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Question

One mole of $${H}_{2}$$, two moles of $${I}_{2}$$ and three moles of $$HI$$ are injected in a $$1$$ litre flask. What will be the concentration of $${H}_{2},{I}_{2}$$ and $$HI$$ at equilibrium at $${490}^{o}C$$? The equilibrium constant for the reaction at $${490}^{o}C$$ is $$45.9$$.


Solution

$$ \displaystyle H_2+I_2 \rightarrow 2HI$$
Since volume of flaskis 1 L, the number of moles is equal to molar concentration.
The initial concentrations are
$$ \displaystyle [H_2]_0=1$$ M
$$ \displaystyle [I_2]_0=2$$ M
$$ \displaystyle [HI]_0=3$$ M
The equilibrium concentrations are
$$ \displaystyle [H_2]=1-x$$ M
$$ \displaystyle [I_2]=2-x$$ M
$$ \displaystyle [HI]=3+2x$$ M
Note:  x M of $$ \displaystyle H_2$$ will react with x M of $$ \displaystyle I_2$$ to form 2x M of $$ \displaystyle HI$$ at equilibrium.
The equilibrium constant
$$ \displaystyle K_c = \dfrac {[HI]^2}{[H_2][I_2]}$$
$$ \displaystyle 45.9 = \dfrac {[3+2x]^2}{[1-x][2-x]}$$
$$ \displaystyle 45.9 = \dfrac {3[3+2x]+2x[3+2x]  }{[2-x]-x[2-x]}$$
$$ \displaystyle 45.9 = \dfrac {9+6x+6x+4x^2]  }{2-x-2x+x^2}$$
$$ \displaystyle 45.9 = \dfrac {9+12x+4x^2  }{2-3x+x^2}$$
$$ \displaystyle 45.9 (2-3x+x^2) = 9+12x+4x^2$$
$$ \displaystyle 91.8-137.7x+45.9x^2 = 9+12x+4x^2$$
$$ \displaystyle 82.8-149.7x+41.9x^2 = 0$$
$$ \displaystyle 41.9x^2-149.7x+ 82.8= 0$$
This is quadratic equation with solution
$$ \displaystyle x = \dfrac {-b \pm \sqrt {b^2-4ac}}{2a}$$
$$ \displaystyle x = \dfrac {-(-149.7) \pm \sqrt {(-149.7)^2-4(41.9)(82.8)}}{2(41.9)}$$
$$ \displaystyle x = \dfrac {149.7 \pm  92.4}{83.8}$$
$$ \displaystyle x =  2.888$$ or $$ \displaystyle x = 0.684$$
The value $$ \displaystyle x =  2.888$$ is discarded as it will lead to negative value of concentration.
$$ \displaystyle x = 0.684$$
The equilibrium concentrations are
$$ \displaystyle [H_2]=1-x=1-0.684=0.316$$ M
$$ \displaystyle [I_2]=2-x=2-0.684=1.316$$ M
$$ \displaystyle [HI]=3+2x=3+2(0.684)=4.36$$ M

Chemistry

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