Consider the following cases of standard enthalpy of reaction ( H r , in kJ mol - 1 )

Question

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

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

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

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

The magnitude of $Δ H_{f C_{2} H_{6} (g)}^{\circ}$ is _____ kJ ${mol}^{- 1}$ (Nearest integer) [2025]

Smart Achievers · Accepted Answer

(95)

$Δ H_{1}^{\circ}, Δ H_{2}^{\circ}$ and $Δ H_{3}^{\circ}$ are enthalpies of combustions of $C_{2} H_{6} (g)$ , C (graphite) and $H_{2} (g)$ respectively.

Enthalpy change for any reaction is $Δ H_{combustion} (reactants) - Δ H_{combustion} (products)$

$Δ H_{f}^{\circ} (C_{2} H_{6} (g))$ is given by enthalpy change of the equation: $2 C (graphite) + 3 H_{2} (g) \to C_{2} H_{6} (g)$

$Δ H_{f}^{\circ} (C_{2} H_{6} (g)) = 2 Δ H_{combustion} (graphite) + 3 Δ H_{combustion} (H_{2}) - Δ H_{combustion} (C_{2} H_{6})$

$= [2 (- 393.5) + 3 (- 286) - (- 1550)] kJ = - 95 kJ$

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Solution

Q. Consider the following cases of standard enthalpy of reaction (ΔHr∘, in kJ mol-1) C2H6(g)+72 O2(g)→2CO2(g)+3H2O(l) ΔH1∘=-1550 C(graphite)+O2(g)→CO2(g) ΔH2∘=-393.5 H2(g)+12 O2(g)→H2O(l) ΔH3∘=-286 The magnitude of ΔHfC2H6(g)∘ is _____ kJ mol-1 (Nearest integer) [2025]

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