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CHEM 122 rate laws Tean Number: Names: versus time, Integrated Dr. Sushilla Knottenbelt): Concentration following table: for the general reaction A orders. products for the different reaction Linear form-what should Half life equation be graphed, y vs x to get a straight line? What will the slope (m) be? The y-axis intercept? Reaction Rate form order integrated form Linear form-what Rate = k 2K x=t m--k I Rote In 2 k tın 2 Rate k[A m-K 2. The concentration of C.Hs, in the reaction 2C Hs)CHl) re varied with time. action was measured at 900 K as it A plot of 1/[CaHe] versus t gave a straight line. a. What is the order of the reaction with respect to Cate? Justify your answer (and check that it is correct before moving on!) b. f the slope of the best-fit line was calculated to be 0.0612 M's and the y axis intercept was 99.368 M, identify the rate constant of the reaction and determine the value of [Alo.

Answer 1

2C_4 H_6 rightarrow C_3 H_12 given: 1/(C_4 H_6) gives a straight time. (a) The order of reaction is second order The integrated rate low for 2^nd order reaction 1/ Vertical-bar A Vertical-bar = Rt + 1/ Vertical-bar A Vertical-bar _0 0This is equation of straight time y = mx + c where y is 1/ Vertical-bar A Vertical-bar x is t (time) m = slope = k c = intercept = 1/[A]_0 (b) Integrated rate law for second order reaction 1/ Vertical-bar A Vertical-bar = Rt + 1/ Vertical-bar A Vertical-bar _0 given: R = 0.0612 m^-1 s^-1, t = 3600 s, [A]_0 = 0.010064 M \r\n 1/ Vertical-bar A Vertical-bar = (0.0612 m^-2 s^-1)(3600 s) + 99.368 m^-1 1/ Vertical-bar A Vertical-bar = 319.688 m^-1 [A] = 1/319.688 m^-1 [A] = 3.13 times 10^-3 M (d) t_12 = 1/R[A]_0 t_12 = 1/(0.0612 m^-1 s^-1)(0.010064 m^-1) t_12 = 1623.66 s (e) initial concentration of C_4 H_6 = 0.010064 M concentration of C_4 H_6 after 3600 s = 3.13 times 10^3 M C_4 H-6 consumed in 3600 s = initial come - final come = 0.010064 M - 3.13 times 10^-3 M C_4 H_6 consumed in 3600 s = 6.94 times 10^-3 M