Please do not block ads on this website.
No ads = no money for us = no free stuff for you!
Examples of Initial Concentration Method to Determine Rate Law
Question 1 : Determining the Rate Law of a Zero Order Reaction
Ammonia, NH_{3(g)}, can be decomposed on the surface of tungsten metal to produce nitrogen, N_{2(g)}, and hydrogen, H_{2(g)}, according to the equation:
2NH_{3(g)} → 3H_{2(g)} + N_{2(g)}
The results of a series of experiments carried out at the same temperature are shown below:
Trial 
Initial [NH_{3}] (M) 
Initial Rate (M/min) 
1 
2.50 × 10^{3} 
6.67 × 10^{7} 
2 
5.00 × 10^{3} 
6.67 × 10^{7} 
Determine the rate law for this reaction.
Solution:
 Determine the change in initial concentration:
The initial concentration of NH_{3} was increased by a factor of
[NH_{3(trial 2)}] [NH_{3(trial 1)}]

= 
5.00 × 10^{3} 2.50 × 10^{3}

= 2

 Determine the change in initial rate as a result of this change in concentration:
The initial rate of the reaction increased by a factor of
rate_{(trial 2)} rate_{(trial 1)}

= 
6.67 × 10^{7} 6.67 × 10^{7}

= 1 
That is, the rate does NOT change when the concentration of NH_{3} changes.
 Determine the relationship between the change in concentration and the change in rate:
Since increasing the concentration by a factor of 2 leads to no increase in the rate, the reaction rate is independent of the concentration of NH_{3}
rate = k
The reaction is zero order.
 Calculate the specific rate constant, k, for the reaction:
k = rate
Using data from Trial 1:
k = 6.67 × 10^{7}
 Write the rate law for this reaction:
rate = 6.67 × 10^{7}
Question 2: Determining the Rate Law of a First Order Reaction
Azomethane, C_{2}H_{6}N_{2}, decomposes to produce ethane, C_{2}H_{6}, and nitrogen, N_{2}.
C_{2}H_{6}N_{2(g)} → C_{2}H_{6(g)} + N_{2(g)}
The results of a series of experiments carried out at the same temperature are shown below:
Trial 
Initial [C_{2}H_{6}N_{2(g)}] (M) 
Initial Rate (M/min) 
1 
2.00 × 10^{2} 
3.00 × 10^{4} 
2 
4.00 × 10^{2} 
6.00 × 10^{4} 
Determine the rate law for this reaction.
Solution:
 Determine the change in initial concentration:
The initial concentration of C_{2}H_{6}N_{2(g)} was increased by a factor of
[C_{2}H_{6}N_{2(g)(trial 2)}] [C_{2}H_{6}N_{2(g)(trial 1)}]

= 
4.00 × 10^{2} 2.00 × 10^{2}

= 2 
 Determine the change in initial rate as a result of this change in concentration:
The initial rate of the reaction increased by a factor of
rate_{(trial 2)} rate_{(trial 1)}

= 
6.00 × 10^{4} 3.00 × 10^{4}

= 2 
 Determine the relationship between the change in concentration and the change in rate:
Since increasing the concentration by a factor of 2 leads to an increase in the rate by a factor of 2, the reaction rate and the concentration of C_{2}H_{6}N_{2(g)} are directly proportional:
rate ∝ [C_{2}H_{6}N_{2(g)}]
So, rate = k[C_{2}H_{6}N_{2(g)}]
The reaction is first order.
 Calculate the specific rate constant, k, for the reaction:
k = 
rate [C_{2}H_{6}N_{2(g)}]

Using data from Trial 1:
k = 
3.00 × 10^{4} 2.00 × 10^{2}

= 0.015 
 Write the rate law for this reaction:
rate = 0.015[C_{2}H_{6}N_{2(g)}]
Question 3: Determining the Rate Law of a Second Order Reaction
Hydrogen iodide decomposes to hydrogen and iodine according to the equation:
2HI → H_{2} + I_{2}
The results of a series of experiments carried out at the same temperature are shown below:
Trial 
Initial [HI] (M) 
Intial Rate (M/min) 
1 
1.50 × 10^{3} 
1.20 × 10^{5} 
2 
3.00 × 10^{3} 
4.8 × 10^{5} 
Determine the rate law for this reaction.
Solution:
 Determine the change in initial concentration:
The initial concentration of HI was increased by a factor of
[HI_{2(g)(trial 2)}] [HI_{2(g)(trial 1)}]

