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A mixture of 0.3 mole of H₂ and 0.3 mole of I₂ is allowed to react in a 10 lit vessel at 500⁰C. If K_Cof $\large {H_2}\, + \,{I_2}\, \Leftrightarrow \,\,2HI$ is 64, the amount of unreacted I₂ at equilibrium is

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0.06 mole

0.15 mole

0.24 mole

0.03 mole

answer is D.

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Detailed Solution

\large \left( {{n_{{H_2}}}} \right) = \left( {{n_{{I_2}}}} \right) = 0.3

Vessel = 10 lit

Let 'x' moles each of H₂ and I₂ reacted at

\large \mathop {{H_2}\left( g \right)}\limits^{1\,mole} +

\large \mathop {{I_2}\left( g \right)}\limits^{1\,mole}

K_C = 64 (given)

\large {K_C} = \frac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}

\large \frac{{\left( {\frac{{2x}}{{10}}} \right)^2}}{{\left( {\frac{{0.3 - x}}{{10}}} \right)}}=64

\large \frac{{2x}}{{\left( {0.3 - x} \right)}} = 8

2x = 2.4 - 8x

\large \boxed{x{\text{ }} = {\text{ }}0.24}

number of moles of I₂ unreacted

= (0.3 - x)

= (0.3 - 0.24)

= 0.06 moles

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