Question

A mixture contained $30.92g\text{ }{{H}_{2}}$,$2.794g\text{ }{{\text{I}}_{2}}$ and $11308.16g\text{ }HI$. Calculate the equilibrium constant for the thermal decomposition of HI.

Answer 1

Hint: First we have to calculate the concentration of hydrogen, iodine and hydrogen iodide in moles per litre by using the formula as; ${{H}_{2}}=\dfrac{n}{V}$ and $n=\dfrac{given\text{ }mass}{molar\text{ }mass}$respectively. Here n represents the number of moles. After finding their concentration , by using the formula as ${{K}_{C}}=\dfrac{[{{H}_{2}}][{{I}_{2}}]}{{{[HI]}^{2}}}$, we can easily find the equilibrium constant for the HI. Now solve it.

Complete answer:
First let's discuss what equilibrium is. By the term equilibrium we mean that at which the number of moles of reactants is equal to equal moles of the products i.e. there is no change in the concentration of both the reactants and the products and no further reaction takes place.
And by thermal decomposition we mean the breakdown of the compound into its constituents in the presence of heat.
Now considering the statement as;
During the thermal decomposition of hydrogen iodide i.e. HI ; the hydrogen iodide breaks into the hydrogen gas and the iodine and this reaction is reversible. The reaction is supposed to occur as;
$2HI\rightleftharpoons {{H}_{2}}+{{I}_{2}}$
Then, the equilibrium constant for the reaction is as;
${{K}_{C}}=\dfrac{[{{H}_{2}}][{{I}_{2}}]}{{{[HI]}^{2}}}-------(1)$
Now, first we have to find the concentration of the reactants and products.
we know that ;
The mass of ${{H}_{2}}=30.92g$ (given)
The mass of ${{I}_{2}}=2.794g$ (given)
The mass of $HI=11308.16g$ (given)
And let V be the volume of the container; then-
${{H}_{2}}=\dfrac{n}{V}$
Here, n represents the number of moles which is;
$n=\dfrac{given\text{ }mass}{molar\text{ }mass}$
Then;
$\begin{align}
  & {{H}_{2}}=\dfrac{given\text{ }mass}{molar\text{ }mass}\times \dfrac{1}{V} \\
 & \text{ =}\dfrac{30.92}{2V} \\
 & \text{ =}\dfrac{15.46}{V} \\
\end{align}$
Similarly;
$\begin{align}
  & {{I}_{2}}=\dfrac{given\text{ }mass}{molar\text{ }mass}\times \dfrac{1}{V} \\
 & \text{ =}\dfrac{2.794}{254V} \\
 & \text{ =}\dfrac{11\times {{10}^{-3}}}{V} \\
\end{align}$
And
$\begin{align}
  & HI\text{=}\dfrac{11308.16}{128V} \\
 & \text{ =}\dfrac{88.345}{V} \\
\end{align}$
Now, put all these values in equation (1), we get;
$\begin{align}
  & {{K}_{C}}=\dfrac{[\dfrac{15.46}{V}][\dfrac{11\times {{10}^{-3}}}{V}]}{{{[\dfrac{88.345}{V}]}^{2}}} \\
 & \text{ = }\dfrac{170.06\times {{10}^{-3}}}{7804.84} \\
 & \text{ =1}\text{.021}\times {{10}^{-3}} \\
\end{align}$
So, thus the equilibrium constant for the thermal decomposition of HI is $\text{1}\text{.021}\times {{10}^{-3}}$.

Note:
Always remember that at equilibrium the rate of forward reaction is always to the rate of the backward reaction and vice-versa and the concentration always remains constant i.e. there is no change in the concentration of the reactants and the products.