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: Understanding the Concept:
The standard enthalpy change of a reaction equals the total formation enthalpy of the products minus that of the reactants, each multiplied by its coefficient. Elements in their standard states, such as \(\text{O}_2\), have zero formation enthalpy.

Step 2: Key Formula or Approach:
\[ \Delta_rH^{\circ} = \sum n\,\Delta_fH^{\circ}(\text{products}) - \sum n\,\Delta_fH^{\circ}(\text{reactants}) \]

Step 3: Detailed Explanation:
Products:
\[ 4(-390) + 6(-285) = -1560 - 1710 = -3270 \text{ kJ} \]
Reactants:
\[ 2(-85) + 7(0) = -170 \text{ kJ} \]
Reaction enthalpy:
\[ \Delta_rH^{\circ} = -3270 - (-170) = -3270 + 170 = -3100 \text{ kJ} \]
The other options are all off by round amounts. For example -3270 alone would result from forgetting the ethane term, and none of the other values follows from a correct application of the formula.

Final Answer:
The standard enthalpy change for the reaction is -3100 kJ, option (B). \[ \boxed{-3100 \text{ kJ (B)}} \]