Question

0 molmol of H2H2 is allowed to react with 0.500 molmol of I2I2 in a 1.00 LL reaction vessel at 700 KK, what is the concentration of I2I2 at eq

Answer 1

`[I_2]=0.105M`

Step-by-step explanation:

Hello.

In this case, when hydrogen and iodine react at equilibrium, the reaction is:

`H_2+I_2\rightleftharpoons 2HI`

So the equilibrium expression is:

`Kc=([HI]^2)/([H_2][I_2])`

Which can be also written in terms of the ICE chart knowing that Kc is 57.0:

`Kc=((2x)^2)/(([H_2]_0-x)([I_2]_0-x))`

Whereas the initial concentrations of both hydrogen and iodine are 0.100 M (0.500mol/1.00L), thus, we write:

`Kc=((2x)^2)/((0.5-x)(0.5-x))\\\\57.0=((2x)^2)/((0.5-x)^2)`

Which can be solved as follows:

`√(57.0) =\sqrt{((2x)^2)/((0.5-x)^2)} \\\\7.55=(2x)/(0.5-x) \\\\7.55*(0.5-x)=2x\\\\3.775-7.55x=2x\\\\x=(3.775)/(9.55)\\ \\x=0.395M`

Therefore, the concentration of iodine at equilibrium is:

`[I_2]=0.5M-0.395M=0.105M`

Best regards!