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Question: On a given condition the equilibrium concentration of\(HI\) , \({H_2}\) and \({I_2}\) are 0.80, 0.10...

On a given condition the equilibrium concentration ofHIHI , H2{H_2} and I2{I_2} are 0.80, 0.10 and 0.10 molelitre1mole\,litr{e^{ - 1}} . The equilibrium constant for the reaction
H2+I22HI{H_2} + {I_2} \rightleftharpoons 2HI will be:
(A) 64
(B) 36
(C) 8
(D) 6

Explanation

Solution

In order to this question, if the equilibrium concentration and the equilibrium reaction is given, then we can find the expression for the equilibrium constant by the formula which is totally dependent on the concentration.

Complete step-by-step solution: First, we will discuss equilibrium reactions, equilibrium concentration and equilibrium constant, then after we will solve the numerical related to equilibrium.
Chemical equilibrium reaction refers to the state of a system in which the concentration of the reactant and the concentration of the products do not change with time and the system does not display any further change in properties.
Equilibrium concentration is a state when the rate of forward reaction in a chemical reaction becomes equal to the rate of backward reaction. At the same time, there is no change in the products and reactants, and it seems that the reaction has stopped. Image will be uploaded soon.
Given:-
The equilibrium reaction is H2+I22HI{H_2} + {I_2} \rightleftharpoons 2HI
A chemical reaction between the reactants that stays in a stable state before and after the completion of the reaction.
The equilibrium concentrations are given below:
[HI]=0.18[HI] = 0.18 ,
[H2]=0.10[{H_2}] = 0.10
[I2]=0.10[{I_2}] = 0.10
The equilibrium constant is the value of the reaction quotient that is calculated from the expression for chemical equilibrium. It depends on the ionic strength and temperature and is independent of the concentrations of reactants and products in a solution.
Now, Let’s find the expression for the equilibrium constant , Kc{K_c} , symbol of equilibrium constant.
\therefore The expression for the equilibrium constant is:
Kc=[HI]2[H2][I2]=0.80×0.800.10×0.10=64{K_c} = \dfrac{{{{[HI]}^2}}}{{[{H_2}][{I_2}]}} = \dfrac{{0.80 \times 0.80}}{{0.10 \times 0.10}} = 64
So, the equilibrium constant for the given reactions is 64.

Hence, the correct option is (A) 64.

Note: Equilibrium is chiefly a thermodynamic property. However, this only tells us how far left or right the reaction is. It does not tell how fast the reaction goes to achieve that. It is also used to measure the volume of one reagent required to react with a measured mass or volume of another reagent.