Consider the following cases of standard enthalpy of reaction ( \(\Delta \mathrm{H}_{\mathrm{r}}^{\mathrm{o}}\) in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) \(\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})+\frac{7}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\ell) \Delta \mathrm{H}_{1}^{\mathrm{o}}=-1550\) \(\mathrm{C}(\) graphite \()+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \Delta \mathrm{H}_{2}^{\mathrm{o}}=-393.5\) \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell) \Delta \mathrm{H}_{3}^{\mathrm{o}}=-286\) The magnitude of \(\Delta \mathrm{H}_{\mathrm{f}_{\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})}}^{\mathrm{o}}\) is \_\_\_\_ \(\mathrm{kJ} \mathrm{mol}^{-1}\) (Nearest integer).

Chemical Thermodynamics — JEE Main · 2025 · 22 Jan 2025 · Shift 2 Question with Solution

Consider the following cases of standard enthalpy of reaction (

Δ H_{r}^{o}

kJ mol^{- 1}

)

C_{2} H_{6} (g) + \frac{7}{2} O_{2} (g) \to 2 CO_{2} (g) + 3 H_{2} O (ℓ) Δ H_{1}^{o} = - 1550

C (

graphite

) + O_{2} (g) \to CO_{2} (g) Δ H_{2}^{o} = - 393.5

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (ℓ) Δ H_{3}^{o} = - 286

The magnitude of

Δ H_{f_{C_{2} H_{6} (g)}}^{o}

is ____

kJ mol^{- 1}

(Nearest integer).

Answer: 95

Solution

The target reaction is the formation of ethane from its elements:

2 C (graphite) + 3 H_{2} (g) ⟶ C_{2} H_{6} (g) Δ H_{f}^{\circ} = ?

This is built from the three given reactions by Hess's law: reverse the combustion of ethane, and add twice the combustion of graphite and three times the formation of water.

Reversing

C_{2} H_{6} (g) + \frac{7}{2} O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (ℓ)

contributes

- Δ H_{1}^{\circ} = + 1550

.

Doubling

C (graphite) + O_{2} (g) \to C O_{2} (g)

contributes

2Δ H_{2}^{\circ} = 2 (- 393.5) = - 787

.

Tripling

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (ℓ)

contributes

3Δ H_{3}^{\circ} = 3 (- 286) = - 858

.

Adding these three legs, all the

O_{2}

and

C O_{2}

cancel, leaving just the formation reaction:

Δ H_{f}^{\circ} = 2Δ H_{2}^{\circ} + 3Δ H_{3}^{\circ} - Δ H_{1}^{\circ} = - 787 - 858 + 1550 = - 95 kJ mo l^{- 1}

The magnitude of the standard enthalpy of formation of

C_{2} H_{6} (g)

is 95 kJ mol⁻¹.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Integer answer

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Chemical Thermodynamics

Solution

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