2.4 ~g coal is burnt in a bomb calorimeter in excess of oxygen at 298 ~K and 1 ~atm pressure. The temperature of the calorimeter rises from 298 ~K to 300 ~K . The enthalpy change during the combustion of coal is -x ~kJ ~mol^-1 . The value of x is \underline\hspace2cm . (Nearest Integer) (Given : Heat capacity of bomb calorimeter 20.0 ~kJ ~K^-1 . Assume coal to be pure carbon)

Thermodynamics PYQ — JEE Main 2022 · 26 Jul · Shift 1 · Q18 with Solution

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Correct answer: 200

Since coal is assumed to be pure carbon,

$\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$

Given:

So, heat absorbed by calorimeter is

$q_{\text{cal}} = C_{\text{cal}}\Delta T = 20.0 \times 2 = 40\,\mathrm{kJ}$

Hence, heat released by combustion of $2.4\,\mathrm{g}$ carbon is

$q = -40\,\mathrm{kJ}$

$n = \frac{2.4}{12} = 0.2\,\mathrm{mol}$

In a bomb calorimeter, combustion occurs at constant volume, so the measured heat is $\Delta U$ .

For $0.2$ mol carbon,

$\Delta U = -40\,\mathrm{kJ}$

Therefore, per mole,

$\Delta U_m = \frac{-40}{0.2} = -200\,\mathrm{kJ\,mol^{-1}}$

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

For the reaction

$\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$

Change in moles of gaseous species:

$\Delta n_g = 1 - 1 = 0$

So,

$\Delta H = \Delta U$

Thus,

$\Delta H = -200\,\mathrm{kJ\,mol^{-1}}$

Hence,

$x = 200$

Stored correct answer = $200$

Our derived answer matches it.