(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

$\frac{1}{2}H_2(g) + \frac{1}{2}F_2(g) \rightarrow HF(g)$

$\Delta H^\circ _f = \Delta H^\circ$

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