given the following thermochemical equations:
4B(s) + 3O2(g) -> 2B2O3(s) deltaH=-2509.1 kJ
2H2(g) + O2(g) -> 2H2O(l) deltaH=-571.70 kJ
B2H6(g) + 3O2(g) -> B2O3(s) + 3H2O(l) deltaH=-2147.5 kJ
calculate the standard enthalpy of formation (in kJ.mol^-1) of B2H6(g)
Use eqn 1 as is.
Multiply eqn 2 by 3 and add to eqn 1.
Multiply eqn 3 by 2, reverse it, add to eqn 1 and 2.
Cancel those materials that appear on both sides of the equation.
If I've not goofed that will give you
4B + 6H2 ==> 2B2H6 but that is twice the eqn you want so divide the dH value by 2 to give you the value for 2B + 3H2 ==> B2H6.....dH = ?
When multiplying an eqn remember to multiply the dH value also.
When reversing an eqn remember to change the sign of the dH value also.
MOST IMPORTANT: I did this in my head and could have made a mistake so put all of this down on paper, make sure all of those extra items such as H2O and O2 cancel and that the equation really is what I found. If I've made an error, you see how it's done. Finally, post your work if you run into trouble and need more help.
To calculate the standard enthalpy of formation of B2H6(g), we can use the concept of Hess's Law. According to Hess's Law, the enthalpy change of a reaction is independent of the pathway taken.
First, we need to identify the relevant thermochemical equations that can be used to obtain the desired equation:
1. 4B(s) + 3O2(g) -> 2B2O3(s) (Equation 1)
2. 2H2(g) + O2(g) -> 2H2O(l) (Equation 2)
3. B2H6(g) + 3O2(g) -> B2O3(s) + 3H2O(l) (Equation 3)
We need to manipulate these equations to obtain the desired equation: B2H6(g) -> B2O3(s) + 3H2O(l)
To do this, we will reverse Equation 1 and multiply it by 3/2 to obtain 3/2B2O3(s) -> 2B(s) + 3/2O2(g).
Next, we will multiply Equation 2 by 2 and Equation 3 by 2/3 to obtain 2H2(g) -> 2H2O(l) and 2/3B2H6(g) + 2O2(g) -> 2/3B2O3(s) + 2H2O(l) respectively.
Now, we can add these equations together:
3/2B2O3(s) + 2H2O(l) + 2H2(g) -> 2/3B2O3(s) + 2B(s) + 3/2O2(g) + 2H2O(l).
Simplifying, we get:
3/2B2O3(s) + 4H2O(l) + 2H2(g) -> 2/3B2O3(s) + 2B(s) + 3/2O2(g).
We can cancel out the water (H2O) molecules to obtain our desired equation:
3/2B2O3(s) + 2H2(g) -> 2/3B2O3(s) + 2B(s) + 3/2O2(g).
The enthalpy change for this combined equation will be equal to the sum of the enthalpy changes for the individual equations:
deltaH = deltaH1 + deltaH2 + deltaH3
Substituting the values given in the question:
deltaH = (-2509.1 kJ) + (-571.70 kJ) + (-2147.5 kJ)
deltaH = -4228.3 kJ
Since we are looking for the standard enthalpy of formation of B2H6(g), we need to consider the stoichiometric coefficients of the reactants and products. In this case, the stoichiometric coefficient of B2H6(g) is 1, so the standard enthalpy of formation of B2H6(g) is:
DeltaHf = deltaH/1 mol
DeltaHf = (-4228.3 kJ) / 1 mol
Therefore, the standard enthalpy of formation of B2H6(g) is -4228.3 kJ/mol.