Barrendero

CHAPTER 14 CHEMICAL KINETICS
PRACTICE EXAMPLES
1A The rate of consumption for a reactant is expressed as the negative of the change in molarity divided by the time interval. The rate of reaction is expressed as the rate of consumption of a reactant or production of a product divided by its stoichiometric coefficient.   A    0.3187 M  0.3629 M  1min rate of consumption of A = =  = 8.93 105 M s 1 t 8.25 min 60 s rate of reaction = rate of consumption of A2 = 1B 8.93  105 M s 1  4.46  105 M s 1 2

We use the rate of reaction of A to determine the rate of formation of B, noting from the balanced equation that 3 moles of B form (+3 moles B) when 2 moles of A react (–2 moles A). (Recall that “M” means “moles per liter.”) 0.5522 M A  0.5684 M A 3moles B rate of Bformation=   1.62 104 M s 1 60s 2 moles A 2.50 min  1min (a)

2A

The 2400-s tangent line intersects the 1200-s vertical line at 0.75 M and reaches 0 M at 3500 s. The slope of that tangent line is thus 0 M  0.75 M slope = = 3.3  104 M s 1 =  instantaneous rate of reaction 3500 s  1200 s The instantaneous rate of reaction = 3.3  104 M s 1 .
At 2400 s,  H 2 O 2  = 0.39 M. At 2450s,  H 2 O 2  = 0.39 M + rate  t
At 2450 s,  H 2 O 2  = 0.39 M +  3.3  10 4 mol H 2 O 2 L1s 1  50s    = 0.39 M  0.017 M = 0.37 M

(b)

2B

We calculate the change in  H 2 O 2  and add it to  H 2 O 2 0 to determine  H 2 O 2 100 .
  H 2 O 2  = rate of reaction of H 2 O 2  t = 15.0  104 M s 1  100 s = 0.15 M

With only the data of Table 14.2 we can use onlythe reaction rate during the first 400 s,   H 2 O 2  /t = 15.0  104 M s 1 , and the initial concentration,  H 2 O 2 0 = 2.32 M.

 H 2O2 100 =  H 2O 2 0 +   H 2O2  = 2.32 M +  0.15 M  = 2.17 M
This value differs from the value of 2.15 M determined in text Example 14-2b because the text used the initial rate of reaction 17.1104 M s 1  , which is a bit faster than theaverage
rate over the first 400 seconds.

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Chapter 14: Chemical Kinetics

3A

We write the equation for each rate, divide them into each other, and solve for n.
R1 = k   N 2 O5 1 = 5.45  105 M s 1 = k  3.15 M     
n n
n

n n
n

R2 = k   N 2 O5  2 = 1.35  105 M s 1 = k  0.78 M     

n k   N 2 O 5 1 k  3.15 M  R1 5.45 105 M s 1 n    3.15    == 4.04 = = =  =  4.04  n n 5 1   R2 1.35  10 M s  0.78  k   N 2 O5  2 k  0.78 M   

We kept an extra significant figure (4) to emphasize that the value of n = 1. Thus, the reaction is first-order in N 2 O5 .
3B
1 2 For the reaction, we know that rate = k  HgCl2   C2 O 4  . Here we will compare Expt. 4 to   Expt. 1 to find the rate. 2

1 2 2      rate 4 k HgCl2  C 2 O 4  rate 4  = 0.025 M   0.045 M  = 0.0214 = = 2 1 2  2 rate1 k  HgCl   C O  1.8 105 M min 1 0.105 M   0.150 M   2  2 4   

2

The desired rate is rate4 = 0.0214  1.8  105 M min 1 = 3.9  107 M min 1 .
4A

We place the initial concentrations and the initial rates into the rate law and solve for k . 2 2 rate = k  A   B = 4.78  102 M s 1 = k 1.12M   0.87 M 

k=

4.78  102 M s 1

1.12 M 

2

0.87 M

= 4.4  102 M 2 s 1
2

4B

1 2 We know that rate = k  HgCl2  C2 O 4  and k = 7.6  103 M 2 min 1 .       Thus, insertion of the starting concentrations and the k value into the rate law yields: 1 2 Rate = 7.6 103 M 2 min 1  0.050 M   0.025 M  = 2.4 107 M min 1

5A

Here we substitutedirectly into the integrated rate law equation. ln A t =  kt + ln A 0 = 3.02  103 s1  325 s + ln 2.80 = 0.982 +1.030 = 0.048

b g

At= e

0.048

= 1.0 M

5B

This time we substitute the provided values into text Equation 14.13.   1.49 M 0.443  H 2O2  t   ln  =  kt = k  600 s = ln = 0.443 = 7.38 104 s 1 k=  2.32 M 600 s  H 2O2  0  
Now we choose  H 2 O2 ...