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Question
Calculate the standard enthalpy of:
\[\ce{N2H_{4(g)} + H_{2(g)} -> 2NH_{3(g)}}\]
If ΔH0(N – H) = 389 kJ mol–1, ΔH0(H – H) = 435 kJ mol–1, ΔH0(N – N) = 159 kJ mol–1.
Solution
\[\begin{array}{cc}
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{......................}\ce{H}\phantom{.....}\\
\phantom{}|\phantom{....}|\phantom{.......................}|\phantom{.....}\\
\phantom{}\ce{N - N(g) + H - H(g) -> 2N - H}\\
\phantom{}|\phantom{....}|\phantom{.......................}|\phantom{.....}\\
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{......................}\ce{H}\phantom{.....}
\end{array}\]
ΔrH0 = ∑ΔH0(reactant) – ∑ΔH0(product)
= [4ΔH0(N – H) + ΔH0(N – N) + ΔH0(H – H)] – [6ΔH0(N - H)]
= ΔH0(N – N) + ΔH0(H – H) – 2ΔH0(N – H)
= 1 × 159 + 1 mol × 435 – 2 × 389
= – 184 kJ
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