Question

For H2(g) + Br2(g) -> 2HBr(g) k=64
<-

Ice table:
H2 Br2 HBr
I 0.10 0.10 0
C -x -x +2x
E

Solve for HBr at equilibrium

I do not know how to start this.

Answer 1

I tryed and got .20. Is this correct?

Answer 2

Where did the 1.0, 10.0 and 10.0 come from?

Answer 3

Those were the values that the teacher put in there. I just assumed that since H2 and Br2 are -x and HBr is +2x that HBr would be 2 X 0.10..

Answer 4

Did the teacher put them in as initial concentrations or are they changes or equilibrium concentrations?

Answer 5

He didn't say, he just posted the question like above.

Answer 6

What are you supposed to do with the numbers? What's the question?