Asked by GeeGee
Nitrosyl bromide, NOBr, readily dissociated according to the following equilibrium equation:
2 NOBr (g) ----> 2 NO(g) + Br2 (g)
NOBr was placed in a 1.00 L flask at 25.0oC and allowed to dissociated. At equilibrium, 0.17 moles of Br2 are
obtained. If NOBr is found to be 34.0% dissociated, what is the value of Kc?
2 NOBr (g) ----> 2 NO(g) + Br2 (g)
NOBr was placed in a 1.00 L flask at 25.0oC and allowed to dissociated. At equilibrium, 0.17 moles of Br2 are
obtained. If NOBr is found to be 34.0% dissociated, what is the value of Kc?
Answers
Answered by
DrBob222
I don't think the question can be answered; there is no quantity given for the initial NOBr. If 1 mol than the following:
..........NOBr ==> 2NO + Br2
E.......1-0.34.....0.34...0.17
Kc = (NO)^2(Br2)/(NOBr)^2
..........NOBr ==> 2NO + Br2
E.......1-0.34.....0.34...0.17
Kc = (NO)^2(Br2)/(NOBr)^2
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