± Linking Equilibrium and Kine5cs Formation of nitrosyl bromide Nitrosyl bromide
ID: 939889 • Letter: #
Question
± Linking Equilibrium and Kine5cs Formation of nitrosyl bromide Nitrosyl bromide, NOBr,s formed in the reaction of nitric oxide, NO, with bromne, Br The generic reaction has the folowing rate laws The reaction rapidy estabishes equilibrian when the reactants are mixeid forward reaction: rate-A reverse, reaction : rate =4/ Part A whoro ke is the rate constant fer the forward reaction and k, is the rate constant for the reverse reaction. A equiibrium, the two rates are equal and so el IAP-a, [B]. The equiibrum oonstart for a reacton Al a certain temperature the intial concentration of NO as0400 M and at of Br was 0235 M.Al equilibrlum the concentration of NOBr was found to be 0250M, What is the value of Ke at this temperature Express your answer munerically is related by the law of mass action to the rale constans or the forward and reverse reactions K,# 25.3 Hints My Aswers Give Up Review Part Part B Al is emperature the rase constant for the reverse reacion in 370. MWhat sa &y; 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 alue Units Hiets My Answers Give Up Review PartExplanation / Answer
1) 2NO + Br2 --> 2NOBr
Initial 0.4 0.235 0
Change -2x -x +2x
Equilibrium 0.4-2x 0.235-x 2x
Equilibrium concentration of NOBr = 2x = 0.250
so x = 0.125
[NOBr] = 0.25
[Br2] = 0.235 - 0.125 = 0.11
[NO] = 0.4-0.25 = 0.15
Kc = [NOBr]^2 / [Br2] [NO]^2 = 0.25 X 0.25 / 0.11 X 0.15 X 0.15 = 25.25
2) we know that Keq = Kf / Kr
Kf = Keq X Kr = 25.25 X 370 = 9342.5 M-2 s-1
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