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Equilibrium

Question
CBSEENCH11006419

The equilibrium constant at 278K for Cu left parenthesis straight s right parenthesis space plus space 2 Ag to the power of plus left parenthesis aq right parenthesis space rightwards harpoon over leftwards harpoon space space space Cu to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space 2 Ag left parenthesis straight s right parenthesis is 2.0 space cross times space 10 to the power of 15. In a solution in which copper has displaced some silver ions from the solution, concentration of Cu2+ ions is 1.8 x 10-2 mol L-1 and the concentration of Ag+ ions is 3·0 ×10–9 mol L–1. Is the system at equilibrium?

Solution
The given reaction in a state of equilibrium may be represented as:
Cu left parenthesis straight s right parenthesis space plus space 2 Ag to the power of plus left parenthesis aq right parenthesis space space space rightwards harpoon over leftwards harpoon space space space Cu to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space 2 Ag left parenthesis straight s right parenthesis
Applying the law of chemical equilibrium
                      straight K space equals space fraction numerator open square brackets Cu to the power of 2 plus end exponent left parenthesis aq right parenthesis close square brackets space open square brackets Ag left parenthesis straight s right parenthesis close square brackets squared over denominator open square brackets Cu left parenthesis straight s right parenthesis close square brackets space open square brackets Ag to the power of plus left parenthesis aq right parenthesis close square brackets squared end fraction
By convention,   [Ag(s)] = 1 and (Cu(s)] = 1
straight K space equals fraction numerator open square brackets Cu to the power of 2 plus end exponent left parenthesis aq right parenthesis close square brackets over denominator open square brackets Ag to the power of plus left parenthesis aq right parenthesis close square brackets squared end fraction
Substituting, open square brackets Cu to the power of 2 plus end exponent left parenthesis aq right parenthesis close square brackets   
  equals space 1.8 space cross times space 10 to the power of negative 2 end exponent space mol space straight L to the power of negative 1 end exponent space and space open square brackets Ag to the power of plus left parenthesis aq right parenthesis close square brackets
equals space 3.0 space cross times space 10 to the power of negative 9 end exponent space mol space straight L to the power of negative 1 end exponent comma
straight K space equals fraction numerator 1.8 space cross times space 10 to the power of negative 2 end exponent space mol space straight L to the power of negative 1 end exponent over denominator left parenthesis 3.0 space cross times space 10 to the power of negative 9 end exponent space mol space straight L to the power of negative 1 end exponent right parenthesis squared end fraction
space space space space equals space fraction numerator 1.8 space cross times space 10 to the power of negative 2 end exponent over denominator left parenthesis 3.0 space cross times space 10 to the power of negative 9 end exponent right parenthesis squared end fraction mol to the power of negative 1 end exponent straight L
space space space space equals space fraction numerator 1.8 over denominator 9.0 end fraction cross times 10 to the power of 16 space mol to the power of negative 1 end exponent straight L
space space space space equals space 2 cross times 10 to the power of 15 space mol to the power of negative 1 end exponent

which is same as for the reaction in equilibrium.
Hence the given system is in equilibrium.