The following reaction was monitored as a function of time: AB→A+BAB→A+B A plot of 1/[AB]1/[AB] versus time yields a straight line with slope 5.2×10−2 (M⋅s)−1(M⋅s)−1 . You may want to reference (Pages 598 - 605) Section 14.5 while completing this problem. Part APart complete What is the value of the rate constant (k)(k) for this reaction at this temperature? Express your answer using two significant figures. kk = 5.2×10−2 M−1s−1M−1s−1 SubmitPrevious Answers Correct Important: If you use this answer in later parts, use the full unrounded value in your calculations. A plot of 1/[AB]1/[AB] versus time produces a straight line according to the problem statement. 1[AB]t = kt + 1[AB]o1[AB]t = kt + 1[AB]o The slope of the graph corresponds to the rate constant for this reaction. Part B Write the rate law for the reaction. Write the rate law for the reaction. Rate=kRate=k Rate=k[AB]Rate=k[AB] Rate=k[AB]2Rate=k[AB]2 Rate=k[AB]3Rate=k[AB]3