JEE Main 2026 — Chemistry Physical Chemistry

Molar mass of ethanol ($C_2H_5OH$C2​H5​OH) $= 2(12) + 6(1) + 16 = 46$=2(12)+6(1)+16=46 g mol$^{-1}$−1

Number of moles of ethanol burnt, $n = \dfrac{3.365}{46} = 0.07315$n=463.365​=0.07315 mol

Since the combustion occurs in a bomb calorimeter (constant volume), the heat produced corresponds to the change in internal energy ($\Delta U_c^\circ$ΔUc∘​).

$\Delta U_c^\circ = -\dfrac{99.472}{0.07315} = -1359.8$ΔUc∘​=−0.0731599.472​=−1359.8 kJ mol$^{-1}$−1

The balanced chemical equation for the combustion of ethanol is:

$$C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$$C2​H5​OH(l)+3O2​(g)→2CO2​(g)+3H2​O(l)

Change in the number of gaseous moles, $\Delta n_g = 2 - 3 = -1$Δng​=2−3=−1

Now, calculating the standard enthalpy of combustion ($\Delta H_c^\circ$ΔHc∘​):

$$\Delta H_c^\circ = \Delta U_c^\circ + \Delta n_g RT$$ΔHc∘​=ΔUc∘​+Δng​RT

$$\Delta H_c^\circ = -1359.8 + (-1) \times (8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}) \times 298.15 \text{ K}$$ΔHc∘​=−1359.8+(−1)×(8.314×10−3 kJ K−1 mol−1)×298.15 K

$\Delta H_c^\circ = -1359.8 - 2.48 = -1362.28$ΔHc∘​=−1359.8−2.48=−1362.28 kJ mol$^{-1}$−1

The standard enthalpy of combustion can also be written in terms of enthalpies of formation:

$$\Delta H_c^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$$ΔHc∘​=∑ΔHf∘​(products)−∑ΔHf∘​(reactants)

$$\Delta H_c^\circ = [2\Delta H_f^\circ(CO_2, g) + 3\Delta H_f^\circ(H_2O, l)] - [\Delta H_f^\circ(C_2H_5OH, l) + 3\Delta H_f^\circ(O_2, g)]$$ΔHc∘​=[2ΔHf∘​(CO2​,g)+3ΔHf∘​(H2​O,l)]−[ΔHf∘​(C2​H5​OH,l)+3ΔHf∘​(O2​,g)]

Substituting the given values (note that $\Delta H_f^\circ$ΔHf∘​ of $O_2(g)$O2​(g) is $0$0):

$$-1362.28 = [2(-393.5) + 3(-285.8)] - \Delta H_f^\circ(C_2H_5OH, l)$$−1362.28=[2(−393.5)+3(−285.8)]−ΔHf∘​(C2​H5​OH,l)

$$-1362.28 = [-787.0 - 857.4] - \Delta H_f^\circ(C_2H_5OH, l)$$−1362.28=[−787.0−857.4]−ΔHf∘​(C2​H5​OH,l)

$$-1362.28 = -1644.4 - \Delta H_f^\circ(C_2H_5OH, l)$$−1362.28=−1644.4−ΔHf∘​(C2​H5​OH,l)

$\Delta H_f^\circ(C_2H_5OH, l) = -1644.4 + 1362.28 = -282.12$ΔHf∘​(C2​H5​OH,l)=−1644.4+1362.28=−282.12 kJ mol$^{-1}$−1

The magnitude is $|\Delta H_f^\circ| = 282.12$∣ΔHf∘​∣=282.12 kJ mol$^{-1} = 2.8212 \times 10^2$−1=2.8212×102 kJ mol$^{-1}$−1

Rounding to the nearest integer, we get $3$3.

Answer: $3$3