Equilibrium Question 878

Question: In the reaction, $ H_2+I_2 $

$ \rightarrow $ $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} $

Options:

A) 20.2

B) 25.4

C) 0.284

D) 11.1

Show Answer

Answer:

Correct Answer: D

Solution:

$ \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}},}}, $

$ \rightarrow $ $\underset{0.50/2}{\mathop{\underset{0.50}{\mathop{2HI}},}}, $

$ K_{c}=\frac{{{[HI]}^{2}}}{[H_2][I_2]}=\frac{{{[ \frac{0.5}{2} ]}^{2}}}{[ \frac{0.15}{2} ],[ \frac{0.15}{2} ]} $

$ =\frac{0.5\times 0.5}{0.15\times 0.15}=11.11 $



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