In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due t

Question

In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ $K^{- 1}$ , the numerical value for the enthalpy of combustion of the gas in kJ $m o l^{- 1}$ is

Answer:

9.00

Solution:

Energy released by combustion of 3.5 g gas

$= 2.5 \times (298.45 - 298) k J$

Energy released by 1 mole of gas $= \frac{2.5 \times 0.45}{\frac{3.5}{28}} = 9 k J m o l^{- 1}$

TestHub · Accepted Answer

Energy released by combustion of 3.5 g gas = 2.5 × ( 298.45 - 298 ) k J Energy released by 1 mole of gas = 2.5 × 0.45 3.5 28 = 9 k J m o l - 1

Chemistry - Thermochemistry Question with Solution | TestHub

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