Question

Predict the signs of ΔHᵣₓₙ and ΔSᵣₓₙ for the following process: 2CH₃OH(g)+H2(g)→C₂H₆(g)+2H₂O(g)

a. Δ Hᵣₓₙ > 0
b. Δ Hᵣₓₙ < 0
c. Δ Sᵣₓₙ > 0
d. Δ Sᵣₓₙ < 0

Answer 1

For the reaction 2CH3OH(g) + H2(g) → C2H6(g) + 2H2O(g), it's predicted that the enthalpy change (ΔHrxn) is negative, indicating an exothermic reaction, and the entropy change (ΔSrxn) is tentatively predicted to be positive, indicating a slight increase in disorder due to more complex molecular structures in the products.

Step-by-step explanation:

To predict the signs of ΔHrxn and ΔSrxn for the process 2CH3OH(g) + H2(g) → C2H6(g) + 2H2O(g), let's examine each component:

• The enthalpy change (ΔHrxn) depends on the bond energies of the reactants and products. Considering that methanol is being converted into ethane and water, both of which are fairly stable compounds, the reaction is likely to be exothermic (ΔHrxn < 0), as energy is released when bonds in stable molecules are formed.
• The entropy change (ΔSrxn) looks at the disorder of the system. Originally, there are 3 moles of gas on the reactants' side and 3 moles of gas on the products' side as well. Since there's no net change in the amount of gas, and no phase change indicating a significant increase or decrease in disorder, predicting the sign of ΔSrxn can be challenging. However, because the products have slightly more complex molecular structures, there might be a very small increase in entropy (ΔSrxn > 0).

Therefore, for this reaction, we would predict ΔHrxn < 0 and tentatively ΔSrxn > 0.