Question

In a constant volume calorimeter, 3.5 g of gas with molecular weight 28 was burnt in excess oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 K to 298.45 K due to the combustion process. Given that the calorimeter constant is 2.5 kJ K⁻¹. Calculate the enthalpy of combustion of the gas in kJ mol⁻¹.

Answer 1

Given, T_i = 298 K

T_f = 298.45 K

k = 2.5 kJ K⁻¹

m = 3.5 g

M_m = 28

heat evolved = k∆T

∆H_C = k (T_f – T_i)

∆H_C = 2.5 kJ K⁻¹ (298.45 – 298) K⁻¹

∆H_C = 1.125 kJ

∆H_C = `1.125/3.5 xx 28` kJ mol⁻¹

∆H_C = 9 kJ mol⁻¹