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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.

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Presentation on theme: "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."— Presentation transcript:

1 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  productsRate = 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

2 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.

3 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.

4 Order of a Reaction - Example Analyze rate data for the reaction SO 2 Cl 2 (g)  SO 2 (g) + Cl 2 (g) at 320°C to prove the reaction is first order and to find the rate constant. time (s)pressure of SO 2 Cl 2 (atm) A plot of P vs. time is not a straight line. I used the “add trendline” feature of Excel. If the rate = constant, then a plot of P vs. t will be a straight line.

5 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 SO 2 Cl 2 (atm) ln P A plot of ln P vs. time is a straight line, so the reaction is first order in SO 2 Cl 2. Why is it okay to use pressure instead of molarity when analyzing the data?

6 Order of a Reaction - Example ln[A] t = - kt + ln[A] o Now we can determine the first order rate constant: time (s) ln Pk (s -1 ) x x x x average k is k = 2.20 x s -1 at T = 320°C. ln P = - kt + lnP o since P o = 1, ln P o = 0 ln P = - kt

7 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.

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

9 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 = - kt ½ [A] o ln ½ = - kt ½ - ln 2 = - kt ½ ln 2 = kt ½ t ½ = 0.693/k = kt ½

10 t ½ of a First Order Reaction - Example We have already determined that the reaction is first order and that k = 2.20 x s -1 at 320°C. t ½ = 0.693/k = 0.693/(2.20 x s -1 ) = 3.15 x 10 4 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: P 2 = M 2 RT = M 2 P 1 M 1 RT M 1 SO 2 Cl 2 (g)  SO 2 (g) + Cl 2 (g) T = 320°C Determine the time required for the pressure of SO 2 Cl 2 to drop from atm to atm. How long does it take for the pressure to drop from atm to atm?

11 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)

12 t ½ of a First Order Reaction - Example Since the reaction is first order, we know ln [A] t = - kt [A] o ln 1.50 = - ktln = = - kt t = 1.344/k 5.75 We get k from the half-life: t ½ = 27.8 days = 0.693/k k = / (27.8 days)k = days -1 t = / k = / = 53.9 days How much time is required for a 5.75 mg sample of 51 Cr to decay to 1.50 mg if it has a half-life of 27.8 days? Radioactive decay is a first order reaction process.


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