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 $
