The following reaction was performed in a sealed vessel at 724*C : H2(g)+I2(g)?2HI(g) Initially, only H2...

The following reaction was performed in a sealed vessel at 724*C :

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

Initially, only H2 and I2 were present at concentrations of [H2]=3.85M and [I2]=2.85M. The equilibrium concentration of I2 is 0.0800M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Express answer numerically

Expert Solution

                            H2(g)+      I2(g)      <--->     2HI(g)
initial:                   3.85           2.85                     0
at equlibrium:       3.85-x          2.85-x                 2x

given at equilibrium, [I2]=0.08 M
so,
2.85-x = 0.08
x= 2.77 M

[H2]= 3.85-x = 3.85 - 2.77 = 1.08 M

[HI]= 2x = 2* 2.77=5.54 M

Kc = [HI]^2/{[H2]*[I2]}
= 5.54^2 / (1.08*0.08)
=355.23

queen_honey_blossom answered 1 year ago

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