In: Chemistry
The generic reaction
2A?B
has the following rate laws:
forward reaction:reverse reaction:rate=kf[A]2rate=kr[B]
where kf is the rate constant for the forward reaction and kr is the rate constant for the reverse reaction. At equilibrium, the two rates are equal and so kf[A]2=kr[B]. The equilibrium constant for a reaction is related by the law of mass action to the rate constants for the forward and reverse reactions:
Kc=[B][A]2=kfkr
Formation of nitrosyl bromide
Nitrosyl bromide, NOBr, is formed in the reaction of nitric oxide, NO, with bromine, Br2:
2NO(g)+Br2(g)?2NOBr(g)
The reaction rapidly establishes equilibrium when the reactants are mixed.
A.
At a certain temperature the concentration of NO was 0.400 M and that of Br2 was 0.265M . At equilibrium the concentration of NOBr was found to be 0.250 M. What is the value of Kc at this temperature?
Express your answer numerically.
2NO(g) + Br2(g) 2NOBr(g)
Initial conc(M) 0.400 0.265 0
Change(M) -2a -a +2a
Equb conc(M) 0.400-2a 0.265-a 2a
So the equilibrium concentration of NOBr is = 2a
But given that at equilibrium the concentration of NOBr was found to be 0.250 M
2a = 0.250 M
a = 0.125 M
The equilibrium concentration of NO is = 0.400-2a
= 0.400-0.250
= 0.150 M
The equilibrium concentration of Br2 is = 0.265 - a
= 0.265 - 0.125
= 0.140 M
Equilibrium constant ,