1. The rate law for a certain reaction is second order with respect to one of th
ID: 626921 • Letter: 1
Question
1. The rate law for a certain reaction is second order with respect to one of the reactants, R. Suppose you study this reaction, observing the absorbance of light at the analytical wavelength for R, and record the data with respect to elapsed time. Also suppose that the concentrations of all the other reactants are in large excess, and that A is the only colored species involved. Explain which absorbance function, A, ln A, or 1/A, would yield a straight-line graph when plotted against elapsed time. 2. Suppose that in the reaction Q + R ? P, only the product, P, is colored. (a) What happens to the absorbance at the analytical wavelength for P of the mixture as the reaction progresses? (b) Suppose you do an experiment involving this reaction in which Q is present in large excess of R. If this reaction is first order with respect to R, would the graph of ln A (for P) versus time be a straight line? Briefly explain. Please help!Explanation / Answer
*****1.*****for a 2nd order reaction rate law is, [(Ao-A)/(Ao*A)]=kt, where Ao=initial conc.=constant, A=conc. of reactant at time t, k is the rate constant.****** Now as time elapses, conc. of reactant, here R , decreases.hence its absorbance will also decrease. From rate law , it becomes clear that a plot of (1/A) vs. time will be straight line (logarithmic function comes into play when reaction is 1st order; but here its mentioned that its 2nd order).*******2.*****(a)****** since P is the product, as time elapses, its conc. increases .Hence its absorbance will also increase, which means , a plot of A vs. time will have positive slope (no further data available to make a more precise statement). *******2(b)******since here Q is present in large excess, so its conc. will remain almost constant during the course of the reaction. So, we'll not worry about Q any more. Now the reaction is 1st order wrt R. Hence overall it becomes a pseudo-unimolecular reaction now. Hence, the graph of ln A (for P) vs. time will be a straight line (recall the 1st order rate law...for product it is ln A=kt, where k is the rate constant. Hence ln A vs. t will give a straight line plot with slope=k and zero intercept.
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