Question

The following data were obtained for the concentration vs. time for a certain chemical reaction. Values were measured at 1.00 s intervals, beginning at 0.00 and ending at 20.00 s. Concentrations in mM are: 10.00, 6.91, 4.98,4.32, 3.55, 3.21, 2.61, 2.50, 2.22, 1.91, 1.80, 1.65, 1.52, 1.36, 1.42, 1.23, 1.20, 1.13, 1.09, 1.00, and 0.92.

A. Decide whether the data best fits zero-order, first-order, or second-order rate kinetics. Make sure to calculate the rate coefficients (with units) for the reaction and write the simplest mechanism you can for the reaction.

B. Describe an experimental method that you think might be used to measure the concentrations that are changing so rapidly. The reaction is over in nearly 20 seconds.

Answer 1

EXPLAINATION :

FIRST PLOT GRAPH BASED ON THE CONCENTRATION GIVEN TO YOU.

6 As we would easily see that plot of Ve versus time is coming as linear. Thus it confirms that reaction is of second order. Now we know that integrated form of second-order is cay = kt And try to relate this with general equation of line- y = mx + c When we plot the graph m =k a slope Rate constant : K = 0.0492 MAN Now, for the simplest mechanism for second order will involue two reactant. The mechanism as followes :- CACOD Doing analysis in real time using rapid mixing system in which two Solutions added with rapid mixing by stürver, followed by spectroscopic reading
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