Question

# Use the data below to determine the standard molar enthalpy of formation of B2H6 (g). 4B(s)...

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

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

To find the ΔHreactiono, 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