Question

K for synthesis of $HI$ is $50$ .What is K for its dissociation.

Answer 1

The state in which both the reactants and products are present in concentrations that exhibit no further tendency to change with time so that there is no observable change in the properties of the system is known as equilibrium. In equilibrium, the forward reaction proceeds at the same rate as the reverse reaction. Although the reaction rates of the forward and backward reactions are generally not zero, they are equal.

Complete answer:
When a reaction is in a state it might appear that there is no reaction taking place. However, the reactants and products are undergoing constant change. To better understand such a condition various parameters were introduced. The equilibrium constant $K$ is one such parameter.
In the above question the reaction for synthesis of Hydrogen Iodide ( $HI$ ) proceeds as follows,
${H_2} + {I_2} \rightleftharpoons 2HI$ ………( $1$ )
Let this be equation number 1.The value equilibrium constant ( $K_1$ ) for this reaction is 50.It’s important to note that this reaction is in equilibrium.By definition of equilibrium of constant we know that,
$K_1 = \dfrac{{[HI]}}{{[{H_2}][{I_2}]}}$ …………….( $2$ )
By considering the reaction of dissociation Hydrogen Iodide we get,
$2HI \rightleftharpoons {H_2} + {I_2}$ ………….( $3$ )
We need to find the equilibrium constant ( $K_2$ ) for reaction number 3.Hence by using its definition,
$K_2 = \dfrac{{[{H_2}][{I_2}]}}{{[HI]}}$ ……………..( $4$ )
Taking inverse of equation 4 we get,
$\dfrac{1}{{K_2}} = \dfrac{{[HI]}}{{[{H_2}][{I_2}]}}$
Substituting value from equation 2,
$\dfrac{1}{{K_2}} = 50$
Taking constant on one side,
$\dfrac{1}{{50}} = K_2$
Solving the equation,
$K_2 = 0.02$
Hence the value of the equilibrium constant for dissociation of Hydrogen Iodide is 0.02.

Note:
Equilibrium constant is used to relate the concentrations of products and reactants in a chemical reaction which are in a state of equilibrium. The equilibrium constant for a dissociation reaction is usually the inverse of the association reaction and vice-versa.