how to calculate delta h for the reaction

2B(s)+3H2(g)arrow B2H6(g)given the following data:
2B(s)+3/2O2(g)arrowB2O3(s) deltaH=-1273kj
B2H6(g)+3O2(g)arrowB2O3(s)+3H2O(g) deltaH=-2035kj
H2(g)+1/2O2(g)arrowH2O(l) deltaH=-286kj
H2O(l)arrowH2O(g) deltaH=+44kj

Why multiply equation 4 by 3?

Use equation 1, reverse equation 2, multiply equation 3 by 3 and multiply equation 4 by 3. Add them. Don't forget to reverse the sign of equations you reverse.

Wouldn't there be a 3/2 left over?

To calculate the ΔH (enthalpy change) for the reaction 2B(s) + 3H2(g) → B2H6(g), you can use the concept of Hess's law. According to Hess's law, the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps or reactions that make up the overall reaction.

Here's how you can use the given data to calculate ΔH for this reaction:

1. Reverse the third equation: H2O(g) → H2(g) + 1/2O2(g) with the enthalpy change ΔH = -44 kJ. By reversing this equation, the sign of ΔH also changes.

2. Multiply the first equation by 3 to balance the number of moles of hydrogen gas. This gives you: 6B(s) + 9/2O2(g) → 3B2O3(s) with the enthalpy change ΔH1 = 3 * (-1273 kJ) = -3819 kJ.

3. Multiply the second equation by 2 to balance the number of moles of B2H6(g). This gives you: 2B2H6(g) + 6O2(g) → 2B2O3(s) + 6H2O(g) with the enthalpy change ΔH2 = 2 * (-2035 kJ) = -4070 kJ.

4. Multiply the third equation by 6 to balance the number of moles of H2(g). This gives you: 6H2(g) + 3O2(g) → 6H2O(l) with the enthalpy change ΔH3 = 6 * (-286 kJ) = -1716 kJ.

5. Add up the equations obtained in steps 1, 2, and 3, and cancel out any common compounds to get the desired equation: 6B(s) + 7H2(g) → 2B2H6(g) + 3O2(g) + 6H2O(l). At this point, you can also add up the enthalpy changes obtained in steps 1, 2, and 3 to find the overall enthalpy change for this equation.

ΔH overall = ΔH1 + ΔH2 + ΔH3 = -3819 kJ + (-4070 kJ) + (-1716 kJ) = -9605 kJ.

Therefore, the ΔH for the reaction 2B(s) + 3H2(g) → B2H6(g) is -9605 kJ.