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: Setup
Use the ideal gas relation $PV = nRT$. At constant $P$ and $T$, $P\Delta V = \Delta n\, RT$.

Step 2: Moles
The reaction takes 2 mol of gas to 1 mol of gas, so $\Delta n = -1$ mol.

Step 3: Compute
$P\Delta V = (-1)(8.314)(300) = -2494.2$ J. Work done by the system is $P\Delta V$, which is negative, so work done on the system is +2494.2 J.

Step 4: Answer
The magnitude of work is 2494.2 J, and it is done on the system because the gas is compressed as moles fall. The other options are not equal to $RT = 2494.2$ J.

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