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Calculating Equilibrium Constants The following reaction was performed in a sealed vessel at 752 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only...

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Chemistry

Calculating Equilibrium Constants

The following reaction was performed in a sealed vessel at 752∘C :
H2(g)+I2(g)⇌2HI(g)
Initially, only H2 and I2 were present at concentrations of[H2]=3.75M and [I2]=3.00M. The equilibrium concentration of I2 is0.0900 M . What is the equilibrium constant, Kc, for the reactionat this temperature?

Answer & Explanation Solved by verified expert
3.7 Ratings (641 Votes)

                            H2(g) +    I2(g)   ⇌      2HI(g)

Initial                  3.75 M    3.00 M             0

at equilibrium       3.75- x     3-x                 2 x

So,

    equilibrium concentration of I2 = 3-x

But given that equilibrium concentration of I2 = 0.0900 M

    Then,

                    3-x = 0.09

                       x = 2.91 M

    Therefore, equilibrium concentrations are

   [H2] = 3-75 -x = 3.75 - 2.91 = 0.84 M

[I2] = 0.09 M

[HI]= 2 x= 2 x 2.91 = 5.82 M

         Kc = [HI]2/ [H2] [I2]

               = (5.82)2 / (0.84) (0.09)

              = 448.05

Therefore, equilibrium constant   Kc = 448.05
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