How does the thermochemical equation S (s) + O2 (g) → SO2 (g), ΔH = –296 kJ, need to be modified to be used as a portion of the data needed to find the change in enthalpy for the chemical reaction 2SO2 (g) + O2 (g) → 2SO3 (g)?

2C(s) + H2(g) ---> C2H2(g)ΔH° = ??? kJ

Given the following thermochemical equations:

C2H2(g) + 5⁄2O2(g) ---> 2CO2(g) + H2O(ℓ)ΔH° = −1299.5 kJ

C(s) + O2(g) ---> CO2(g)ΔH° = −393.5 kJ

H2(g) + 1⁄2O2(g) ---> H2O(ℓ)ΔH° = −285.8 kJ

Solution:

1) Determine what we must do to the three given equations to get our target equation:

a) first eq: flip it so as to put C2H2 on the product side

b) second eq: multiply it by two to get 2C

c) third eq: do nothing. We need one H2 on the reactant side and that's what we have.

2) Rewrite all three equations with changes applied:

2CO2(g) + H2O(ℓ) ---> C2H2(g) + 5⁄2O2(g)ΔH° = +1299.5 kJ

2C(s) + 2O2(g) ---> 2CO2(g)ΔH° = −787 kJ

H2(g) + 1⁄2O2(g) ---> H2O(ℓ)ΔH° = −285.8 kJ

Notice that the ΔH values changed as well.

3) Examine what cancels:

2CO2 ⇒ first & second equation

H2O ⇒ first & third equation

5⁄2O2 ⇒ first & sum of second and third equation

4) Add up ΔH values for our

+1299.5 kJ + (−787 kJ) + (−285.8 kJ) = +226.7 kJ

Explanation:

fe 55.85 g/mol and h2o 18.02 g/mol.