Question

Help please! I keep messing up w/ my math somewhere....
At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3 At this temp, 0.3000 mol of H2 and 0.300 mol of I2 were placed in a 1L container to react. What concentration of HI is present at equilibrium?
H2+I2<-> 2HI

Answer 1

so you made and ICE table and wrote the equilibrium expression?

Answer 2

yep I got .471 this time, I keep getting different #'s cuz im unsure if I need to square 2x when I plug everything in the equation

Answer 3

or .2355 if I don't multiply it by 2 at the end...

Answer 4

you gotta multiply x by 2 at the end to get [HI], but not square it.

Answer 5

so I'd end up with a quadratic on the denominator?

53.3= 2x/(.300-x)(.300-x)

Answer 6

you'd square it when calculating x

so, 53.3= (2x)^2/(.300-x)(.300-x)

but not at the end when finding [HI]=2x

Answer 7

oh yeah, that's what I meant by squaring when I plugged it into the equation, and when I solved for x, I multiplied it by 2

Answer 8

so I'm guessing I did it right :)

Answer 9

quick Q, why do we square the top? because the other two were on the bottom, or is the top always squared?

Answer 10

it depends on the coefficients of the balanced equation: H2+I2<-> 2HI

if you had "A + 3 B <->2 C", then \(K=\dfrac{[C]^2}{[A][B]^3}\)