Question

The work done during combustion of 9 x $10^{-2}$ kg of ethane, $C_2H_{6(g)}$ at 300 K is ____________________. (Given R = 8.314 J $K^{-1}$ $mol^{-1}$, atomic mass C = 12, H = 1)

• A. 6.236 kJ
• B. -6.236 kJ
• C. 18.71 kJ
• D. -18.71 kJ

Answer 1

Answer: (D)

-18.71 kJ

Answer 2

We first write the combustion reaction for ethane (with water as liquid):

$C_2H_6(g) + \tfrac{7}{2}\,O_2(g) \to 2\,CO_2(g) + 3\,H_2O(l)$

Step 1. Find the net change in moles of gas (Δn):

• Reactant gases: $1$ (ethane) $+\tfrac{7}{2} = 1 + 3.5 = 4.5$ moles
• Product gases: $2$ (CO₂) (water is liquid)

So,

$\Delta n = \text{moles of gas products} - \text{moles of gas reactants} = 2 - 4.5 = -2.5$

Step 2. Work at constant pressure:

Work done by the system is given by:

$w = -\Delta n\,RT$

For one mole of ethane, at $T = 300\,K$ and $R=8.314\,J\,K^{-1}\,mol^{-1}$:

$w_{\text{per mole}} = -(-2.5) \times 8.314 \times 300 = 2.5 \times 8.314 \times 300 \approx 6235.5\,J \approx 6.236\,kJ$

Since $\Delta n$ is negative, the reaction involves a contraction so the system does not do work but rather work is done on it. In thermochemistry the “work done by the system” is taken with the sign from the above formula. However, because the process is a contraction, the work done by the system is actually $-6.236\,kJ$ per mole.

Step 3. Apply to the given mass of ethane:

Molar mass of ethane, $C_2H_6$, is:

$2(12) + 6(1) = 24 + 6 = 30\,g/mol$

Given mass = $9 \times 10^{-2}\,kg = 90\,g$, so moles of ethane:

$\text{moles} = \frac{90}{30} = 3\,mol$

Total work done by the system:

$w_{\text{total}} = 3 \times (-6.236\,kJ) \approx -18.71\,kJ$

Thus, the correct answer is option (D) $-18.71\,kJ$.