InterviewSolution
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InterviewSolution
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Calcualte the work down when 11.2 g of iron dissolves in hydrochloric acid in (i) a closed vessel (ii) an open beaker at `25^(@)C` ( Atomic meass of `Fe = 56u)` |
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Answer» Iron reacts with HCl acid to produce `H_(2)` gas as `Fe(s) + 2HCl(aq) rarr FeCl_(2) (aq) + H_(2)(g)` Thus, 1 mole of Fe, i.e., 56 g Fe produce `H_(2)` gas `= 1 `mole `:. 11.2 g` Fe will produce `H_(2)` ga `= (1)/( 56) xx 11.2 = 0.2 `mole. (i) If the reaction is carried out in a closed vessel , `Delta V = 0` `:. w = - P_(ext)DeltaV = 0` (ii) If the reaction is carried out in open beaker ( external pressure being 1 atm) Final volume occupied by 0.2 moles of `H_(2) ` at `25^(@)C` and 1 atm pressure can be calculated as follows `:` `PV = nRT` `:. V = ( nRT)/( P) = ( 0.2 mol xx 0.0 821L atm K^(-1) xx 298 K)/( 1 atm) = 4.89 L` `:. DeltaV = V _("final") - V_("initial") = 4.89 L` `w= - P_(ext)DeltaV = -1 atm xx 4.89 L atm` `= - 4.89 L atm` `= - 4.89 xx 101.3 J = - 495.4 J` |
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