Question

Use the data below to determine the standard molar enthalpy of formation of B2H6 (g).

4B(s) + 3 O2(g) --> 2 B2O3................................deltaHrxn= -2543.9 kj/mol

H2(g) + 1/2 O2(g) --> H2O(g).............................deltaHrxn= -241.8 kj/mol

B2H6(g) + O2(g) --> B2O3(s) + 3 H2O(g)..........deltaHrxn= -2032.9kj/mol

Answer 1

B2H6(g) + O2(g) --> B2O3(s) + 3 H2O(g)..........deltaHrxn= -2032.9kj/mol

To find the ΔH_reaction^o, use the formula for the standard enthalpy change of formation:

ΔHoreaction=∑ΔHof(products)−∑ΔHof(Reactants)

The relevant standard enthalpy of formation values are:

• O₂(g): 0 kJ/mol
• H₂O(g): -241.8 kJ/mol
• B₂O₃(s): -2543.9kJ/mol

Putting these values into the formula above gives the following:

ΔHoreaction= (1mol)(-2543.9 kJ/mol + 3* -241.8)−[(1mol)(X kJ/mol)+(0mol kJ/mol]

-2032.9kj/mol = (-3269.3kj/mol) − X

standard molar enthalpy of formation of B2H6 (g) = −1236. 4 kJ/mol