The following reaction was performed in a sealed vessel at 720 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only...

The following reaction was performed in a sealed vessel at 720 ∘C :

H2(g)+I2(g)⇌2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.50M and [I2]=2.90M. The equilibrium concentration of I2 is 0.0600 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Expert Solution

H2(g) +   I2(g) <----> 2HI(g)
3.50            2.90                  0    (initial)
-x                  -x                     2x (change)
3.50-x       2.90-x                2x   (at equilibrium)

given equilibrium concentration of I2 = 0.0600 M

so,
2.90-x = 0.06
x = 2.84 M

Kc = (2x) / {(3.50-x) (2.90-x) }
       = (2*2.84)^2 / {(3.50-2.84) (2.90-2.84) }
       = 5.68 / {(0.66) (0.06)}
        = 814.7

Answer: 814.7

queen_honey_blossom answered 2 years ago

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