Given the following equations and ${ΔH}^{\circ}$ values, determine the enthalpy of reaction at 298 K for the reaction:
$C_{2} H_{4} (g) + 6 F_{2} (g) ⟶ 2 {CF}_{4} (g) + 4 HF (g)$
$\begin{array}{lc} H_{2} (g) + F_{2} (g) ⟶ 2 HF (g); & Δ H_{1}^{\circ} = - 537 kJ \\ C (s) + 2 F_{2} (g) ⟶ {CF}_{4} (g); & Δ H_{2}^{\circ} = - 680 kJ \\ 2 C (s) + 2 H_{2} (g) ⟶ C_{2} H_{4} (g); & Δ H_{3}^{\circ} = 52 kJ \end{array}$

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answer is B.

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Detailed Solution

Given,
$\begin{array}{lc} H_{2} (g) + F_{2} (g) ⟶ 2 HF (g); & Δ H_{1}^{\circ} = - 537 kJ \\ C (s) + 2 F_{2} (g) ⟶ {CF}_{4} (g); & Δ H_{2}^{\circ} = - 680 kJ \\ 2 C (s) + 2 H_{2} (g) ⟶ C_{2} H_{4} (g); & Δ H_{3}^{\circ} = 52 kJ \end{array}$
Change in enthalpy for the given reaction,
$Δ H^{\circ} = - 537 \times 2 - 680 \times 2 + 52 = - 2382$