Nitrosyl bromide, NOBr, is formed in the reaction of nitric oxide, NO, with brom

Question

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.

Part A

At a certain temperature the initial concentration of NO was 0.400 M and that of Br2 was 0.255 M . At equilibrium the concentration of NOBr was found to be 0.250 M. What is the value of Kc at this temperature?

Kc = 21.4

PART B

At this temperature the rate constant for the reverse reaction is 340. M2s1 . What is kf for the reaction?

Express your answer to three significant figures and include the appropriate units. Include an asterisk to indicate mulitplication in compound units, for example to write the units for a second order rate constant type M-1*s-1.

I need help with Part B

Explanation / Answer

Given that; the rate constant for the reverse reaction is 340. M2s1 .

2NOBr(g) 2NO(g)+Br2(g)

The rate law is then...

Rate = Kf * [NO]^2[Br2]

At a certain temperature the initial concentration of NO was 0.400 M and that of Br2 was 0.255 M

Rate = 340. M2s1 * 0.400 M^2 * 0.255 M

Rate = 13.8 M s1

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