JEE Main 2025 Jan 22 Shift 2, Chemistry Q24: Hess’s Law and Bond Energy

JEE Main2025Jan 22, Shift 2Chemistry

Consider the following cases of standard enthalpy of reaction $(Δ H_{r}^{o} in kJ mol^{- 1})$

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

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

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

The magnitude of $Δ H_{f, C_{2} H_{6} (g)}^{o}$ is ______ kJ mol $^{- 1}$ (Nearest integer).

Your answer

Solution

The formation reaction we want is:

$2 C (graphite) + 3 H_{2} (g) \to C_{2} H_{6} (g) Δ H_{f}^{o} = ?$

Use Hess's law: combine the three known reactions so that the target appears.

$Δ H_{f}^{o} (C_{2} H_{6}) = 2 Δ H_{2}^{o} + 3 Δ H_{3}^{o} - Δ H_{1}^{o}$

$= 2 (- 393.5) + 3 (- 286) - (- 1550)$

$= - 787 - 858 + 1550$

$= - 95 kJ mol^{- 1}$ .

Magnitude $= 95 kJ mol^{- 1}$ .

Concept behind this question

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