= 
3.00 × 10^{3} 1.50 × 10^{3}

= 2 
 Determine the change in initial rate as a result of this change in concentration:
The initial rate of the reaction increased by a factor of
rate_{(trial 2)} rate_{(trial 1)}

= 
4.8 × 10^{5} 1.20 × 10^{5}

= 4 
= 2^{2} 
 Determine the relationship between the change in concentration and the change in rate:
Since increasing the concentration of HI by a factor of 2 leads to an increase in the rate by a factor of 2^{2}, the rate is proportional to the square of the concentration of HI,
rate ∝ [HI]^{2}
So, rate = k[HI]^{2}
The reaction is second order.
 Calculate the specific rate constant, k, for the reaction:
k = rate/[HI]^{2}
Using data from Trial 1:
k = 
1.20 × 10^{5} [1.50 × 10^{3}]^{2}

= 
1.20 × 10^{5} 2.25 × 10^{6}

= 5.33 
 Write the rate law for this reaction:
rate = 5.33[HI]^{2}
Examples of Graphical Method for Determining Rate Law
Question 1: Determining the Rate Law of a First Order Reaction
The decomposition of gaseous dinitrogen pentoxide, N_{2}O_{5}, was investigated at a particular temperature.
The results of the experiment are shown below:
Time (min) 
0.00 
5.00 
10.00 
15.00 
20.00 
25.00 
[N_{2}O_{5}] (M) 
1.30 
1.08 
0.90 
0.75 
0.62 
0.52 
Show that the reaction is first order and determine the rate law for this reaction.
Solution:
 Calculate log_{10}[N_{2}O_{5}]
Time (min) 
0.00 
5.00 
10.00 
15.00 
20.00 
25.00 
[N_{2}O_{5}] (M) 
1.30 
1.08 
0.90 
0.75 
0.62 
0.52 
log_{10}[N_{2}O_{5}] 
0.11 
0.03 
0.05 
0.12 
0.21 
0.28 
 Plot log_{10}[N_{2}O_{5}] against time (shown below)
Straight line graph means the reaction is first order.
Rate is directly proportional to N_{2}O_{5} concentration.
rate ∝ [N_{2}O_{5}]
rate = k[N_{2}O_{5}]
 Calculate the slope of the line:
from the graph (shown in blue):
slope = 
0.16 10.00 
= 0.016 
mathematically:
slope = 
y_{1}  y_{2} x_{1}  x_{2} 
= 
0.11  (0.28) 0.00  25.00 
= 
0.39 25.00 
= 
0.016 
 Calculate the specific rate constant, k, for this reaction:
slope 
= 
k 2.303 
2.303 × slope 
= 
k 
2.303 × slope 
= 
k 
Substitute the value for the slope into the equation and solve to find the specific rate constant, k
k = slope × 2.303
= (0.016) × 2.303
= 0.037
 Write the rate law for this reaction:
rate = 0.037[N_{2}O_{5}]
Question 2: Determining the Rate Law of a Second Order Reaction
An investigation into the decomposition of NO_{2} at a particular temperature produced the following results:
Time (s) 
0.00 
5.00 
10.00 
15.00 
20.00 
[NO_{2}] (M) 
0.10 
0.017 
0.0089 
0.0062 
0.0047 
Use the data to verify that that reaction is second order and determine the rate law for the reaction.
Solution:
 Calculate the reciprocal concentrations of NO_{2}
Time (s) 
0.00 
5.00 
10.00 
15.00 
20.00 
[NO_{2}] (M) 
0.10 
0.017 
0.0089 
0.0062 
0.0047 
1/[NO_{2}] 
10 
59 
112 
161 
213 
 Plot 1/[NO_{2}] against time (shown below):
Straight line graph means the reaction is second order.
rate is proportional to the square of the concentration
rate ∝ [NO_{2}]^{2}
rate = k[NO_{2}]^{2}
 Calculate the slope of the line:
from the graph (shown in blue):
mathematically:
slope = 
(y_{1}  y_{2}) (x_{1}  x_{2})

