Question

Standard heat of formation values for $ \text{C}_2\text{H}_6(g), \, \text{CO}_2(g), \, \text{and} \, \text{H}_2\text{O}(l) $ are -21.1, -94.1, and -68.3 kcal mol$^{-1}$, respectively. Indicate the standard heat of combustion value of $ \text{C}_2\text{H}_6(g) $ involving the above data.

• A. -188.2 kcal/mole
• B. -372 kcal/mole
• C. -204.9 kcal/mole
• D. -183.5 kcal/mole

Hint:

For combustion reactions, use Hess's law to calculate the heat of combustion by considering the heats of formation of the products and reactants.

Answer 1

The Correct Option is B

Step 1: Write the combustion reaction.
The combustion reaction of ethane (\( \text{C}_2\text{H}_6 \)) is: \[ \text{C}_2\text{H}_6(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \]
Step 2: Use Hess's Law.
The heat of combustion of a compound is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Using Hess's Law: \[ \Delta H_{\text{comb}} = \left[ 2(\Delta H_f (\text{CO}_2)) + 3(\Delta H_f (\text{H}_2\text{O})) \right] - \left[ \Delta H_f (\text{C}_2\text{H}_6) + \Delta H_f (\text{O}_2) \right] \] Since \( \Delta H_f (\text{O}_2) = 0 \), the equation simplifies to: \[ \Delta H_{\text{comb}} = \left[ 2(-94.1) + 3(-68.3) \right] - (-21.1) \]
Step 3: Perform the calculation.
\[ \Delta H_{\text{comb}} = \left[ -188.2 - 204.9 \right] + 21.1 = -372 \, \text{kcal/mol} \]
Step 4: Conclusion.
Therefore, the standard heat of combustion of ethane is \( -372 \, \text{kcal/mol} \).