Question

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?

Answer 1

The change in enthalpy (ΔH°) for the given reaction is -1452.8 kJ per mole. If multiplied by 2, the ΔH° becomes -2905.6 kJ per mole. Reversing the reaction gives a ΔH° of +1452.8 kJ per mole, and changing the product to water vapor would result in a different ΔH° value.

Step-by-step explanation:

The student asked about the change in enthalpy (ΔH°) for the reaction 2CH3OH(l) + 3O2(g) → 4H2O(l) + 2CO2(g), which is given as -1452.8 kJ mol⁻¹. There are three parts to the question:

1. Multiplying the equation by 2: Enthalpy change is an extensive property, so if the coefficients in a balanced chemical equation are multiplied by a factor, the ΔH° is also multiplied by that factor. Therefore, if the reaction is multiplied by 2, the value of ΔH° becomes -1452.8 kJ mol⁻¹ × 2 = -2905.6 kJ mol⁻¹.
2. Reversing the reaction: When the direction of a chemical reaction is reversed, the sign of ΔH° is also reversed. So, if the products become reactants and vice versa, the ΔH° for the reverse reaction would be +1452.8 kJ mol⁻¹.
3. Using water vapor instead of liquid water: The enthalpy change depends on the physical states of the reactants and products. Changing the product from liquid water (H2O(l)) to water vapor (H2O(g)) alters the ΔH°. For this reaction, the ΔH° of forming water vapor would be different because liquid water and water vapor have different enthalpies of formation.

To find the exact ΔH° when water vapor is formed, one would need the standard enthalpy of formation for water vapor and perform a Hess's Law calculation using that new value to recalculate ΔH° for the reaction.

Answer 2

a. ΔH = -2905.6kJ/mol

b. ΔH = +1452.8kJ/mol

c. ΔH = -1276.8kJ/mol

Step-by-step 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

ΔH = -1276.8kJ/mol