Question

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

Answer 1

Step 1 Given data:

Mass of gas is $3.5\mathrm{g}$

The molecular mass of gas is $28\mathrm{g}$

Heat capacity $\left(\mathrm{C}\right)$ is $2.5\mathrm{kJ}{\mathrm{K}}^{-1}$

Temperature change is $\left(\mathrm{\Delta T}\right)$ is $298.45-298.0=0.45\mathrm{K}$

Step 2 Formula used:

The energy released due to the combustion of a gas at a constant volume, use the formula $\mathrm{C}\times \mathrm{\Delta T}$

Step 3 Energy released due to combustion of $3.5\mathrm{g}$ of a gas:

The heat produced from the combustion of $3.5\mathrm{g}$ of compound rises the temperature of the calorimeter by $0.45\mathrm{K}$.

Therefore, the energy released due to the combustion of $3.5\mathrm{g}$ of a gas at a constant volume is,

$$\begin{gathered} \mathrm{C}\times \mathrm{\Delta T}=2.45\times 0.45 \\ \mathrm{C}\times \mathrm{\Delta T}=1.1025\mathrm{kJ} \end{gathered}$$

Step 4 Energy released due to the combustion of $28\mathrm{g}$ of a gas:

The energy released due to the combustion of $3.5\mathrm{g}$ of a gas at a constant volume is $1.1025\mathrm{kJ}$.

Thus, the energy released by the combustion of $28\mathrm{g}$ (1 mole) of gas is $1.1025\times \frac{28}{3.5}=8.82\mathrm{kJ}{\mathrm{mol}}^{-1}$.

Therefore, the energy released by the combustion of $28\mathrm{g}$ (1 mole) of gas will be $8.82\mathrm{kJ}{\mathrm{mol}}^{-1}$.