Question

Using the Hess's law calculate the enthalpy change for the systhesis of dibroane from its elements accordingbto the equaruon 2B(s) +3H2(g) = B2H6(g) by using the following data,
1: 2B(s) + 3/2O2(g) = B2O3(s) -1273kj/mol
2: B3H6(g) + 3O2(g) = B2O3(s) + 3H2O(g)
-2035kj/mol
3: H2(g) + 1/2O2(g) = H2O (l) -286kj/mol
4: H2O(l) = H2O(g) 44kj/mol
Show calculation step by step for full marks .

Answer 1

To calculate the enthalpy change for the synthesis of diborane from its elements, we need to first write the overall equation by adding the given equations together in a way that cancels out the intermediates.

Here is the overall equation we need to solve for:
2B(s) + 3H2(g) = B2H6(g)

Given equations:
1: 2B(s) + 3/2O2(g) = B2O3(s) -1273 kJ/mol
2: B3H6(g) + 3O2(g) = B2O3(s) + 3H2O(g) -2035 kJ/mol
3: H2(g) + 1/2O2(g) = H2O(l) -286 kJ/mol
4: H2O(l) = H2O(g) 44 kJ/mol

Now, we can manipulate the given equations to get the overall equation:
- Reverse equation 3:
H2O(g) = H2(g) + 1/2O2(g) +(-44) kJ/mol

- Multiply equation 1 by 3:
6B(s) + 9/2O2(g) = 3B2O3(s) -3819 kJ/mol

- Multiply equation 2 by 2:
2B3H6(g) + 6O2(g) = 2B2O3(s) + 6H2O(g) -4070 kJ/mol

Adding the manipulated equations together:
6B(s) + 9/2O2(g) + 2B3H6(g) = 3B2O3(s) + 6H2O(g) -7889 kJ/mol

Now, we can compare this newly formed equation with the desired overall equation:
2B(s) + 3H2(g) = B2H6(g)

It is clear that we need to manipulate the equations to get the desired overall equation.

- Divide the overall equation by 2:
B(s) + 3/2H2(g) = 1/2B2H6(g)

Now, by multiplying the manipulated equations by the appropriate coefficients, we can obtain the desired equation:
6B(s) + 9H2(g) + 2B3H6(g) = 3B2H6(g) + 6H2O(g) -15778 kJ/mol

Therefore, the enthalpy change for the synthesis of diborane from its elements is -15778 kJ/mol.