Question:
12g of pure graphite is burnt. Temperature rises from 298 K to 308 K. Heat capacity of the calorimeter is $20.7~kJ~K^{-1}$. Enthalpy change for combustion of 1 mole of graphite $(kJ~mol^{-1})$ is ________.
12g of pure graphite is burnt. Temperature rises from 298 K to 308 K. Heat capacity of the calorimeter is $20.7~kJ~K^{-1}$. Enthalpy change for combustion of 1 mole of graphite $(kJ~mol^{-1})$ is ________.
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Combustion always releases heat (negative $\Delta H$).
Updated On: Jun 26, 2026
- -2070
- -207
- +2070
- +207
- +2.07
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Calculate total heat ($q$) released and relate to moles.
Step 2: Meaning
$q = C \times \Delta T$. $\Delta T = 308 - 298 = 10 K$. $C = 20.7 kJ/K$.
Step 3: Analysis
$q = 20.7 \times 10 = 207 kJ$. 12g graphite $= 1 \text{ mole}$.
Step 4: Conclusion
Combustion is exothermic, so $\Delta H = -207 kJ/mol$. Final Answer: (B)
Calculate total heat ($q$) released and relate to moles.
Step 2: Meaning
$q = C \times \Delta T$. $\Delta T = 308 - 298 = 10 K$. $C = 20.7 kJ/K$.
Step 3: Analysis
$q = 20.7 \times 10 = 207 kJ$. 12g graphite $= 1 \text{ mole}$.
Step 4: Conclusion
Combustion is exothermic, so $\Delta H = -207 kJ/mol$. Final Answer: (B)
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