Consider the reaction at 300K
$H_{2} (g) + {Cl}_{2} (g) ⟶ 2 HCl (g); Δ H^{\circ} = - 185 kJ$
If 2 mole of $H_{2}$ completely react with 2 mole of ${CI}_{2}$ to form HCL What is $Δ U^{\circ}$ for this reaction?

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370 kJ

-370 kJ

-185 kJ

answer is D.

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Detailed Solution

$\begin{array}{l} H_{3}^{+} {Cl}_{2} \to 2 HCL, {ΔH}^{\circ} = - 18 kJ \\ 1 mol 1 mol 2 mol \\ 3 mol 3 mol 6 mol \\ Δng = 0 \\ {ΔH}^{\circ} = {ΔU}^{6} \\ \Rightarrow {ΔU}^{6} = {ΔH}^{\circ} = - 185 KJ \end{array}$

For2molesofthereaction, $Δ U = - 185 \times 2 = - 370 kJ$

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