Question

What is the value of \(\Delta H\) for the formation of ethanol from ethene gas and liquid water from the following data?
(i) \(\text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H = -1368 \, \text{kJ}\)
(ii) \(\text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(g) \quad \Delta H = 1410 \, \text{kJ}\)

• A. -1326.0 kJ
• B. -4188.0 kJ
• C. -42.0 kJ
• D. -2778.0 kJ

Hint:

Hess’s Law allows us to add or subtract reactions to find the enthalpy change for a desired reaction.

Answer 1

The Correct Option is C

Step 1: Understanding the enthalpy of formation.
The reaction of ethanol formation involves the following steps: 1. The enthalpy of formation of ethanol from ethene and water involves subtracting the enthalpies of the combustion of ethanol and water. 2. Using Hess’s Law, we add or subtract the given reactions to get the formation enthalpy of ethanol.
Step 2: Analyzing the reactions and enthalpy values.
We have the following reactions: \[ \text{(i)} \quad \text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H = -1368 \, \text{kJ} \] \[ \text{(ii)} \quad \text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(g) \quad \Delta H = 1410 \, \text{kJ} \] By combining the reactions appropriately (subtracting reaction i from reaction ii), we get the enthalpy of ethanol formation: \[ \Delta H = -42.0 \, \text{kJ} \] Step 3: Conclusion.
The correct answer is (C) -42.0 kJ.