Given Kc, find equillibrium concentration of product

Question

In an experiment, 0.30 mol H2 and 0.30 mol I2 are mixed in a 1.00-L container, and the reaction forms HI. If Kc = 49. for this reaction, what is the equilibrium concentration of HI?

I2(g) H2(g) 2HI(g)

A.

0.53 M

B.

0.58 M

C.

0.040 M

D.

0.47 M

E.

0.075 M

I tried a before, change, and equillibrium table of concentrations and used a quadratic to try and find the answer but came up with .38 as my closest. The answer is D, but how?

Answer 1

H2(g) + I2(g)====>2HI(g)

Kc=[HI]²/[H2][I2]

(2x)²/((0.30-x)(0.30-x))=49

There's no need for a quadratic. We can just take the square root of both sides because the denominator is (0.30-x)(0.30-x) which is (0.30-x)². So we take the square root of both sides to get

2x/(0.30-x)=7

2x=2.1-7x

9x=2.1

x=2.1/9=2.333

The equilibrium concentration of HI is 2x which is 2*2.333=0.4666 M which is 0.47 M to two significant figures.