Question

In the reaction, ${{H}_{2}}+{{I}_{2}} \rightarrow 2HI$, in a 2 litre flask 0.4 mole of${{H}_{2}}$ each and${{I}_{2}}$ are taken. At equilibrium 0.5 mol of $HI$ are formed. What will be the value of equilibrium constant ${{K}_{c}}$?
A) 20.2
B) 25.4
C) 0.284
D) 11.1

Answer 1

The equilibrium constant $({{K}_{c}})$ expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit. And first we find their concentration and then use the relationship between concentration and rate constant to find rate constant.

Complete Solution :
In the lower classes of chemistry, we have studied various concepts that deal with the equilibrium conditions in a reversible reaction and calculation of the equilibrium constants and this is mainly in the reaction dynamics chapter which is included in the physical chemistry part.
- Now, let us see how we can calculate the equilibrium constant on the basis of the given sata.
Given equation:
${{H}_{2}}+{{I}_{2}} \rightarrow 2HI$
Also given that:
Initial moles of both ${{H}_{2}}$ and ${{I}_{2}}$ is 0.4 mole of each. And at equilibrium 0.5 mole of $HI$ are formed.
Let’s make a table:
${{H}_{2}}+{{I}_{2}} \rightarrow 2HI$

Initial moles	0.4	0.4	0
At equilibrium	0.4-(0.50/2)=0.15	0.4-(0.50/2)=0.15	0.50

We know the equilibrium constant in terms of concentration, and we know that at equilibrium rate of forward reaction equal to rate of backward reaction.
Let’s assume the given reaction is an elementary reaction.
So, ${{k}_{1}}[{{H}_{2}}][{{I}_{2}}]={{k}_{2}}{{[HI]}^{2}}$
By rearranging:
$\dfrac{{{k}_{1}}}{{{k}_{2}}}=\dfrac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}$
And we know that
${{K}_{c}}=\dfrac{{{k}_{1}}}{{{k}_{2}}}$
Where,${{k}_{1}}$ is the forward reaction rate constant and ${{k}_{2}}$ is the backward reaction rate constant.
So,
${{K}_{c}}=\dfrac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}$
Now we will put all given values in the above equation:
${{K}_{c}}=\dfrac{{{[0.15]}^{2}}}{[0.15][0.15]}=11.11$
So, the correct answer is “Option D”.

Note: In this question equilibrium constant is unit less, but this thing is not possible in every problem. Because the units of forward rate constant and backward rate constant can be different.