Chemistry Q71 – JEE-MAIN 2026

Home Chemistry Chemical ThermodynamicsQuestion #299

If 3.365g of ethanol

(l)

is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The

| Δ H_{f}^{\circ} |

of ethanol at 298.15 K is _______

\times 10^{2}

kJ mol

^{- 1}

. (Nearest integer) Given: Standard enthalpy for combustion of graphite =

- 393.5

kJ mol

^{- 1}

Standard enthalpy of formation of water

(l)

- 285.8

kJ mol

^{- 1}

Molar mass in g mol

^{- 1}

of C, H, O are 12, 1 and 16 respectively

Enter Calculated Numerical Value:

Clue 1 of 3: Stoichiometry and Thermochemistry

First, determine the moles of ethanol combusted and the molar enthalpy of combustion.

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Step 1: Calculate Molar Mass and Moles of Ethanol✦ Active

The molar mass of ethanol ( $C_{2} H_{5} O H$ ) is calculated from the atomic masses of C (12), H (1), and O (16):

M = (2 \times 12) + (6 \times 1) + (1 \times 16) = 24 + 6 + 16 = 46 g/mol

Moles of ethanol burnt can then be determined from the given mass:

n = \frac{3.365 g}{46 g/mol} = 0.073152 mol

Step 2: Determine Molar Enthalpy of Combustion○ Expand

The heat produced is $99.472$ kJ. Since heat is produced (exothermic reaction), the enthalpy change for combustion is negative. The molar enthalpy of combustion ( $Δ H_{c}^{\circ}$ ) is:

Δ H_{c}^{\circ} = \frac{- 99.472 kJ}{0.073152 mol} = - 1359.89 kJ/mol

💡 Teacher's Secret Hint

Remember that heat produced implies a negative enthalpy change for the system.

Step 3: Apply Hess's Law to find Enthalpy of Formation○ Expand

The balanced combustion reaction for ethanol is:

C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)

Using Hess's Law, the enthalpy of combustion is related to the standard enthalpies of formation of products and reactants:

Δ H_{c}^{\circ} = [2 Δ H_{f}^{\circ} (C O_{2}) + 3 Δ H_{f}^{\circ} (H_{2} O)] - [Δ H_{f}^{\circ} (C_{2} H_{5} O H) + 3 Δ H_{f}^{\circ} (O_{2})]

Given $Δ H_{f}^{\circ} (C O_{2}) = - 393.5$ kJ/mol (standard enthalpy of combustion of graphite) and $Δ H_{f}^{\circ} (H_{2} O) = - 285.8$ kJ/mol. The standard enthalpy of formation of $O_{2} (g)$ is 0 kJ/mol. Substituting the values:

- 1359.89 = [2 (- 393.5) + 3 (- 285.8)] - [Δ H_{f}^{\circ} (C_{2} H_{5} O H) + 3 (0)]

- 1359.89 = [- 787.0 - 857.4] - Δ H_{f}^{\circ} (C_{2} H_{5} O H)

- 1359.89 = - 1644.4 - Δ H_{f}^{\circ} (C_{2} H_{5} O H)

Δ H_{f}^{\circ} (C_{2} H_{5} O H) = - 1644.4 + 1359.89 = - 284.51 kJ/mol

The absolute value is $| Δ H_{f}^{\circ} (C_{2} H_{5} O H) | = 284.51$ kJ/mol. To express this in the format $\times 10^{2}$ kJ mol $^{- 1}$ :

284.51 = X \times 10^{2} ⟹ X = \frac{284.51}{100} = 2.8451

Rounding to the nearest integer, $X = 3$ .

💡 Teacher's Secret Hint

Ensure correct signs for enthalpy changes and remember that elements in their standard states have zero enthalpy of formation.

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Chemistry Question 71 – JEE-MAIN 2026

Clue 1 of 3: Stoichiometry and Thermochemistry

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