Question

The equilibrium constant, K, for the following reaction is 1.80×10^-2 at 698 K.

2HI(g) --> H₂(g) + I₂(g)

An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.322 M HI,   4.33×10^-2 M H₂ and 4.33×10^-2 M I₂. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.213 mol of HI(g) is added to the flask?

[HI]	=	M
[H2]	=	M
[I2]	=	M

Answer 1

for the reaction

2HI(g) -->   H₂(g) +   I₂(g)

0.322 M   4.33×10^-2 M    4.33×10^-2 M

after addition

  0.535 M - 2 x   4.33×10^-2 M + x    4.33×10^-2 M + x

therefore

Keq = ( 4.33×10^-2 M + x )^2 / (  0.535 M - 2 x )^2

1.80×10^-2 =   ( 4.33×10^-2 M + x )^2 / (  0.535 M - 2 x )^2

0.134 =  ( 4.33×10^-2 M + x ) / (  0.535 M - 2 x )

0.072 - 0.268 x = 0.0433 + x

1.268 x = 0.0287

x = 0.023 M

and

therefore

[HI]	= 0.535 - 0.045 = 0.489	M
[H2]	= 0.0663	M
[I2]	= 0.0663	M