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Chapter 14: Problem 60

The following substrate concentration [S] versus time data were obtained during an enzyme-catalyzed reaction: \(t=0 \min ,[\mathrm{S}]=1.00 \mathrm{M} ; 20 \mathrm{min}, 0.90 \mathrm{M}; 60 \min , 0.70 \mathrm{M} ; 100 \mathrm{min}, 0.50 \mathrm{M} ; 160 \mathrm{min}, 0.20 \mathrm{M}.\) What is the order of this reaction with respect to \(\mathrm{S}\) in the concentration range studied?

Short Answer

Expert verified
The reaction is first order with respect to S.

Step by step solution

01

Identify the type of reaction

The types of reactions typically studied in kinetics are zero order, first order, and second order. We know a reaction is zero order if the rate is independent of the concentration of the reactant. The reaction is first order if the rate is directly proportional to the concentration of the reactant. It's a second order reaction if the rate is proportional to the square of the concentration of the reactant. Based on these principles, we will check for each type of reaction.
02

Checking if it's a zero order reaction

In a zero order reaction, concentration [S] will decrease linearly with time. From the provided set of data, we see that the concentration is not decreasing linearly over time, hence it is not a zero order reaction.
03

Checking if it's a first order reaction

In a first order reaction, assuming \([S]\) at \(t=0\) is \(S_0\), and at time \(t\) is \([S]\), we would use the equation, \(\ln([S]/S_0) = -kt\). Choose two data points, for example, \((t=0 min, [S]= 1.00M)\) and \((t=20min, [S]=0.90M)\) , if we substitute these in our equation, \(\ln{(0.90/1)} = -k * 20\) and solve for \(k\), repeat this for multiple pairs and if \(k\) value is nearly constant we can say it is a first order reaction.
04

Checking if it's a second order reaction

The approach is similar to the first order reaction but with a different equation: \(1/[S] - 1/S_0 = kt\). If after substituting pairs of \([S]\) and \(t\) in above equation, we get nearly constant \(k\), it's a second order reaction. However, given previous analysis we wouldn't need to do this calculation.

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Most popular questions from this chapter

According to collision theory, chemical reactions occur through molecular collisions. A unimolecular elementary process in a reaction mechanism involves dissociation of a single molecule. How can these two ideas be compatible? Explain.

For the reaction \(A \longrightarrow\) products, the data tabulated below are obtained. (a) Determine the initial rate of reaction (that is, \(-\Delta[\mathrm{A}] / \Delta t)\) in each of the two experiments. (b) Determine the order of the reaction. $$\begin{array}{ll} \hline \text { First Experiment } & \\ \hline[\mathrm{A}]=1.512 \mathrm{M} & t=0 \mathrm{min} \\ \begin{array}{l} | \mathrm{A}\rfloor=1.490 \mathrm{M} \\ {[\mathrm{A}]=1.469 \mathrm{M}} \end{array} & \begin{array}{l} t=1.0 \mathrm{min} \\ t=2.0 \mathrm{min} \end{array} \\ \hline & \\ \hline \text { Second Experiment } & \\ \hline[\mathrm{A}]=3.024 \mathrm{M} & t=0 \mathrm{min} \\ {[\mathrm{A}]=2.935 \mathrm{M}} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=2.852 \mathrm{M}} & t=2.0 \mathrm{min} \\ \hline \end{array}$$

For the reaction \(A \longrightarrow\) products, derive the integrated rate law and an expression for the half-life if the reaction is third order.

For the reaction \(A+2 B \longrightarrow 2 C\), the rate of reaction is \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) at the time when \([\mathrm{A}]=0.3580 \mathrm{M}.\) (a) What is the rate of formation of \(\mathrm{C}\) ? (b) What will \([\mathrm{A}]\) be 1.00 min later? (c) Assume the rate remains at \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) How long would it take for \([\mathrm{A}]\) to change from 0.3580 to \(0.3500 \mathrm{M} ?\)

The half-lives of both zero-order and second-order reactions depend on the initial concentration, as well as on the rate constant. In one case, the half- life gets longer as the initial concentration increases, and in the other it gets shorter. Which is which, and why isn't the situation the same for both?
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