Question

Pentaborane, BsHs, was once considered a potential rocket fuel. BsH burns in the presence o oxygen to form B20s and water vapor.

2BsH_9 (g) + 120_2 (g) → 5B0_3 (g) + 9H_20 (g)
At 298K, the enthalpy change for the reaction above is -8686.6 kJ/mol.Calculate the change in enthalpy when 0.600 moles of pentaborane are consumed. Express your answer in kJ.

Answer 1

The change in enthalpy when 0.600 moles of pentaborane are consumed is -2606.0 kJ.

Step-by-step explanation:

The question asks us to calculate the change in enthalpy when 0.600 moles of pentaborane (B5H9) are burned in the presence of oxygen to form B2O3 and water vapor, using the provided enthalpy change for the reaction. Given that the enthalpy change for the reaction is -8686.6 kJ/mol, we can use this information along with stoichiometry to find the total enthalpy change for 0.600 moles. The reaction consumes two moles of B5H9 for every -8686.6 kJ, so the enthalpy change for 0.600 moles is calculated as follows:

(-8686.6 kJ/mol × 0.600 moles) / 2 moles = -2606.0 kJ