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    chemical equilibrium :0000 cashmash

    does any budy know how to do he 2013 kc question 9 part c i need help on it

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      Fe CNS Fe(CNS)

      I 1x10^-3 1x10^-3 0

      C -(1.1x10^-4) -(1.1x10^-4) +(1.1x10^-4)

      E 8.9x10^-4 8.9x10^-4 1.1x10^-4

      1.1x10*-4/(8.9x10*-4)^2 = 138.87

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      kc= concentration of products divided by the reactants therefore it's Fe(CNS) divided by (Fe x CNS)

      So take your equation and leave three lines under it.

      Fe + CNS ------> Fe(CNS)

      Initial: 1x10 to power of -3 1x10 to power of -3 0 (as isnt present at beginning)

      Change minus 1.1x10 to power -4 for both these add 1.1x10 to power -4 therefore subtract on other side

      @equilibrium 0.00089 0.00089 We are told 1.1x10 to power of -4

      So products divided by reactants= 1.1x10 to power of -4 over (0.00089) squared gives you 138.87 which is Kc

      Hope that helps xo

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