Question

For the combustion of one mole acetic acid, work done at 298 K is

• A. 0.0 J
• B. -2.0 J
• C. 2.0 J
• D. -1.5 J

Answer 1

Answer: (A)

0.0 J

Answer 2

The balanced combustion reaction of acetic acid is:

$$\ce{CH3COOH(l) + 2O2(g) -> 2CO2(g) + 2H2O(l)}$$

• Reactant gases: $2$ moles (only $\ce{O2}$ gas)

• Product gases: $2$ moles (only $\ce{CO2}$ gas)

So, the change in moles of gas is:

$$\Delta n = 2 - 2 = 0$$

At constant pressure, the work done is given by:

$$w = -\Delta nRT$$

Since $\Delta n = 0$, we have:

$$w = 0$$