Chemistry - Thermodynamics - 1 Question with Solution | TestHub

$\begin{array}{ccccccc}\text{Fe}& +& 2\text{HCl}& \to & {\text{FeCl}}_2& +& {\text{H}}_2\\ \text{1 mol}& & & & & & \text{1 mol}\\ \frac{112}{56}=\text{2 mol}& & & & & & \text{2 mol}\end{array}$

Now,${\text{V}}_2={\text{V}}_{{\text{H}}_2}$and${\text{V}}_1=0$(For solid + aqueous state)

Work done$=-{\text{p}}_{\text{ext}}\Delta \text{V}$

$=-\text{Δ}(\text{P}\text{V})=-{\text{n}}_{{\text{H}}_2}\text{R}\text{T}$

$=-2\times 2\times 300$

$=-\text{1200 cals}(Work\; of\; expansion)$