Question

Given $15$ moles of ${{H}_{2}}$ and $5.2$ moles of ${{I}_{2}}$ are mixed and then allowed to attain equilibrium at $500{{\text{ }}^{\circ }}C$. At equilibrium the concentration of $HI$ is found to be $10$ moles. The equilibrium constant for the formation of $HI$is:
(a) $50$
(b) 15
(c) $100$
(d) $25$

Answer 1

Hint: By the equilibrium constant, we means the ratio of the concentration of the products to the reactants and first we have to find the moles of the reactants and products at equilibrium and then, we can find the equilibrium constant by applying the formula as; ${{K}_{c}}=\dfrac{{{[AB]}^{c}}}{{{[A]}^{a}}{{[B]}^{b}}}$. Now solve it.

Complete answer:
First of all , let’s discuss what equilibrium is. By the equilibrium constant we mean the ratio of the concentration of the products to the reactants raised to the power of their coefficient. It is denoted by the symbol as :${{K}_{c}}$
Consider the general reaction at equilibrium as;
$aA+bB\to cAB$
Then, the equilibrium constant of the given reaction is as;
${{K}_{c}}=\dfrac{{{[AB]}^{c}}}{{{[A]}^{a}}{{[B]}^{b}}}$
Here, square brackets represent the concentration of the reactants and the products.
Now considering the statement;
The reaction occurs as;
$${{H}_{2}}+{{I}_{2}}\to 2HI$$
The equilibrium constant ${{K}_{c}}$for the reaction is as;
$${{K}_{c}}=\dfrac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}$$--------------(1)
First, we will calculate the moles of the reactants and the products at equilibrium and then, after that we will find the equilibrium constant of the reaction;
Initial moles of the reaction is :
$$ \begin{matrix}
   {{H}_{2}} \\
   15 \\
\end{matrix}+\begin{matrix}
   {{I}_{2}} \\
   5.2 \\
\end{matrix}\to \begin{matrix}
   2HI \\
   0 \\
\end{matrix}$$
Change in the number of the moles during the reaction is :
Initial moles of the reaction is :
$$ \begin{matrix}
     {{H}_{2}} \\
  15 \\
  x \\
\end{matrix}+\begin{matrix}
   {{I}_{2}} \\
   5.2 \\
   x \\
\end{matrix}\to \begin{matrix}
   2HI \\
   0 \\
   2x \\
\end{matrix}$$
No of the moles at equilibrium is:
Initial moles of the reaction is :
$$ \begin{matrix}
     {{H}_{2}} \\
  15 \\
  x \\
  15-x \\
\end{matrix}+\begin{matrix}
   {{I}_{2}} \\
   5.2 \\
   x \\
   5.2-x \\
\end{matrix}\to \begin{matrix}
   2HI \\
   0 \\
   2x \\
   2x=10
\end{matrix}$$
So, the number of moles of $HI$ at equilibrium is;
$\begin{align}
  & HI=2x=10 \\
 & \text{ =x=}\dfrac{10}{2} \\
 & \text{ =x =5} \\
\end{align}$
so, $x=5$
Now, from this x we can calculate the number of moles of the ${{H}_{2}}$ and ${{I}_{2}}$at equilibrium as;
Number of moles of ${{H}_{2}}$at equilibrium is:
$\begin{align}
  & {{H}_{2}}=15-x \\
 & \text{ =15-5} \\
 & \text{ =10} \\
\end{align}$
Similarly, the Number of moles of ${{I}_{2}}$at equilibrium is: $0.2$
Now, put these value of in equation (1), we get;
$\begin{align}
  & {{K}_{c}}=\dfrac{{{[10]}^{2}}}{[10][0.2]} \\
 & \text{ =}\dfrac{1000}{20} \\
 & \text{ =50} \\
\end{align}$
So, thus $15$ moles of ${{H}_{2}}$and $5.2$ moles of ${{I}_{2}}$ are mixed and then allowed to attain equilibrium at $500{{\text{ }}^{\circ }}C$. At equilibrium the concentration of $HI$is found to be $10$ moles. The equilibrium constant for the formation of $HI$ is:$50$.

Hence, option (a) is correct.

Note:
If the reaction is at equilibrium only, then we can find the equilibrium constant of the given reaction and if the given reaction is not at equilibrium, then the equilibrium constant of the reaction can not be found. So, it means that the equilibrium is the necessary condition for the equilibrium constant.