Question

4.5 moles each of hydrogen and iodine are heated in a sealed 10 litre vessel. At equilibrium 3 moles of hydrogen iodine was found. The equilibrium constant for \[{H_{2\left( g \right)}} + {I_{2\left( g \right)}} \rightleftharpoons 2H{I_{\left( g \right)}}\]
a.) 1
b.) 10
c.) 5
d.) 0.33

Answer 1

Hint: Before finding the equilibrium constant one must first identify concentration hydrogen and iodine and also remember to use the formula of equilibrium constant i.e. ${K_c} = \dfrac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}$, using this information will help you to approach the solution of the question.

Complete step by step answer:
According to the given information, we know that when hydrogen and iodine of 4.5 moles each are heated in a sealed 10 litre vessel gives 3 moles of hydrogen iodine
So, the given reaction is \[{H_{2\left( g \right)}} + {I_{2\left( g \right)}} \rightleftharpoons 2H{I_{\left( g \right)}}\]
Let moles of the given elements after reaction be x
And we know that the ratio of the given equilibrium is given as 1 : 1 : 2
Now let’s find the concentration of the hydrogen and iodine
\[\begin{gathered}
  {\text{Given reaction: }}{H_{2\left( g \right)}}{\text{ }} + {\text{ }}{I_{2\left( g \right)}}{\text{ }} \rightleftharpoons {\text{ }}2H{I_{\left( g \right)}} \\
  {\text{Initial moles: 4}}{\text{.5 4}}{\text{.5 0}} \\
  {\text{At equilibrium: }}\left( {4.5 - x} \right){\text{ }}\left( {4.5 - x} \right){\text{ }}2x \\
\end{gathered} \]

We know that at equilibrium moles of hydrogen and iodine is equal to 3
Therefore, 2x = 3
\[ \Rightarrow \]\[x = \dfrac{3}{2}\]or x = 1.5
The number of moles at equilibrium will be
\[{H_{2\left( g \right)}}\] = 4.5 – x
Substituting the value of x we get
\[{H_{2\left( g \right)}}\] = 4.5 – 1.5 = 3
\[{I_{2\left( g \right)}}\]= 4.5 – 1.5
Substituting the value of x we get
\[{I_{2\left( g \right)}}\]= 4.5 – 1.5 = 3
We know that formula of concentration is given as; $\left[ A \right] = \dfrac{{{\text{no of moles of reactant}}}}{{{\text{volume of reactant}}}}$ here $\left[ A \right]$represents the concentration of A which can be either reactant or product.

Substituting the values in the above formula to find the concentration of reactants and product
For \[{H_{2\left( g \right)}}\]:
$\left[ {{H_2}} \right] = \dfrac{3}{{10}}$
$ \Rightarrow $$\left[ {{H_2}} \right] = 0.3$

For \[{I_{2\left( g \right)}}\]:
$\left[ {{I_2}} \right] = \dfrac{3}{{10}}$
$ \Rightarrow $$\left[ {{I_2}} \right] = 0.3$

For \[HI\]
$\left[ {HI} \right] = \dfrac{3}{{10}}$
$ \Rightarrow $$\left[ {HI} \right] = 0.3$

We know that for an equilibrium reaction equilibrium constant ($aA + bB \rightleftharpoons cC$) is given as i.e. ${K_c} = \dfrac{{{{\left[ C \right]}^c}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}$here A and B are reactant and C is the product and a, b, c are the stoichiometric coefficients.

Substituting the values in the above formula we get
 ${K_c} = \dfrac{{{{\left[ {0.3} \right]}^2}}}{{\left[ {0.3} \right]\left[ {0.3} \right]}}$
$ \Rightarrow $${K_c} = \dfrac{{0.3 \times 0.3}}{{0.3 \times 0.3}}$
$ \Rightarrow $${K_c} = 1$
Therefore, the equilibrium constant i.e. ${K_c}$ is equal to 1.
So, the correct answer is “Option A”.

Note: In the above solution we came across the term “equilibrium reaction” which can be explained as chemical reaction where rate of forward reaction is equal to rate of reverse reaction which means when a reaction is said to be equilibrium then there is no change in amount or concentration of the reactant and the products of the given reaction doesn’t changes.