JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Topic Test - Chemical Equilibrium (6-5-21)

In the reaction, \[{{H}_{2}}+{{I}_{2}}\]\[\rightleftharpoons \]\[2HI\]. In a 2 litre flask 0.4 moles of each \[{{H}_{2}}\] and \[{{I}_{2}}\] are taken. At equilibrium 0.5 moles of \[HI\] are formed. What will be the value of equilibrium constant, \[{{K}_{c}}\]

A) 20.2

B) 25.4

C) 0.284

D) 11.1

Correct Answer: D

Solution :

[d] \[\underset{0.4-0.25\ =\ 0.15}{\mathop{\underset{0.4}{\mathop{{{H}_{2}}}}\,}}\,\]\[+\] \[\underset{0.4-0.25\ =\ 0.15/2}{\mathop{\underset{0.4}{\mathop{{{I}_{2}}}}\,}}\,\]\[\rightleftharpoons \] \[\underset{0.50/2}{\mathop{\underset{0.50\,\,\,\,}{\mathop{2HI\,\,\,\,}}\,}}\,\]

\[{{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}=\frac{{{\left[ \frac{0.5}{2} \right]}^{2}}}{\left[ \frac{0.15}{2} \right]\,\left[ \frac{0.15}{2} \right]}\] \[=\frac{0.5\times 0.5}{0.15\times 0.15}=11.11\]

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