ThermochemistryHard

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

$2B(s) + \frac{3}{2}O_{2}(g) \rightarrow B_{2}O_{3}(s)$

$\Delta_{r}H^{o} = \Delta_{f}H_{B_{2}O_{3}(s)}^{o} = 2 \times \Delta_{C}H_{B(s)}^{o}$

Create a free account to view solution

View Solution Free
Topic: Thermochemistry·Practice all Thermochemistry questions

More Thermochemistry Questions