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 increases from 298.0 K to 98.45 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 mol -1 is

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 increases from 298.0 K to 98.45 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 mol -1 is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Temperature rise =T2-T1=298.45-298=0.45 K q= heat capacity × Δ T=2.5 × 0.45=1.125 kJ ⇒ Heat produced per mole =(1.125/3.5) × 28=9 kJ

Temperature rise =T2​−T1​=298.45−298=0.45K q= heat capacity ×ΔT=2.5×0.45=1.125kJ ⇒ Heat produced per mole =3.51.125​×28=9kJ

Temperature rise $= T_{2} - T_{1} = 298.45 - 298 = 0.45 K$
$q =$ heat capacity $\times Δ T = 2.5 \times 0.45 = 1.125 k J$
$\Rightarrow$ Heat produced per mole $= \frac{1.125}{3.5} \times 28 = 9 k J$