Question

1. (a) Plot the concentrations listed below as a function of time. Then, plot any other graphs required to verify that the data is consistent with a first-order reaction. time concentration (in minutes) (in moles / liter) 0.00 1.0000 1.00 0.8607 2.00 0.7408 3.00 0.6376 4.00 0.5488 5.00 0.4724 6.00 0.4066 7.00 0.3499 8.00 0.3012 9.00 0.2592 10.00 0.2331 (b) Determine the value of the rate constant k. Write an expression for the rate law. (c) Determine the time at which the concentration of [A] = 0.500 mol/L. Determine the time at which [A] = 0.250 mol / L. Determine the time at which [A] = 0.125 mol/L. (d) The half-life for a reaction is the time required for the concentration at some time to decrease by a factor of one- half. The half-life for a first-order reaction is constant, and can be calculated from the value of k. The relationship is tu2 = (0.693 /k). Calculate the half-life from the value of k that you determined in (b) above. Can you see that this value is related to the times that you calculated in part (c) above? Explain.

Answer 1

Answer a)

1.2 time Concentration 0 1 1 1 0.8607 2 00 0.7408 0.6376 3 Concentration (moles/L) 4 0.5488 0.6 5 0.4 6 7 0.2 0.4724 0.4066 0.3499 0.3012 0.2592 0.2331 8 9 10 0 2 8 LO 12 6 Time (minutes)

As the plot of concentration v/s time shows that concentration is decreasing with time.

time Time (minutes) 0 0 1 2 4 6 8 10 12 -0.2 2 3 In(Concentration) 0 -0.150009267 -0.300024596 -0.450044152 -0.600021203 -0.749929195 -0.899925378 -1.05010788 -0.4 4 -0.6 5 6 In(Concentration) -0.8 7 -1 8 y = -0.148x-0.006 9 -1.199980783 -1.350155315 -1.456287733 -1.2 10 -1.4 -1.6

Further plot of natural logarithm of concentration v/s time is a straight line curve with negative slope which confirms that it is a first order reaction.

Answer b)

Slope of ln(concentration) v/s time graph is equal to rate constant.

Slope = change in y-axis/change in x-axis

From ln(concentration) at time = zero and time = 2 minutes,

change in ln(concentration) =- 0.3000 -(0.00) = -0.3000

change in time = 2 minutes

Slope = -0.3000/2 minutes

= -0.1500 minutes^-1

This is approximately equal to that from best fit line of the curve.

Negative sign indicates that concentration of reactants is decreasing with time

Rate law = k[A] i.e. rate = (0.1500 minutes^-1) [A]

Answer c)

Integrated rate law for first order reaction is ln[Ao] = ln[A] + kt

Here, [Ao] = initial concentration i.e. 1.00 mol/L

[A] = final concentration

k is rate constant = 0.1500 minutes^-1

t is time

Case 1: [A] = 0.500 mol/L

ln(1.00) = ln(0.500) + (0.1500 minutes^-1)t

0.00 = -0.693 + 0.1500t

t = 4.62 minutes

Case 2: [A] = 0.250 mol/L

ln(1.00) = ln(0.250) + (0.1500 minutes^-1)t

0.00 = -1.386 + 0.1500t

t = 9.24 minutes

Case 3: [A] = 0.125 mol/L

ln(1.00) = ln(0.125) + (0.1500 minutes^-1)t

0.00 = -2.080 + 0.1500t

t = 13.86 minutes

Answer d)

Half life = 0.693/k

= 0.693/0.1500 minutes^-1

= 4.62 minutes

Yes, the values are related to times calculated in c)

It is so because, when concentration drops to 0.500 M, it is halved and takes one half-life.

when concentration drops to 0.250 M, it is one-fourth and takes two half-lives.

when concentration drops to 0.125 M, it is one-eighth and takes four half-lives.