Question

# In the reaction, $${ H }_{ 2 }(g)+{ I }_{ 2 }(g)\rightleftharpoons 2HI(g)\quad$$, in a $$2$$ litre flask $$0.4$$ mole of each $${H}_{2}$$ and $${I}_{2}$$ are taken. At equilibrium $$0.5$$ mol 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

Solution

## The correct option is D $$11.1$$In the reaction, $${ H }_{ 2 }(g)+{ I }_{ 2 }(g)\rightleftharpoons 2HI(g)\quad$$, in a $$2$$ litre flask $$0.4$$ mole of each $${H}_{2}$$ and $${I}_{2}$$ are taken. At equilibrium $$0.5$$ mol of $$HI$$ are formed. The value of equilibrium constant $${K}_{c}$$ will be $$11.1$$.0.5 moles of HI will be obtained from 0.25 moles of $$\displaystyle H_2$$ and 0.25 moles of $$\displaystyle I_2$$. $$\displaystyle 0.4-0.25=0.15$$ moles of $$\displaystyle H_2$$ and $$\displaystyle 0.4-0.25=0.15$$ moles of $$\displaystyle I_2$$ will remain.The equilibrium constant $$\displaystyle K_c = \dfrac { [HI]^2}{ [H_2][I_2] }$$$$\displaystyle K_c = \dfrac { (\dfrac { \text { 0.5 mol } }{ \text { 2 L } })^2 }{ (\dfrac { \text { 0.15 mol } }{ \text { 2 L } }) \times (\dfrac { \text { 0.15 mol } }{ \text { 2 L } })}$$$$\displaystyle K_c=11.1$$Chemistry

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