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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
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B
25.4
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C
0.284
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D
11.1
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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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