Question

The standard reaction of enthalpy for the hydrogenation of
propene, CH2CHCH3 (g) + H2 (g) → CH3CH2CH3 (g) is -124 kJ/mol.
The standard reaction enthalpy for the combustion of propane,
CH3CH2CH3 (g) + 5O2 (g) → 3CO2 (g) + 4H2O(l) is -2220 kJ/mol.
Calculate the standard enthalpy of combustion of propene. The
standard enthalpy of formation of liquid water from hydrogen gas
and oxygen gas is -286 kJ/mol.

Answer 1

To calculate the standard enthalpy of combustion of propene (ΔHc), we need to use Hess's law.

First, let's write the balanced equation for the combustion of propene:
C3H6(g) + (9/2)O2(g) -> 3CO2(g) + 3H2O(l)

Now, we can use the given reactions and their enthalpies to calculate the enthalpy of combustion for propene.

1. Multiply the first reaction by 3 and reverse the second reaction to match the products:
3(CH2CHCH3(g) + H2(g) -> CH3CH2CH3(g)) -> 3CO2(g) + 3H2O(l)

2. Multiply the second reaction by 2/5 and reverse the first reaction:
(2/5)(CH3CH2CH3(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)) -> (2/5)(CH2CHCH3(g) + H2(g))

3. Add these two reactions together:
3(CH2CHCH3(g) + H2(g) -> CH3CH2CH3(g)) + (2/5)(CH3CH2CH3(g) + 5O2(g) -> 3CO2(g) + 4H2O(l))
= (7/5)(CH2CHCH3(g) + H2(g)) -> (7/5)(CH3CH2CH3(g))

4. Cancel out the common terms:
CH2CHCH3(g) + H2(g) -> CH3CH2CH3(g)

5. Using the given enthalpies of the desired reaction and the formation of water:
Reaction 1: ΔH1 = -124 kJ/mol
Reaction 2 (reversed): ΔH2 = 2220 kJ/mol
Formation of water: ΔH3 = -286 kJ/mol

6. Apply Hess's law and calculate the enthalpy of combustion for propene:
ΔHc = (7/5)ΔH1 + (7/5)ΔH2 + ΔH3
= (7/5)(-124 kJ/mol) + (7/5)(2220 kJ/mol) + (-286 kJ/mol)
≈ 2519 kJ/mol

Therefore, the standard enthalpy of combustion for propene is approximately 2519 kJ/mol.