= 
(10  213) (0  20)

= 
203 20 
= 
10.2 
 Calculate the specific rate constant, k, for this reaction:
slope = k = 10.2
 Write the rate law for this reaction:
rate = 10.2[NO_{2}]^{2}
Examples of HalfLife Method for Determining Rate Law
Question 1: Using a Graph to Determine HalfLife
The results of the decomposition of gaseous dinitrogen pentoxide, N_{2}O_{5}, at various times are tabulated below.
Time (min) 
0.00 
10 
20 
30 
40 
50 
60 
[N_{2}O_{5}] (M) 
348.8 
247 
185 
140 
105 
78 
58 
Use the data to confirm that this reaction is first order and find the specific rate constant at this temperature.
Solution:
 Plot [N_{2}O_{5}] against time as shown below:
 Use the graph to obtain halflife values:
[N_{2}O_{5}]_{0} (M) 
Graph Reference 
t_{½} (min) 
300 
a to b 
24 
250 
c to d 
24 
200 
e to f 
24 
150 
b to g 
24 
 Halflife is independent of [N_{2}O_{5}]_{0} so the reaction is first order.
That is, as [N_{2}O_{5}]_{0} changes, the halflife, t_{½} is constant
 Calculate the specific rate constant, k.
For a first order reaction:
t_{½} 
= 
0.693 k 
k 
= 
0.693 t_{½} 

= 
0.693 24 

= 
0.029 
Question 2: Using the Exponential Equation to Determine HalfLife
The decomposition of gaseous N_{2}O_{3} to form NO_{2} and NO is first order with k = 3.2 × 10^{4}s^{1}.
If the initial concentration of N_{2}O_{3} is 1.00 M, how long will it take for its concentration to be decreased to 0.125 M ?
Solution:
 The relationship between concentration and time is :
log_{10}[A]_{t} 
= 
kt 2.303 
+ log_{10}[A]_{0} 
 Extract the data from the question:
[N_{2}O_{3}]_{0} = 1.00 M
[N_{2}O_{3}]_{t} = 0.125 M
k = 3.2 × 10^{4}s^{1}
t = ?
 Substitute the values into the equation:
log_{10}[A]_{t} 
= 
kt 2.303 
+ log_{10}[A]_{0} 
log_{10}[0.125] 
= 
3.2 x 10^{4}t 2.303 
+ log_{10}[1.00] 
0.903 
= 
3.2 x 10^{4}t 2.303 
+ 0 
0.903 × 2.303 3.2 x 10^{4} 
= 
t 

6499 s 
= 
t 

Alternatively:
[N_{2}O_{3}]_{t} 
= 
[N_{2}O_{3}]_{0}e^{kt} 
ln( [N_{2}O_{3}]_{t} ) [N_{2}O_{3}]_{0} 
= 
kt 
ln([N_{2}O_{3}]_{t}/[N_{2}O_{3}]_{0} k 
= 
t 
ln(0.125/1.00) (3.2 × 10^{4}) 
= 
t 
6498 s 
= 
t 
Alternatively:
 Recognize that the time for 1.00 M to reduce to 0.125 M is equivalent to 3 halflives:
Time 
[N_{2}O_{3}] 
0 
1.00 
1 halflife 
0.500 M 
2 halflives 
0.250 M 
3 halflives 
0.125 M 
 For a first order reaction:
t_{½} 
= 
0.693 k 

= 
0.693 3.2 × 10^{4} 

= 
2166 s 
 3 halflives = 3 × 2166 = 6498 s
1. If you are looking for information about radioactive halflife, go to the Halflife Tutorial