First slide
Fundamentals of thermodynamics

The work done then 6.5 g of zinc reacts with dil.HC1 is an open beaker at 298 K is 


         Zn+2HClZnCl2+H2(g)                     …(1)

           1mol                  1mol   

   Since H2 gas drives back the atmosphere in an open beaker, work is done by the gas which is negative.         

    W=Pext ×ΔV

 Since initial volume zero  so  ΔV=  Final volume 

         initial volume; V = Final volume 

But           ΔV=nRTPext 

      W=Pext ×nRTPext =nRTn= no. of mol of Zn= wt. of Zn at. wt. of Zn=6.565=110

From equation (1), 

          1 mol of Zn1 mol of H2;110 mol  of 

               Zn110mol of H2W=nRT;W=110mol×8.314JKmol×298K=-247.76J 

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