Calculate the work done when `56g` of iron reacts with hydrochloric acid in `(a)` a closed vessel of fixed volume and `(b)`an open beaker at `25^(@)C`.

Explanation

a. Vessel is of fixed volume, hence `DeltaV = 0`. No work is done.
b. The gas driver back the atmosphere hence.
`W =- P_(ex) DeltaV`
Also, `DeltaV = V_("final") - V_("initial") = V_("final") ( :. V_("initial") = 0)`
`:. DeltaV = (nRT)/(P_(ex))` or `W =- P_(ex) =- nRT`
where `n` is the number of mole of `H_(2)`gas obtained from `n` mole of `Fe(s)`.
`Fe(s) +2HCI(aq) rarr FeCI_(2)(aq) +H_(2)(g)`
1mol, 1mol
`:. n = (56)/(56) = 1mol`
`:. w =- 1 xx 8.314 xx 298 =- 2477.57J`
The raaction mixture in the given system does `2.477kJ` of work driving back to atmosphere.