \quad \Delta H = -393.5\ \text{kJ mol}^{-1}$
Molar mass of $\text{CO}_2 = 12 + 2 \times 16 = 44\ \text{g mol}^{-1}$
Number of moles of $\text{CO}_2$ formed $= \dfrac{35.2}{44} = 0.8\ \text{mol}$
Enthalpy change for the formation of $0.8\ \text{mol}$ of $\text{CO}_2 = 0.8 \times (-393.5\ \text{kJ}) = -314.8\ \text{kJ}$
Rounding to the nearest integer, the heat released (enthalpy change) is approximately $-315\ \text{kJ}$.
Answer: $-315\ \text{kJ}$
Hence, the correct answer is option: 'B'.