Calculate the work done when 50 g of iron is dissolved in HCl at `25^(@)C` in (i) a cloed vessel and (ii) an open beaker when the atmospheric pressure is 1atm.

Explanation

(i) When the reactiion is carried in a closed vessel, the change in volume is zero. Hence, the work done by the system will be zero. (ii) when iron dissolves in HCl, hydrogen is produced.
`undersetunderset(50g)(56g)(Fe)+2HCltoFeCl_(2)+undersetunderset((1)/(56)xx50"Mole")(1" mole")(H_(2))`
Volume of hdyrogen produced at `25^(@)C`
`=(nRT)/(P)=(50)/(56)xx(0.0821xx298)/(1)=21.84L`
this is equal to volme change when the reaction is carried in open beaker.
Work done by the system `=-PDeltaV=-1.0xx21.84`
`=-21.84"Litre"-atm =-2212.89J`