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Which of the following reactions will define $\Delta H^\circ _f$?

(a) $C(diamond)+O_2(g) \rightarrow CO_2(g)$
(b) $\frac{1}{2}H_2(g) + \frac{1}{2}F_2(g) \rightarrow HF(g)$
(c) $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
(d) $CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$

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Answer: $\frac{1}{2}H_2(g) + \frac{1}{2}F_2(g) \rightarrow HF(g)$
Standard enthalpy of formation is equal to enthalpy change for the preparation of 1 mol of a compound from its standard state.
Therefore, for the reaction
$\frac{1}{2}H_2(g) + \frac{1}{2}F_2(g) \rightarrow HF(g)$
$\Delta H^\circ _f = \Delta H^\circ$
answered Feb 28, 2014 by mosymeow_1

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