Question

Using the hess's law to calculats the enthalpy change for the synthesis of diborane from its element according to equation
2B(s) + 3H2(g) = B2H6(g) by using the following data:
1: 2B(s) + 3/2O2(g) = B2O3(s) -1273kj/mol
2: B2H6(g) + 3O2(g) = B2O3(s) + 3H2O(g) -2035kj/mol
3: H2(g) +1/2O2(g) = H2O(l) - 286kj/mol
4: H2O(l) = H20(l) 44kj/mol
Calculate step by step.

Answer 1

To find the enthalpy change for the synthesis of diborane (B2H6) from its elements, we need to calculate the enthalpy change for the reaction 2B(s) + 3H2(g) = B2H6(g) using the given data.

First, we need to use the given reactions to create the overall reaction of interest:
2B(s) + 3/2O2(g) = B2O3(s) ΔH1 = -1273 kJ/mol
B2H6(g) + 3O2(g) = B2O3(s) + 3H2O(g) ΔH2 = -2035 kJ/mol
H2(g) +1/2O2(g) = H2O(l) ΔH3 = -286 kJ/mol
H2O(l) = H2O(l) ΔH4 = 44 kJ/mol

Multiply reaction 1 by 2, reaction 2 by 2, and reaction 3 by 3, so the oxygen cancels out in the overall reaction:
4B(s) + 3O2(g) = 2B2O3(s) ΔH1' = -2546 kJ/mol
2B2H6(g) + 6O2(g) = 2B2O3(s) + 6H2O(g) ΔH2' = -4070 kJ/mol
3H2(g) +3/2O2(g) = 3H2O(l) ΔH3' = -858 kJ/mol

Now, add the above three equations to get the desired reaction:
4B(s) + 2B2H6(g) + 3H2(g) = 3H2O(l) ΔH_total = -6474 kJ/mol

Therefore, the enthalpy change for the synthesis of diborane from its elements is -6474 kJ/mol.