The reaction of bromine and nitric oxide (shown below) was followed at 273 °C un
ID: 550580 • Letter: T
Question
The reaction of bromine and nitric oxide (shown below) was followed at 273 °C under pseudo-first order conditions with [Br] 0.005 M and excess nitric oxide. A plot of the observed rate constant versus the nitric oxide concentration squared ([NO]) was linear with a slope of 1.2x10* Ms and an intercept of zero Br2(g) + 2 NO(g) 2 NOBr(g) Based on the information given, what can be deduced about the order of reaction for each of the reacting molecules? Describe how the conclusion was reached. a) [4 marks] b) Determine the rate law for this reaction. 4 marks]Explanation / Answer
answer. a) let assume rate =k[NO]a[Br2]b
now we have to determine a &b
If a=0 and b=1(over all first order), then r=k[Br2] this means that is not at all acceptable.
If a=1, b=1(over all second order), then r=k[NO][Br2], doubling the concentration of both the reactants will increase the rate by a factor of 4,
If a=0 and b=2( overall second order),then r=k[Br2]2 i.e not acceptable because doubling the concentration of [Br2]2 will increase the rate by a factor of 4.
If a=1 & b=2 (over all third order), then r= k[NO][Br2]2, doubling the concentration of both reactants will increase the rate by a factor of 8 but doubling the concentration of concentration of Br2 alone will increase the rate by a factor of 4.
If a=2 &b=1 then rate=[NO]2 [Br2], then we can come to the conclusion that the rate reaction is homogeneous and the rate of the reaction is doubled when the Bromine concentration is double but increase by factor of eight when the concentration of both the reactants are doubled.
Hence overall the reaction is of the third order, being second order in NO and First order in Br2.
Answer b) the rate for the reaction ias r=k[NO]2[Br2].
same as which has been written for answer a.
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