Question

Calculate the work done for the following reaction at \(27\,^{\circ}\text{C}\)
\(\text{C}_2\text{H}_{4(g)}+\text{H}_{2(g)}⟶\text{C}_2\text{H}_{6(g)}\) (\(R = 8.314\,\text{JK}^{-1}\text{mol}^{-1}\))

• A. \(2494.2\) J
• B. \(124.71\) J
• C. \(3741.3\) J
• D. \(187.07\) J

Hint:

Find the change in moles of gas, then use w = -(delta n) RT for work done on the system.

Answer 1

The Correct Option is A

Step 1: Understand the concept
For a gaseous reaction at constant pressure, the work done is \(w = -P\Delta V = -\Delta n_g RT\), where \(\Delta n_g\) is the change in moles of gas.

Step 2: Find \(\Delta n_g\)
Reactants: 1 mol \(\text{C}_2\text{H}_4\) + 1 mol \(\text{H}_2\) = 2 mol gas. Product: 1 mol \(\text{C}_2\text{H}_6\). So \(\Delta n_g = 1 - 2 = -1\).

Step 3: Substitute
Temperature: \(27^\circ\text{C} = 300\) K.
\[ w = -(-1)(8.314)(300) = +2494.2\ \text{J} \]

Step 4: Interpret
The sign is positive because the gas volume shrinks and the surroundings do work on the system. The magnitude is 2494.2 J, matching option (A). The other values (124.71, 3741.3, 187.07) do not equal \(RT\) for this temperature and are wrong multiples.

Final Answer:
Work done on the system is 2494.2 J. This is option (A). \[ \boxed{\text{(A) }2494.2\ \text{J}} \]