For the combustion of one mole acetic acid, work done at $298 \mathrm{~K}$ is

For the combustion of one mole acetic acid, work done at $298 \mathrm{~K}$ is
  1. -2.0 J
  2. -1.5 J
  3. 2.0 J
  4. 0.0 J

Solution

$\begin{aligned} & \mathrm{CH}_3 \mathrm{COOH}(l)+2 \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \\ & \Delta n_g=2-2=0 \\ & W=-\Delta n_g R T \\ & =-0 \times R T=0.0 \mathrm{~J}\end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 2)

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