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**Using Rate Laws to Predict Concentrations at Time = t**

Unit 3 - Kinetics Part 2 Using Rate Laws to Predict Concentrations at Time = t If our experimental study of a reaction shows that its rate is first order, we can calculate the concentrations of reactant and products at a time t. Process: A products Rate = k[A] Using the form for the instantaneous rate - d[A] = k [A] rearranging: d[A] = -k dt dt [A] integrating from 0 to t gives: ln[A]t – ln[A]o = -kt which rearranges to: ln[A]t = - kt + ln[A]o

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**Using Rate Laws to Predict Concentrations at Time = t**

For any first order reaction A products the concentration of A at time t is given by ln[A]t = - kt + ln[A]o where [A]o is the initial concentration and k is the rate constant. ln[A] is the natural logarithm of A and that is the power to which e ( ) must be raised to get A.

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**Plots of First Order Reactions**

ln[A]t = - kt + ln[A]o y = mx + b A plot of ln[A]t versus time will be a straight line with slope = -k and y-intercept = ln[A]o.

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**Order of a Reaction - Example**

Analyze rate data for the reaction SO2Cl2(g) SO2(g) + Cl2(g) at 320°C to prove the reaction is first order and to find the rate constant. time (s) pressure of SO2Cl2 (atm) 1.000 2500 0.947 5000 0.895 7500 0.848 10000 0.803 If the rate = constant, then a plot of P vs. t will be a straight line. A plot of P vs. time is not a straight line. I used the “add trendline” feature of Excel.

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**Order of a Reaction - Example**

When we plotted P versus t, we did not get a straight line. So we calculate ln P and plot ln P vs t. time (s) pressure of SO2Cl2 (atm) ln P 1.000 2500 0.947 5000 0.895 7500 0.848 10000 0.803 Why is it okay to use pressure instead of molarity when analyzing the data? A plot of ln P vs. time is a straight line, so the reaction is first order in SO2Cl2.

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**Order of a Reaction - Example**

Now we can determine the first order rate constant: ln[A]t = - kt + ln[A]o time (s) ln P k (s-1) 2500 2.18 x 10-5 5000 2.26 x 10-5 7500 2.16 x 10-5 10000 ln P = - kt + lnPo since Po = 1, ln Po = 0 ln P = - kt average k is k = 2.20 x 10-5 s-1 at T = 320°C.

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**Half-Life of a Reaction**

The half-life of a reaction is a convenient way to describe the speed of a reaction. The half-life is the time it takes for the concentration of a reactant to drop to one-half of its initial value. If a reaction has a short half-life, it is a fast reaction. The equation that gives the half-life depends on the rate law for the reaction. We will focus on first order reactions.

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**Half-Life of a First Order Reaction**

The half-life (t½) is the time it takes for the concentration of a reactant A to drop to one-half of its initial value ( ½ [A]o). For a first order reaction, ln[A]t = - kt + ln[A]o which can be rearranged: ln [A]t = - kt [A]o

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**Half-Life of a First Order Reaction**

ln[A]t = - kt + ln[A]o is the same as ln [A]t = - kt [A]o Now we put in the requirements for half-life: ln ½[A]o = - kt½ 0.693 = kt½ t ½ = 0.693/k ln ½ = - kt½ - ln 2 = - kt½ ln 2 = kt½

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**t½ of a First Order Reaction - Example**

SO2Cl2(g) SO2(g) + Cl2(g) T = 320°C Determine the time required for the pressure of SO2Cl2 to drop from atm to atm. How long does it take for the pressure to drop from atm to atm? We have already determined that the reaction is first order and that k = 2.20 x 10-5 s-1 at 320°C. t½ = 0.693/k = 0.693/(2.20 x 10-5 s-1) = 3.15 x 104 s = 8.75 hr The answer to BOTH questions is 8.75 hr, because in both cases the final pressure is one-half the initial pressure. We did not have to convert pressure to molarity because half-lives apply to the ratio of two pressures: P2 = M2RT = M2 P1 M1RT M1

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**Half-Life of a First Order Reaction**

The half-life applies to ANY drop in concentration (or pressure) of one-half. [A]t = [A]o exp(- kt)

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**t½ of a First Order Reaction - Example**

How much time is required for a 5.75 mg sample of 51Cr to decay to 1.50 mg if it has a half-life of 27.8 days? Radioactive decay is a first order reaction process. Since the reaction is first order, we know ln [A]t = - kt [A]o ln 1.50 = - kt ln = = - kt t = 1.344/k 5.75 We get k from the half-life: t½ = 27.8 days = 0.693/k k = / (27.8 days) k = days-1 t = / k = / = 53.9 days

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NOTES 14-3 obj 14.4. 1.) Using calculus to integrate the rate law for a first-order process gives us A.) INTEGRATED RATE LAWS ln [A] t [A] 0 = −kt Where.

NOTES 14-3 obj 14.4. 1.) Using calculus to integrate the rate law for a first-order process gives us A.) INTEGRATED RATE LAWS ln [A] t [A] 0 = −kt Where.

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