A sample of -octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ . As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K. The magnitude of the heat of combustion

Question

A sample of $n$ -octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ $K^{- 1}$ . As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K. The magnitude of the heat of combustion of octane at constant volume is ____ kJ ${mol}^{- 1}$ (nearest integer). [2025]

Smart Achievers · Accepted Answer

(2500)

$Number of moles of octane (n) = \frac{Mass of octane}{Molar mass of octane} = \frac{1.14}{114} mol = 0.01 mol$

$Heat evolved (Q) = C Δ T = 5 {kJ K}^{- 1} \times 5 K = 25 kJ$

$Magnitude of heat of combustion = \frac{Q}{n} = \frac{25 kJ}{0.01 mol} = 2500 kJ/mol$

Solution

Ans.
(2500)

$Number of moles of octane (n) = \frac{Mass of octane}{Molar mass of octane} = \frac{1.14}{114} mol = 0.01 mol$

$Heat evolved (Q) = C Δ T = 5 {kJ K}^{- 1} \times 5 K = 25 kJ$

$Magnitude of heat of combustion = \frac{Q}{n} = \frac{25 kJ}{0.01 mol} = 2500 kJ/mol$

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Solution

Ans. (2500) Number of moles of octane (n)=Mass of octaneMolar mass of octane=1.14114 mol=0.01 mol Heat evolved (Q)=CΔT=5kJ K-1×5 K=25 kJ Magnitude of heat of combustion=Qn=25 kJ0.01 mol=2500 kJ/mol

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Ans.
(2500)

$Number of moles of octane (n) = \frac{Mass of octane}{Molar mass of octane} = \frac{1.14}{114} mol = 0.01 mol$

$Heat evolved (Q) = C Δ T = 5 {kJ K}^{- 1} \times 5 K = 25 kJ$

$Magnitude of heat of combustion = \frac{Q}{n} = \frac{25 kJ}{0.01 mol} = 2500 kJ/mol$