Chapter 1 - Solutions



CHM 1046. Chapter 13 Homework Solutions.

Problems: 6, 8, 26, 30, 32, 38, 42, 46, 49, 50, 55, 60, 64, 68, 72, 74, 94

6) Zeroth order rate = k

If you double the concentration of A the rate of reaction is unchanged.

First order rate = k [A]

If you double the concentration of A the rate of reaction also doubles.

Second order rate = k [A]2

If you double the concentration of A the rate of reaction goes up by a factor of four.

8) The overall order of the reaction is equal to the sum of the individual reaction orders.

For example, consider a reaction that depends on three reactant concentrations.

Rate = k [A]p [B]q [C]r .

Then p, q, and r are the reaction orders with respect to A, B, and C, respectively. The overall order of reaction is equal to p + q + r.

26) a) rate = - ([N2O] ; rate = ([N2] ; rate = ( [O2]

(t (t (t

b) The concentration of O2 produced is

( [O2] = 0.018 mol = 0.072 mol/L

0.250 L

average rate = 0.072 mol/L = 0.0072 mol/L.s

10 s

c) For every mole of O2 that appears, 2 moles of N2O disappear.

So - ( [N2O] = 2 (0.0072 mol/L.s) = 0.0144 mol/L.s

(t

30) a) Between 10 and 20 seconds

average rate = - ( [NO2] = - ( 0.904 - 0.951) mol/L = 0.0047 mol/L.s

(t (20 - 10) s

Between 50 and 60 seconds

average rate = - ( [NO2] = - ( 0.740 - 0.778) mol/L = 0.0038 mol/L.s

(t (20 - 10) s

b) For every mole of NO2 that disappears, 1/2 mole of O2 appears. So, between 50 and 60 seconds

( [O2] = 1/2 (0.0038 mol/L.s) = 0.0019 mol/L.s

(t

32)

a) Data can be read from the figure above.

t = 10 s [H2O2] = 0.76 mol/L

t = 20 s [H2O2] = 0.56 mol/L

average rate = - (0.56 - 0.76) mol/L = 0.020 mol/L.s

(20 - 10) s

b) The instantaneous rate at t = 30 s is equal to the absolute value of the slope of the tangent line in the above figure. Pick two points on the tangent line to find the slope. I picked the following two points: (20 s, 0.51 M) and (40 s, 0.29 M).

instantaneous rate = |slope| = 0.22 mol/L = 0.011 mol/L.s

20 s

c) The instantaneous rate for the disappearance of H2O2 at t = 50 s is equal to the absolute value of the slope of the tangent line (to avoid confusion the tangent line is not shown).

instantaneous rate of disappearance of H2O2 = 0.14 mol/L = 0.007 mol/L.s

20 s

For every 2 mole of H2O2 that disappears 1 mole of O2 appears, so

instantaneous rate of formation of O2 = 1/2 (0.007 mole/L.s) = 0.0035 mol/L.s

d) Data can be read from the figure above

t = 0 s [H2O2] = 1.00 mol/L

t = 50 s [H2O2] = 0.22 mol/L

The number of moles of H2O2 that have disappeared is

# moles H2O2 = 1.5 L (1.00 - 0.22) mol/L = 1.17 mol

For every 2 moles of H2O2 that disappears 1 mole of O2 appears, so

# moles O2 formed = 1/2 (1.17 mol) = 0.585 mol

38) a) rate = k [B]1/2 [C]2

b) overall order = 1/2 + 2 = 5/2

c) If the concentration of A is doubled there is no change in the rate of reaction.

d) If the concentration of B is doubled the rate of reaction increases by 21/2 ( 1.41

e) If the concentration of C is doubled the rate of reaction increases by 22 = 4.

f) If the concentrations of all three reactants are doubled the rate of reaction increases by 21/2 22 = 25/2 = 5.66

42) This is a method of initial rates problem. The easiest (though not the only) way to use the method is to compare experiments where one initial concentration is changed and other initial concentrations are held constant. Note that we will not use units in the numbers in the calculation when we have ratios, where units automatically cancel.

Let rate = k [CH3Cl]p [Cl2]q p, q are the reaction orders

Experiment 1 and 2

rate 2 = k [CH3Cl]2p [Cl2]2q = [CH3Cl]2p [Cl2]2q

rate 1 = k [CH3Cl]1p [Cl2]1q [CH3Cl]1p [Cl2]1q

0.029 = (0.100)p (0.050)q

0.014 (0.050)p (0.050)q

2.07 = (2.00)p therefore p = 1

Experiment 2 and 3

rate 3 = k [CH3Cl]3p [Cl2]3q = [CH3Cl]3p [Cl2]3q

rate 2 = k [CH3Cl]2p [Cl2]2q [CH3Cl]2p [Cl2]2q

0.041 = (0.100)p (0.100)q

0.029 (0.100)p (0.050)q

1.41 = (2.00)q therefore q = 1/2

So the rate law is rate = k [CH3Cl] [Cl2]1/2

k = (rate)

[CH3Cl] [Cl2]1/2

We can use any of the experiments to find k (in a real series of experiments we would calculate k for each experiment and then take an average).

k = (0.014 mol/L.s) = 1.25 L1/2/mol1/2.s

(0.050 mol/L) (0.050 mol/L)1/2

46) The general procedure used in these type of problems is as follows

1) plot ln (concentration) vs time. A straight line means a 1st order reaction.

2) plot 1/(concentration) vs time. A straight line means a 2nd order reaction.

3) plot concentration vs time. A straight line means a 0th order reaction.

