The equilibrium constant for the following reaction is 1.80x102 at 698K 2HI(8)H2( ) +I2(g) If an...

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The equilibrium constant for the following reaction is 1.80x102 at 698K 2HI(8)H2( ) +I2(g) If an equilibrium mixture of the t

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Answer 1

Answer #1

2 HI(g) $\rightleftharpoons$ H2(g) + I2(g) Kc = 1.80 x10^-2 = 0.0180M

equilibrium concentration of HI = 2.35x10^-2M = 0.0235M

equilibrium concentration of H2 = 2.63x10^-2M = 0.0263M

Kc = [H2] [I2] / [ HI]^2

0.0180 = 0.0263 x [I2] / ( 0.0235)^2

[I2] = 3.78 x10^-4 M

Equilibrium concentration of I2 = 3.78 x10^-4M

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Answer 2

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