Question

Consider the following reaction: H2(g)+I2(g)⇌2HI(g) A reaction mixture in a 3.74 L flask at a certain temperature initially contains 0.767 g H2 and 96.8 g I2. At equilibrium, the flask contains 90.2 g HI.calculate the equillibrium constant (Kc) for the reaction at this temperature.

please show work

Answer 1

Volume of flask =3.74L

    H2          +             I₂     ---------->   2HI

0.767/2               96.8/254                 0         initial moles

0.3835              =0.3811                     0            "

0.3835-x              0.3811-x                 2x       moles at equilibrium

                  Given [HI] = 90.2/256=0.3523

    Thus x= 0.3523/2 = 0.1761 moles

0.2074                0.205                     0.3523   moles at equilibrium

0.2074/3.74       0.205/3.74            0.3523/3.74   equilibrium concentration

    Hence Kc = [HI]² / [H2][I2]  

by substituting the above equilibrium concentrations and calculating we get

       Kc = 2.9191