Question

Calculate the standard enthalpy change for the following reaction,
\(2\text{C}_2\text{H}_6\text{(g)}+7\text{O}_2\text{(g)}⟶4\text{CO}_2\text{(g)}+6\text{H}_2\text{O}\text{(l)}\)
Given, \(\Delta _fH^{\circ}(\text{C}_2\text{H}_6) = -85 \text{kJ mol}^{-1}\)
\(\Delta _fH^{\circ}(\text{CO}_2) = -390 \text{kJ mol}^{-1}\)
\(\Delta _fH^{\circ}(\text{H}_2\text{O}) = -285 \text{kJ mol}^{-1}\)

• A. \(-2900 \text{kJ}\)
• B. \(-3100 \text{kJ}\)
• C. \(-3000 \text{kJ}\)
• D. \(-3200 \text{kJ}\)

Hint:

Use delta H = sum of formation enthalpies of products minus reactants. O2 has zero formation enthalpy.

Answer 1

The Correct Option is B

Step 1: Think of a two-leg path:
The reaction as written burns 2 mol of ethane. The heat released relates to breaking the reactants down into elements and then building the products from the elements.

Step 2: Two legs:
Leg 1: Decompose 2 mol of ethane into elements. This costs $+2\times 85 = +170$ kJ.
Leg 2: Form products from elements: $4\times(-390) = -1560$ kJ for $\text{CO}_2$, and $6\times(-285) = -1710$ kJ for water.

Step 3: Add:
$\Delta H = +170 - 1560 - 1710 = -3100$ kJ.

Step 4: Check:
The value is large and negative, as expected for a combustion. Forgetting the factor 2 on ethane would give $-3185$, and forgetting the sign of the 170 would give $-3440$, neither of which is an option.

Final Answer:
$\Delta H^{\circ} = -3100$ kJ, option (B). \[ \boxed{-3100 \text{ kJ (B)}} \]