equilibrium - stoichiometry problem - chemistry

Question

Initially, a reaction mixture contains stoichiometric amounts of H2 and I2. The reaction mixture initially contains no HI. At a temperature of 160°C, the conversion rate for the next reaction is equal to 35%. H2+I2=(arrow goes both ways)2HI The numerical value of the equilibrium constant, Ke , at 160°C is equal to:

(the answer is 1.16)

the answer description is,

initial;

x (for H2)

x (for I2)

none (for 2HI)

reaction;

-0.35x (for H2)

-0.35x (for I2)

+0.7x (for 2HI)

at equilibrium; (initial minus reaction)

0.65 (for H2)

0.65 (for I2)

0.7x (for 2HI)

therefore,

Ke= (0.7x)^2/(0.65)^2=1.16...

my question is, if to get equilibrium is initial minus reaction, how did they get 0.65? would it not be x-0.35x for H2 and I2?

Answer 1

Set up an ICE table and it's easier to see what's going on. But first, you need the correctly balanced equation for the reaction.

H₂(g) + I₂(g) ==> 2HI(g)

1.............1...............0........Initial

-0.35......-0.35.......+0.70....Change

0.65.......0.65.........0.70.....Equilibrium

K = (HI)² / (H₂)(I₂) = (0.70)² / (0.65)(0.65)

K = 1.16

You don't want to call initial amounts of H₂ and I₂ x, because then you'll have 2 unknowns, i.e. H₂/I₂ and Keq. So, just give the initial H₂ and I₂ any value (I chose 1), and the use 0.35 (35%) of that value to subtract.