A chemist does a reaction rate analysis on the following reaction:
2CO (g) + O2 (g) → 2CO2 (g)
She collects the following data:
Trial Initial Concentration of
CO (M) Initial Concentration
of O2 (M) Instantaneous
Reaction Rate (M/s)
1 0.150 0.150 0.113
2 0.300 0.150 0.226
3 0.300 0.300 0.904
What is the rate equation for this reaction?
The rate equation is rate = k(CO)^x(O2)^y.
Divide rate 2 by rate 1.
0.226/0.113 = 2 and
2 = (0.300)x(0.150)y/(0.150)x(0.150)y
Note that 0.150y cancels with 0.150y and leaves just
2= (2)x and x must be 1.
Do the same thing for rate 3 divided by rate 2 and determine y. You should get 2. Then substitute x = 1 and y = 2 into any of the three rate equations to determine k, then write the equation. If you get stuck, post your work in a new post at the top of the page (this post is close to the bottom when I read it) and I can help you through it.
To determine the rate equation for a reaction, we need to analyze the effect of the initial concentrations on the instantaneous reaction rate.
Let's compare Trials 1 and 2, where the concentration of CO is different but the concentration of O2 is the same. In Trial 2, the concentration of CO doubles, and the reaction rate also doubles. This indicates that the rate is directly proportional to the concentration of CO, and we can represent this relationship as follows:
Rate = k[CO]^a
Now, let's compare Trials 2 and 3, where the concentration of O2 is different but the concentrations of CO are the same. In Trial 3, the concentration of O2 doubles, and the reaction rate quadruples. This suggests that the rate is directly proportional to the concentration of O2 to the power of 2, and we can represent this relationship as follows:
Rate = k[O2]^b
Based on the given data, we can determine the values of a and b:
For Trials 1 and 2, [CO] doubles and the rate doubles:
0.113 M/s = k(0.150 M)^a
0.226 M/s = k(0.300 M)^a
Dividing the second equation by the first equation, we get:
2 = (0.300 M/0.150 M)^a
2 = 2^a
Therefore, a = 1.
For Trials 2 and 3, [O2] doubles and the rate quadruples:
0.226 M/s = k(0.150 M)^1
0.904 M/s = k(0.150 M)^1
Dividing the second equation by the first equation, we get:
4 = (0.300 M/0.150 M)^b
4 = 2^b
Therefore, b = 2.
Thus, we can write the rate equation for this reaction as:
Rate = k[CO]^1[O2]^2
Simplifying, we have:
Rate = k[CO][O2]^2
Therefore, the rate equation for this reaction is Rate = k[CO][O2]^2.
To determine the rate equation for the given reaction, we need to analyze the effect of the initial concentrations of CO and O2 on the instantaneous reaction rate.
The rate equation is typically expressed as:
rate = k [CO]^m [O2]^n
In this equation, k is the rate constant, [CO] and [O2] are the concentrations of CO and O2 respectively, and m and n represent the orders of the reaction with respect to CO and O2.
To determine the orders of the reaction with respect to CO and O2, we can compare the rate of reaction under different conditions.
Let's consider the first and second trials:
For Trial 1, [CO] = 0.150 M, [O2] = 0.150 M, and rate = 0.113 M/s.
For Trial 2, [CO] = 0.300 M, [O2] = 0.150 M, and rate = 0.226 M/s.
Since the concentration of CO doubled while the concentration of O2 remained the same, we can conclude that the rate is directly proportional to the concentration of CO. Therefore, the order of the reaction with respect to CO is 1.
Next, let's compare the second and third trials:
For Trial 2, [CO] = 0.300 M, [O2] = 0.150 M, and rate = 0.226 M/s.
For Trial 3, [CO] = 0.300 M, [O2] = 0.300 M, and rate = 0.904 M/s.
In this case, the concentration of O2 doubled while the concentration of CO remained the same. As the rate increased by a factor of 4, we can deduce that the rate is directly proportional to the square of the concentration of O2. Therefore, the order of the reaction with respect to O2 is 2.
Putting it all together, the rate equation for the given reaction is:
rate = k [CO]^1 [O2]^2
Simplifying:
rate = k [CO] [O2]^2
Therefore, the rate equation for this reaction is rate = k [CO] [O2]^2.