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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 concentrations of reactant and products at a time t . Process: A  products Rate = k[A]. Using the form for the instantaneous rate

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slide1

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

slide2

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 (2.71828) must be raised to get A.

slide3

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.

slide4

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.

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.

slide5

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.

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.

slide6

Order of a Reaction - Example

Now we can determine the first order rate constant:

ln[A]t = - kt + ln[A]o

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.

slide7

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.

slide8

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

slide9

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

[A]o

0.693 = kt½

t ½ = 0.693/k

ln ½ = - kt½

- ln 2 = - kt½

ln 2 = kt½

slide10

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 1.000 atm to 0.500 atm. How long does it take for the pressure to drop from 0.050 atm to 0.025 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

slide11

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)

slide12

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 0.261 = -1.344 = - kt t = 1.344/k

5.75

We get k from the half-life: t½ = 27.8 days = 0.693/k

k = 0.693 / (27.8 days) k = 0.02493 days-1

t = 1.344 / k = 1.344 / 0.02493 = 53.9 days