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:

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.