xercise 14.54 Enhanced with Feedback Part A The following reaction was monitored
ID: 487350 • Letter: X
Question
xercise 14.54 Enhanced with Feedback Part A The following reaction was monitored as a function of time: What is the value of the rate constant (k) for this reaction at this temperature? AB A B Aplot of 1/ AB) versus time yields a straight line Express your answer using two significant figures. with slope 5.9x10-2 (M. s) You may want to reference (Ua.pages 635-641) section 14.4 while completing this problem. M 1 s 1 Submit My Answers Give up Part B Write the rate law for the reaction. Rate A Rate k AB. Rate k AB, Rate k ABExplanation / Answer
A. A straight line of 1/[AB] versus t plot indicates that you have second order reaction. The slope is equal to the rate constant, hence,
k = 5.9×10² (M.s)¹
B. Rate law is rate = -d[AB]/dt = k[AB]²
C. The half-life of a second-order reaction depends on the initial concentration, t½ = 1/(k[AB])
= 1/(5.9×10²(Ms)¹ 0.57M) = 297.35 s = 3.0 x 102 s
D. the reaction mixture initially contains no products, hence, [A] = [B] = [AB] - [AB]
Reactant concentration after 80 sec is
[AB] = 1/(1/[AB] + kt)
= 1/[1/0.220M + 5.9×10²(Ms)¹80s]
= 0.108 M
So the product concentrations are:
[A] = [B] = [AB] - [AB] = 0.220M - 0.108 M = 0.112M
Therefore, [A], [B] = 0.11M, 0.11M
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