As part of a kinetics experiment, a student plots three graphs: [C_4H_6], M vs.
ID: 531500 • Letter: A
Question
As part of a kinetics experiment, a student plots three graphs: [C_4H_6], M vs. Time, s ln [C_4H_6] vs. Time, s 1/[C_4H_6], 1/M vs. Time, s and determines that the plot of 1/[C_4H_6], 1/M vs. Time, s has the greatest correlation coefficient (has data points that correspond the most to a linear response). The equation of this plot is y = 0.0614x + 100.0 a. What order is this reaction with respect to the butadiene molecule? b. What is the k-value for this reaction? And the units? c. What is the rate law of this reaction with respect to the butadiene molecule? d. What is the rate of a reaction (related to this experiment) given a butadiene concentration of 0.00350 M?Explanation / Answer
(a) The plot of 1/concentrattion vs time gives the most linear response it means the reaction is second order with respect to C4H6.
(b) Given equation of the plot is, y = 0.0614X + 100.0
k = slop and in the given equation the slop is 0.0614
so, the value of rate constant k is 0.0614.
The rate law for second order reaction in general is, rate = k[A]2
where [A] stands for the concentration of the reactant.
rate is change in concentration dx per unit time dt. so,.....
dx/dt = k[A]2
let's take molarity, M for the concentration and second, s for the time.
M/s = k[M]2
k = (M/s)/M2
k = M.s-1/M2 = M-1.s-1
(c) rate law with respect to butadiene would be as...
rate = k[butadiene]2
(d) butadiene concentration is given as 0.00350 M.
rate = k [butadiene]2
rate = 0.0614 M-1.s-1(0.00350 M)2
rate = 7.52 x 10-7 M.s-1
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