Question

Calculate \(\Delta \text{H}\) for the following reaction at \(300 \text{K}\)
\(2\text{C}_{(s)}+3\text{H}_{2(g)}⟶\text{C}_2\text{H}_{6(g)}\) if \(\Delta \text{U}\) for the reaction is \(-80 \text{kJ}\) (\(\text{R} = 8.314 \text{JK}^{-1}\text{mol}^{-1}\))

• A. \(-85.00 \text{kJ}\)
• B. \(-43.00 \text{kJ}\)
• C. \(-128.00 \text{kJ}\)
• D. \(-170.00 \text{kJ}\)

Hint:

Use delta H = delta U + delta n(g) R T, counting only gaseous moles.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Enthalpy and internal energy are related by \(\Delta H = \Delta U + \Delta n_g RT\), where \(\Delta n_g\) is moles of gaseous products minus moles of gaseous reactants.

Step 2: Key Formula or Approach:
Reaction: \(2\text{C}(s) + 3\text{H}_2(g) \to \text{C}_2\text{H}_6(g)\). Solids are ignored when counting gas moles.

Step 3: Detailed Explanation:
\(\Delta n_g = 1 - 3 = -2\).
\[ \Delta n_g RT = (-2)(8.314 \times 10^{-3}\ \text{kJ K}^{-1}\text{mol}^{-1})(300\ \text{K}) = -4.99\ \text{kJ} \]
\[ \Delta H = -80 + (-4.99) = -84.99 \approx -85.00\ \text{kJ} \]
Options B, C and D are far from this value, because the \(\Delta n_g RT\) correction is only about 5 kJ here.

Final Answer:
\(\Delta H\) is about \(-85.00\) kJ, option (A). \[ \boxed{-85.00\ \text{kJ}} \]