Question

D'Youville College Chemistry 115 - Problem Solving Chemistry Updated Spring 2016 Page 8 Half-lifes and order of reaction. The "reaction order" describes the dependence of rate on the concetration of a reactant. If the reaction is oth order, it's rate is independent of concentration, if it is 1st order, the rate is directly related to the concentration (ie. if the concentration is doubled, the rate will double), and if it is 2nd order, the rate has a squared relationship with the concentration (ie. if the concentration is doubled, the rate will increase by a factor of 22 4). It is often useful to determine the reaction order using graphs! Grahical data can be fit to one of the following equations of a line. These correspond to integrated rate laws for first and second order reactions. Therefore, if you find which graph is linear, you know the reaction order! Integrated Rate Laws: 1st order: In [A]t =-kt + In [A]o 2nd order: İAT-k + 4.0 3.5 T 3.0 3.0 160 140 120 100 80 60 3.5 -4.0 2.0 1.5 르1.0 0.5 4.5 5.5 20 Time (s) x 103 Time ls) × 103 Time (s) x 10 The half-life of a reaction (tı2) can also be determined using graphical data. The half-life is defined as the times required for concentration of a reactant to reach half of its initial value. [Ale [AJo From the first graph above, at time 0, the cocentration of N20s is 3.6x10-3 M. Determine the half life of this reaction. a. b. Which graph (left, right, or middle) results in a straight line? Using the information given above, what does that tell you about the order of the reaction. The m.ddle because t's a si ine and tsa st order be t cirecila related 4he c a d'CA (eotosn、Q、w result3 n a Staant Estimate the k for this-reaction .

Estimate the k for this reaction. (Last question on sheet)

Answer 1

from the leftmost and middle graph, it shows that the reaction is first-order reaction

for first-order reaction

lnAt = -Kt + ln A0

so slope of the second graph gives the K value

slope = - (-4.7 + 3.6 )/ (3-1)*10^3

slope = 0.55*10^-3 S^-1

so K for the reaction = 5.5*10^- 4 S^-1