If none of these gives a straight line, then the reaction order is different than 1st, 2nd, or 0th order. (Note that there are more general methods that can be used to determine the experimental order, but they have not been discussed in the book and so will not be discussed here).

time [N2O5] ln [N2O5] 1/[N2O5]

0. 1.000 0.000 1.000

25. 0.822 - 0.196 1.217

50. 0.677 - 0.390 1.477

75. 0.557 - 0.585 1.795

100. 0.458 - 0.781 2.183

125. 0.377 - 0.976 2.653

150. 0.310 - 1.171 3.226

175. 0.255 - 1.366 3.922

200. 0.210 - 1.561 4.762

I will give all of the plots discussed above, though usually we would plot them in order, as suggested above, and stop if a linear plot was found.

Note that of the three plots the only one that gives a linear relationship is the plot of ln [N2O5] vs time. Therefore the reaction is first order. So k = - (slope)

Based on fitting the data in the plot of ln [N2O5] vs time, I get

slope - 0.0078 s-1

So k = - (slope) = 0.0078 s-1

At t = 250. s,

ln [N2O5] = (- 0.0078 s-1) (250. s) = - 1.95

[N2O5] = e-1.95 = 0.142 mol/L

49) a) The plot of ln [A] vs time gives a straight line. Therefore, the reaction is first order, and so k = - (slope) = 0.0045 s-1.

b) rate = k [A] also [A]t = [A]0 e-kt for concentration vs time

c) For a first order reaction t1/2 = ln(2)/k.

So t1/2 = ln(2)/0.0045 s-1 = 154. s.

d) [A] = (0.250 mol/L) exp[ - (0.0045 s-1) (225. s) ]

= (0.250 mol/L) e-1.012 = 0.091 mol/L

50) a) The plot of 1/[A] vs time gives a straight line. Therefore, the reaction is second order, and so k = 0.055 L/mol.s.

b) rate = k [A]2 also [A]t = [A]0

1 + kt [A]0

c) For a second order reaction t1/2 = 1/k[A]0.

So t1/2 = 1 . = 33.1 s

(0.055 L/mol.s) (0.55 mol/L)

d) [AB] = (0.250 mol/L) = 0.123 mol/L

1 + (0.055 L/mol.s) (75 s) (0.250 mol/L)

The concentration of AB that has disappeared is (0.250 - 0.123) mol/L = 0.127 mol/L.

The initial concentrations of A and B are zero. Since for every AB that disappears an A and a B are produced, then it follows that [A] = [B] = 0.127 mol/L.

55)

60) To find A and Ea (pre-exponential factor and activation energy) we plot ln k vs 1/T. The data for the plot are given below.

T (K) k (s-1) 1/T (K-1) ln k

300. 0.0134 0.003333 - 4.313

310. 0.0407 0.003226 - 3.202

320. 0.114 0.003125 - 2.172

330. 0.303 0.003030 - 1.194

340. 0.757 0.002941 - 0.278

The best fitting line for the above data is

ln k = ( - 10280 K)/T + 29.97

Ea = - R (slope) = - (8.314 x 10-3 kJ/mol.K) ( - 10280. K) = 85.5 kJ/mol

ln A = 29.97 , so A = e29.97 = 1.0 x 1013 s-1

64) a) The form of the Arrhenius equation we need to use is

ln(k2/k1) = - (Ea/R) [ (1/T2) - (1/T1) ] k1 = 0.000122 s-1 T1 = 300. K

k2 = 0.228 s-1 T2 = 350. K

So Ea = - R ln(k2/k1) = - (8.314 x 10-3 kJ/mol.K) ln (0.228/0.000122)

[ (1/T2) - (1/T1) ] [ (1/350. K) - (1/300. K) ]

= 131.5 kJ/mol

b) We can use the same equation to find the value for k at 17. (C = 290. K. We now know Ea, so

k1 = 0.000122 s-1 T1 = 300. K

k2 = ? T2 = 290. K

ln(k2/k1) = - 131500. J/mol [ (1/290. K) - (1/300. K) ] = - 1.818

(8.314 J/mol.K)

Taking the inverse logarithm of both sides gives

k2/k1 = e-1.818 = 0.162

k2 = k1 (0.162) = (0.000122 s-1) (0.162) = 2.0 x 10-5 s-1

68) Reaction b. No matter where the oxygen atom strikes the N2 molecule in a it will be near an N atom it can form a bond with. Reaction b will likely occur only if the NO molecule is oriented so the O side strikes the Cl2 molecule, and not if the N strikes the Cl2 molecule.

72) a) Overall reaction can be found by adding the individual elementary reactions.

2 NO2(g) + Cl2(g) ( 2 ClNO2(g)

b) The Cl(g) atom is a reaction intermediate, as it does not appear as either a reactant or a product.

c) Overall rate = rate of slow step, so

rate = k1 [NO2] [Cl2] First order in NO2, first order in Cl2, second order overall.

74)

94) Let us first find the value for k, the rate constant for radioactive decay.

t1/2 = ln(2)/k , so k = ln(2)/t1/2 = ln(2)/(4.5 x 109 yr) = 1.54 x 10-10 yr-1

Now

[U]t = [U]0 e-kt Divide by [U]0

( [U]t/[U]0 ) = e-kt Take the inverse logarithm

ln ( [U]t/[U]0 ) = - kt Divide by kk

t = - (1/k) ln ( [U]t/[U]0 ) = - (1/1.54 x 10-10 yr-1) ln (0.832) = 1.19 x 109 yr

= 1.19 billion years

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