Consider the reaction : $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ carried out at constant temperature and pressure. If $∆$ H and $∆$ U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

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$ΔH = 0$

$ΔH = ΔU$

$ΔH > ΔU$

$ΔH < ΔU$

answer is B.

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Detailed Solution

$\begin{array}{l} ΔH = ΔU + ΔnRT for N_{2} + 3 H_{2} ⟶ 2 {NH}_{3} \\ {Δn}_{g} = 2 - 4 = - 2 \\ ∴ ΔH = ΔU - 2 RT or ΔU = ΔH + 2 RT ∴ ΔU > ΔH \end{array}$

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