Question:
The enthalpy of combustion of glucose (molecular weight = 180 g mol$^{-1}$) is $-2840$ kJ mol$^{-1}$. Then the amount of heat evolved when $0.9$ g of glucose is burnt, will be}
The enthalpy of combustion of glucose (molecular weight = 180 g mol$^{-1}$) is $-2840$ kJ mol$^{-1}$. Then the amount of heat evolved when $0.9$ g of glucose is burnt, will be}
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Always convert mass into moles before using enthalpy values given per mole. Also, negative enthalpy indicates heat is evolved.
Updated On: May 8, 2026
- $14.2$ kJ
- $14.2$ J
- $28.4$ kJ
- $1420$ kJ
- $142$ kJ
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The Correct Option is A
Solution and Explanation
Concept:
Enthalpy of combustion represents the heat released when 1 mole of a substance is completely burnt. The heat released is directly proportional to the number of moles of substance burnt:
\[
q = n \times \Delta H
\]
where $n = \frac{\text{given mass}}{\text{molar mass}}$.
Step 1: Calculate number of moles of glucose. \[ n = \frac{0.9}{180} = 0.005 \text{ mol} \]
Step 2: Use enthalpy relation. \[ \Delta H = -2840 \text{ kJ/mol} \]
Step 3: Calculate heat evolved. \[ q = 0.005 \times (-2840) = -14.2 \text{ kJ} \]
Step 4: Interpretation. Negative sign indicates heat is released (exothermic reaction). Hence magnitude: \[ \boxed{14.2 \text{ kJ}} \]
Step 1: Calculate number of moles of glucose. \[ n = \frac{0.9}{180} = 0.005 \text{ mol} \]
Step 2: Use enthalpy relation. \[ \Delta H = -2840 \text{ kJ/mol} \]
Step 3: Calculate heat evolved. \[ q = 0.005 \times (-2840) = -14.2 \text{ kJ} \]
Step 4: Interpretation. Negative sign indicates heat is released (exothermic reaction). Hence magnitude: \[ \boxed{14.2 \text{ kJ}} \]
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