JEE Chemistry 2023 Question with Solution

Given:

• Mass of ethane = $0.3 \, \text{g}$
• Temperature rise = $0.5 \, \text{K}$
• Heat capacity of calorimeter system = $20 \, \text{kJ K}^{-1}$
• $R = 8.3 \, \text{J K}^{-1} \, \text{mol}^{-1}$
• Temperature = $27^\circ \text{C} = 300 \, \text{K}$

Find: Heat evolved during combustion of ethane at constant pressure in $\text{kJ mol}^{-1}$.

Bomb calorimeter works at constant volume, so the measured heat gives $\Delta U$.

Heat released for the sample:

$$q = C_{cal} \Delta T = 20 \times 0.5 = 10 \, \text{kJ}$$

Molar mass of ethane $C_2H_6$ is:

$$2 \times 12 + 6 \times 1 = 30 \, \text{g mol}^{-1}$$

So, heat released by combustion of $1$ mole:

$$\Delta U = -\left(\frac{10}{0.3}\right) \times 30 = -1000 \, \text{kJ mol}^{-1}$$

Combustion reaction:

$$C_2H_6(g) + \frac{7}{2} O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$$

Change in moles of gaseous species:

$$\Delta n_g = 2 - \left(1 + \frac{7}{2}\right) = -\frac{5}{2}$$

Using the relation:

$$\Delta H = \Delta U + \Delta n_g RT$$

Substituting values as used in the solution:

$$\Delta H = -1000 + \left(-\frac{5}{2}\right)(8.3)(300) \times 10^{-3}$$ $$\Delta H = -1000 - 6.225$$ $$\Delta H \approx -1006 \, \text{kJ mol}^{-1}$$

Therefore, the heat evolved at constant pressure is $1006 \, \text{kJ mol}^{-1}$, so the required numerical answer is 1006.

Given: Bomb calorimeter measures heat at constant volume.

At constant volume, the combustion heat corresponds to $\Delta U$, not $\Delta H$. Since the temperature of the calorimeter rises, the reaction releases heat, so the thermodynamic sign is negative for the system.

First calculate the heat absorbed by the calorimeter:

$$q_{cal} = 20 \times 0.5 = 10 \, \text{kJ}$$

Hence for the reacting sample:

$$q_{rxn} = -10 \, \text{kJ}$$

Now convert sample mass into per mole basis. For ethane:

$$M = 30 \, \text{g mol}^{-1}$$

So,

$$0.3 \, \text{g} = \frac{0.3}{30} = 0.01 \, \text{mol}$$

Therefore,

$$\Delta U = \frac{-10}{0.01} = -1000 \, \text{kJ mol}^{-1}$$

Next convert $\Delta U$ to $\Delta H$ using:

$$\Delta H = \Delta U + \Delta n_g RT$$

From

$$C_2H_6(g) + \frac{7}{2} O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$$

only gaseous moles are counted, hence:

$$\Delta n_g = 2 - \left(1 + \frac{7}{2}\right) = -\frac{5}{2}$$

Using $T = 300 \, \text{K}$ and $R = 8.3 \, \text{J K}^{-1} \, \text{mol}^{-1}$:

$$\Delta n_g RT = \left(-\frac{5}{2}\right)(8.3)(300) \times 10^{-3} = -6.225 \, \text{kJ mol}^{-1}$$

Thus,

$$\Delta H = -1000 - 6.225 = -1006.225 \, \text{kJ mol}^{-1}$$

Nearest integer heat evolved = 1006.

Note: the provided the solution reaches the same final answer 1006, although its printed intermediate value for $\Delta n_g$ appears inconsistent; the final numerical conclusion matches $\Delta H \approx -1006 \, \text{kJ mol}^{-1}$.