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

Answer:

$[I_2]=0.105M$

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:

$Kc=\frac{[HI]^2}{[H_2][I_2]}$

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

$Kc=\frac{(2x)^2}{([H_2]_0-x)([I_2]_0-x)}$

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

$Kc=\frac{(2x)^2}{(0.5-x)(0.5-x)}\\\\57.0=\frac{(2x)^2}{(0.5-x)^2}$

Which can be solved as follows:

$\sqrt{57.0} =\sqrt{\frac{(2x)^2}{(0.5-x)^2}} \\\\7.55=\frac{2x}{0.5-x} \\\\7.55*(0.5-x)=2x\\\\3.775-7.55x=2x\\\\x=\frac{3.775}{9.55}\\ \\x=0.395M$

Therefore, the concentration of iodine at equilibrium is:

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

Best regards!