Asked by Hannah
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.
<-
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.
Answers
Answered by
Hannah
I tryed and got .20. Is this correct?
Answered by
DrBob222
Where did the 1.0, 10.0 and 10.0 come from?
Answered by
Hannah
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..
Answered by
DrBob222
Did the teacher put them in as initial concentrations or are they changes or equilibrium concentrations?
Answered by
Hannah
He didn't say, he just posted the question like above.
Answered by
DrBob222
What are you supposed to do with the numbers? What's the question?
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