Question

Pentaborane B5H9(s) burns vigorously in O2 to give B2O3(s) and H2O(l). Calculate for the combustion of 1mol of B5H9.

B2O3(s) = -1273.5 kJ/mol
B5H9(s) = 73.2 kJ/mol
H2O(l) = -285.8 kJ/mol

Give balance equation and show work.

Answer 1

To calculate the enthalpy of combustion for 1 mol of B5H9, we use the enthalpy of formation values for B2O3 and H2O, and the given enthalpy of combustion value for B5H9. The balanced combustion equation is 2B5H9 + 12O2 -> 5B2O3 + 9H2O. By substituting the values into the equation and performing calculations, the enthalpy of combustion is found to be -8385.4 kJ/mol.

Step-by-step explanation:

To calculate the enthalpy of combustion for 1 mol of B5H9, we need to use the enthalpy of formation values for B2O3 and H2O, and the given enthalpy of combustion value for B5H9.

The balanced combustion equation for the reaction is:

2B5H9 + 12O2 -> 5B2O3 + 9H2O

Using the enthalpy of formation values, we can calculate the enthalpy of combustion as follows:

ΔH = (5 x ΔHf[B2O3]) + (9 x ΔHf[H2O]) - ΔHc[B5H9]

= (5 x -1273.5 kJ/mol) + (9 x -285.8 kJ/mol) - 73.2 kJ/mol

= -5740 kJ/mol - 2572.2 kJ/mol - 73.2 kJ/mol

= -8385.4 kJ/mol