Question

A mixture of 0.3 mole of ${{H}_{2}}$ and 0.3 mole of ${{I}_{2}}$ is allowed to react in a 10 lit. evacuated flask at $500{}^\circ C$. The reaction is ${{H}_{2}}+{{I}_{2}}\rightleftharpoons 2HI$, the ${{K}_{c}}$ is found to be 64. The amount of unreacted ${{I}_{2}}$ at equilibrium is:
A. 0.15 mole
B. 0.06 mole
C. 0.03 mole
D. 0.2 mole

Answer 1

${{K}_{c}}$for any reaction in equilibrium is the equilibrium constant, that tells the concentration of product upon reactants raised to their stoichiometric values.

Formula used:
${{K}_{c}}=\dfrac{{{[products]}^{a}}}{{{[reac\tan ts]}^{b}}}$

Complete answer: We have been given a mixture of hydrogen and iodine gas with concentration of 0.3 moles each, kept in a 10 L flask at $500{}^\circ C$. We have to find the unreacted iodine gas at equilibrium with ${{K}_{c}}$= 64
For this we will use the initial concentration of gases ${{H}_{2}}$ and ${{I}_{2}}$ as 0.3 and their concentration at equilibrium as 0.3-x. So, the concentration of $HI$ at equilibrium will be 2x. now putting these entities at equilibrium into the formula of ${{K}_{c}}$, we will have,
${{K}_{c}}=\dfrac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}$
64 = $\dfrac{{{\left[ \dfrac{2x}{10} \right]}^{2}}}{\left[ \dfrac{0.3-x}{10} \right]\left[ \dfrac{0.3-x}{10} \right]}$
Therefore, $60{{x}^{2}}-38.4x+5.76=0$
Solving the equation for x, we will get, x = 0.24.
As, number of moles of ${{I}_{2}}$ are 0.3, so moles unreacted will be,
Moles of ${{I}_{2}}$ left = 0.3 – 0.24
Moles of ${{I}_{2}}$ left = 0.06 moles
Hence, the unreacted concentration of ${{I}_{2}}$ is 0.06 moles.

So, option B is correct.

Note: As the concentration is calculated in the volume of the flask which is 10 L, so, $\dfrac{moles}{volume}$ gives us the concentration in 10 L of the mixture.