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: Count gas moles:
Gaseous reactant moles $= 3$ (hydrogen); gaseous product moles $= 1$ (ethane).
So $\Delta n_g = -2$.

Step 2: Compute:
$RT = 8.314 \times 300 = 2494.2$ J $= 2.494$ kJ.
$\Delta H = \Delta U + \Delta n_g RT = -80 - 2(2.494) = -84.99$ kJ.
This rounds to $-85.00$ kJ.

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