The standard heat of combustion of solid boron is equal to:

Δ_{f} H^{\circ} (B_{2} O_{3})

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\frac{1}{2} Δ_{f} H^{\circ} (B_{2} O_{3})

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2 Δ_{f} H^{\circ} (B_{2} O_{3})

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- \frac{1}{2} Δ_{f} H^{\circ} (B_{2} O_{3})

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Solution

The correct option is B $\frac{1}{2} Δ_{f} H^{\circ} (B_{2} O_{3})$
The balanced chemical equation for the formation of $B_{2} O_{3}$ from B and oxygen is as shown below.
$2 B + \frac{3}{2} O_{2} ⟶ B_{2} O_{3}; Δ_{f} H^{\circ} o f B_{2} O_{3}$
The balanced chemical equation for the combustion of 1 mole of B is as shown below.
$(B + \frac{3}{4} O_{2} ⟶ \frac{1}{2} B_{2} O_{3}); Δ_{c} H^{\circ} o f B$
Hence, $: Δ_{c o m b} H^{\circ} o f b o r o n = \frac{1}{2} Δ_{f} H^{\circ} o f B_{2} O_{3}$
Thus, standard heat of combustion of solid boron is equal to one half the standard heat of formation of $B_{2} O_{3}$ .

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