Question

At ${{490}^{o}}$C, the equilibrium constant for the synthesis of HI is 50, the value of K for the dissociation of HI will be

Answer 1

Hint: The equilibrium constant is given by the value of the reaction quotient which is calculated from the expression for chemical equilibrium. It depends on the ionic strength and temperature but independent from the concentrations of reactants and products in a solution.

Complete Step by step solution: Let us take the example of the following chemical reaction$aA+bB\to cC+dD$. In this equation equilibrium constant which is represented by ${{K}_{c}}$ is calculated by using its coefficients raise to the power of molar concentration which can be shown as:
${{K}_{c}}=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$. Where [A], [B], [C], [D] are the molar concentrations of A, B, C, D and a, b, c, d, are the coefficients of chemical equations. The equilibrium constant is a dimensionless quantity i.e. it has no units. Value of dissociation of HI will be:
$\implies \dfrac{1}{2}{{H}_{2}}+\dfrac{1}{2}{{I}_{2}}\to HI$
$\implies K=\dfrac{[HI]}{{{[{{H}_{2}}]}^{\dfrac{1}{2}}}{{[{{I}_{2}}]}^{\dfrac{1}{2}}}}$
K = 50 (Given), But we have to find the value of dissociation of HI so reverse the reaction
By reversing the reaction it comes as:
$\implies HI\to \dfrac{1}{2}{{H}_{2}}+\dfrac{1}{2}{{I}_{2}}$
$\implies K'=\dfrac{{{[{{H}_{2}}]}^{\dfrac{1}{2}}}{{[{{I}_{2}}]}^{\dfrac{1}{2}}}]}{[HI]}$
Thus we can consider that $K'=\dfrac{1}{K}=\dfrac{1}{50}=0.02$

Thus the value of K for the dissociation of HI will be 0.02.

Note: If the value of is very large then the equilibrium favors the reaction in the right direction which means there are more products than reactants and the reaction is said to be quantitative or complete. But if the value for the equilibrium constant is small then the equilibrium favors the reaction in the right direction which states that there are more reactants than products. If the value of approaches is zero the reaction may not occur.