For the reactionH2F2(g)→H2(g)+F2(g)ΔU=−59.6kJmol−1 at 27∘CThe enthalpy change for the above reaction is (–) ___ kJ mol–1 [nearest integer] Given : R = 8.314 JK—1 mol–1.

ΔH=ΔU+ΔngRTΔH=−59.6+1×8.314×300×10−3=−57.10

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For the reaction
$\begin{array}{l} H_{2} F_{2} (g) \to H_{2} (g) + F_{2} (g) \\ ΔU = - 59 .6 kJ {mol}^{- 1} at 27^{\circ} C \end{array}$
The enthalpy change for the above reaction is (–) ___ kJ mol^–1 [nearest integer] Given : R = 8.314 JK^—1 mol^–1.

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answer is 57.

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$\begin{array}{l} ΔH = ΔU + {Δn}_{g} RT \\ ΔH = - 59 .6 + 1 \times 8 .314 \times 300 \times 10^{- 3} = - 57 .10 \end{array}$

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