Question

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...

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.

Expert Solution

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 ,

queen_honey_blossom answered 3 weeks ago

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