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$$\displaystyle [HI]_0=3$$ MThe equilibrium concentrations are$$\displaystyle [H_2]=1-x$$ M$$\displaystyle [I_2]=2-x$$ M$$\displaystyle [HI]=3+2x$$ MNote:  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$$\displaystyle [HI]=3+2x=3+2(0.684)=4.36$$ MChemistry

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