Calculate the work done when 156g of potassium reacts with water in an open container of water at 300K.

$K\left(s\right)+H_{2}O\left(l\right)\to KOH\left(aq\right)+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.H_2\left(g\right)$

1 mole of K gives - - - - - ½ mole of ${\mathrm{H}}_2$

$\frac{156}{39}=4\text{ moles of }K\text{ gives - - - - - - }2\text{ moles of }{H}_2$

$\begin{array}{l}\text{ PV work } W=-P_{ext}\Delta V\\ =-P_{\text{ext }}\left(V_f-V_i\right)\\ =-P_{\text{ext }}\left(V_f-0\right)\\ =-P_{\text{ext }}{V}_f\left(\because V_i=0\right)\\ =-nRT\\ =-2\times 8.314J\times 300K\\ =-4988.4J\\ \approx -5KJ\end{array}$