Consider the following reaction: 2CH30H (/) + 302(g) --+ 4H20(/) + 2C02(g) l:irH0 = - 1452.8 kJ mol-1 What is the value of l:irH0 if (a) the equation is multiplied throughout by 2, (b) the direction of the reaction is reversed so that the products become the reactants and vice versa, and (c) water vapor instead of liquid water is the product?

a. ΔH = -2905.6kJ/mol

b. ΔH = +1452.8kJ/mol

c. ΔH = -1276.8kJ/mol

Explanation:

Based on the reaction:

2CH₃OH(l) + 3O₂(g) → 4H₂O(l) + 2CO₂(g) ΔH = -1452.8kJ/mol

There are released -1452.8kJ/mol

a. Multiplying the reaction by 2:

4CH₃OH(l) + 6O₂(g) → 8H₂O(l) + 4CO₂(g) ΔH = -2905.6kJ/mol

The energy released is twice the initial energy.

b. The reverse reaction:

4H₂O(l) + 2CO₂(g) → 2CH₃OH(l) + 3O₂(g) ΔH = +1452.8kJ/mol

The energy absorbed is the same released in the reverse reaction.

c. Using Hess's law:

2CH₃OH(l) + 3O₂(g) → 4H₂O(l) + 2CO₂(g) ΔH = -1452.8kJ/mol

4 H₂O(l) → 4 H₂O(g) ΔH = 176kJ/mol

The sum of both reactions produce:

2CH₃OH(l) + 3O₂(g) → 4H₂O(g) + 2CO₂(g) ΔH = -1452.8kJ/mol + 176kJ/mol