ThermochemistryMedium
Question
The standard enthalpy of combustion of solid boron is numerically equal to
Options
A.$\frac{1}{2}\Delta_{f}H_{B_{2}O_{3}(s)}^{o}$
B.$\Delta_{f}H_{B_{2}O_{3}(s)}^{o}$
C.$- \Delta_{f}H_{B_{2}O_{3}(s)}^{o}$
D.$- \frac{1}{2}\Delta_{f}H_{B_{2}O_{3}(s)}^{o}$
Solution
$2B(s) + \frac{3}{2}O_{2}(g) \rightarrow B_{2}O_{3}(s)$
$\Delta_{r}H^{o} = \Delta_{f}H_{B_{2}O_{3}(s)}^{o} = 2 \times \Delta_{C}H_{B(s)}^{o}$
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