The standard enthalpies of formation of CO2(g),H2O(l) and glucose(s) at 25∘C are −400kJ/mol,−300kJ/mol and −1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25° is

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The standard enthalpies of formation of ${CO}_{2} (g), H_{2} O (l) and$ ${glucose(s) at 25}^{\circ} C a r e - 400 kJ / mol, - 300 kJ / mol a n d - 1300$ kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25° is

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+2900kJ

-—2900kJ

-16.11kJ

+16.11kJ

answer is C.

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Detailed Solution

The standard enthalpy of the combustion of glucose can be calculated by the eqn.
$C_{6} H_{12} O_{6} (s) + 6 O_{2} (g) \to 6 {CO}_{2} (g) + 6 H_{2} O (l) {ΔH}_{C}^{o} = 6 \times {ΔH}_{f} ({CO}_{2}) + 6 \times {ΔH}_{f} (H_{2} O) - {ΔH}_{f} [C_{6} H_{12} O_{6}] {ΔH}_{C}^{o} = 6 (- 400) + 6 (- 300) - (- 1300) {ΔH}_{C}^{o} = - 2900 kJ / mol$
For one gram of glucose, enthalpy of combustion
$Δ H^{\circ} = - \frac{2900}{180} = - 16 .11 kJ / gm$