Question

From the following data, determine ΔfH⦵ for diborane, B2H6(g), at 298 K:

(1) B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g) ΔrH⦵ = −1941 kJ mol−1

(2) 2 B(s) + 3/2 O2(g) → B2O3(s) ΔrH⦵ = −1271.9 kJ mol−1

(3) H2(g) + 1/2 O2(g) → H2O(g) ΔrH⦵ = −241.8 kJ mol−1

Answer 1

To determine ΔfH⦵ for diborane, B2H6(g), at 298 K, we can use the given reactions and their enthalpy changes. From reaction (2), we can determine the enthalpy change for the formation of 1 mole of B2O3(s). From reaction (3), we can determine the enthalpy change for the formation of 3 moles of H2O(g). Using these values, we can calculate the enthalpy change for the formation of 1 mole of B2H6(g).

Step-by-step explanation:

To determine ΔfH⦵ for diborane, B2H6(g), at 298 K, we need to use the given reactions and their enthalpy changes. From reaction (2), we know that the enthalpy change for the formation of 1 mole of B2O3(s) is -1271.9 kJ. From reaction (3), we know that the enthalpy change for the formation of 3 moles of H2O(g) is -241.8 kJ. Since reaction (1) shows the formation of 1 mole of B2O3(s) and 3 moles of H2O(g) from 1 mole of B2H6(g), we can calculate the enthalpy change for the formation of 1 mole of B2H6(g) using the given enthalpy changes: ΔfH⦵ for B2H6(g) = [-1941 kJ - (3*-241.8 kJ + -1271.9 kJ)] / 1 = -1193.5 kJ/mol. Therefore, ΔfH⦵ for diborane, B2H6(g), at 298 K is -1193.5 kJ/mol.