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.
6 answers
I tryed and got .20. Is this correct?
Where did the 1.0, 10.0 and 10.0 come from?
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..
Did the teacher put them in as initial concentrations or are they changes or equilibrium concentrations?
He didn't say, he just posted the question like above.
What are you supposed to do with the numbers? What's the question?
4 answers