given the thermochemical equations,
B2O3 + 3H2O -> 3O2 + B2H6 delta h = -2035 kj
H2O -> H2O delta h = 44 kj
H2 + 1/2O2 -> H2O delta h = -286 kj
2B + 3H2 -> B2H6 delta h = 36 kj
what is the delta hf for 4B + 3O2 -> 2B2O3?
To find the ΔHf (the standard enthalpy of formation) for the reaction 4B + 3O2 -> 2B2O3, we need to use the given thermochemical equations and apply Hess's Law.
Hess's Law states that the overall enthalpy change of a reaction is independent of the pathway taken as long as the initial and final conditions are the same.
To calculate ΔHf, we need to manipulate the given equations to match the desired reaction.
We'll start by rearranging the given equation 2B + 3H2 -> B2H6 to isolate B2H6:
2B + 3H2 -> B2H6 ΔH = 36 kJ
Now, let's rearrange the equation B2O3 + 3H2O -> 3O2 + B2H6 to get B2O3:
B2O3 + 3H2O -> 3O2 + B2H6 ΔH = -2035 kJ
Divide the entire equation by 2 to balance the number of boron atoms:
(1/2)B2O3 + (3/2)H2O -> (3/2)O2 + (1/2)B2H6 ΔH = -1017.5 kJ
Since we want to calculate the ΔHf for the reaction 4B + 3O2 -> 2B2O3, we need to multiply the equation we just derived by 4:
2B2O3 + 6H2O -> 6O2 + 2B2H6 ΔH = -4063 kJ
Finally, we combine the desired equation with the rearranged equation by aligning and summing up the relevant chemicals:
4B + 3O2 -> 2B2O3
+
2B2O3 + 6H2O -> 6O2 + 2B2H6
____________________________________________________
4B + 3O2 -> 2B2O3
Now, sum up the ΔH values:
ΔHf (4B + 3O2 -> 2B2O3) = ΔH (4B + 3O2 -> 2B2O3) + ΔH (2B2O3 + 6H2O -> 6O2 + 2B2H6)
ΔHf (4B + 3O2 -> 2B2O3) = 0 kJ + (-4063 kJ)
ΔHf (4B + 3O2 -> 2B2O3) = -4063 kJ
Therefore, the ΔHf for 4B + 3O2 -> 2B2O3 is -4063 kJ.