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Question: \({I_2} + {I^ - } \rightleftharpoons I_3^ - \).This reaction is set up in aqueous medium. We start w...

I2+I−⇌I3−{I_2} + {I^ - } \rightleftharpoons I_3^ - .This reaction is set up in aqueous medium. We start with 1 mole of I2{I_2} and 0.5 mole of I−{I^ - } in 1L flask. After equilibrium, the excess of AgNO3AgN{O_3} gave 0.25 moles of yellow ppt. then the equilibrium constant is:
A. 1.33
B. 2.66
C. 2.00
D. 3.00

Explanation

Solution

According to the law of mass action we know that, rate of reaction at a given temperature is directly proportional to the product of active masses of reactants, raised to the power equal to the stoichiometric coefficient of the balanced equation at a particular instant of reaction.

Complete step by step solution:
Let us write a balanced equation of the reaction being carried about, it is
I−+AgNO3→AgI+NO3−{I^ - } + AgN{O_3} \to AgI + NO_3^ -
In the question we have been asked about equilibrium constant, therefore in an equilibrium the rate of forward reaction is equal to the rate of backward reaction which is given by kfandkb{k_f} and {k_b} respectively. The ratio of both the rates gives us the equilibrium constant. Mathematically it can be written as,
K=kfkbK = \dfrac{{{k_f}}}{{{k_b}}}
In the reaction given the equilibrium constant is given by
K=[I3−][I2][I−]K = \dfrac{{[I_3^ - ]}}{{[{I_2}][{I^ - }]}} (These are the concentration or volume of the ions)
We are given that 0.25 mole of silver nitrate is added after attaining equilibrium which reacts with iodine ions with initial concentration of 0.5 moles, reacting with 0.25 moles of ions to form 0.25 moles of iodate ions.
Also 1 mole of I2{I_2} reacts with 0.25 moles hence,
1-0.25=0.75 moles of iodine remains at equilibrium. Since this takes place within a 1L flask , the number of moles becomes equal to the molar concentration. Therefore substituting the values in the equilibrium constant for the reaction we get,
K=[I3−][I2][I−]=0.250.75×0.25=1.33K = \dfrac{{[I_3^ - ]}}{{[{I_2}][{I^ - }]}} = \dfrac{{0.25}}{{0.75 \times 0.25}} = 1.33

Hence the correct option is A.

Note: To write the equilibrium constant expression, the concentration of pure liquid and pure solid assumed to be unity, as the concentration of such substance remain constant,i.e., Concentration=mole/litre