The decompostion of dinitrogen pentoxide has been studied in carbon tetrachloride (CCl4) at a certain temperature: 2N2O5→4NO2+O2
[N2O5] Inital rate (M/s)
0.92 ...... 0.95x10^-5
1.23 ...... 1.20x10^-5
1.79 ...... 1.93x10^-5
2.00 ...... 2.10x10^-5
2.21 ...... 2.26x10^-5
Determine graphically the rate law for the reaction and calculate the rate constant.
I am so lost with this problem. If you could also explain how you get your answer that would be great because I don't even know where to start. Thanks!
We can't draw diagrams/graphs on this forum.
Plot the data, as the problem asks, zero, first, second order reactions look different. Here are some images.
https://www.google.com/search?q=graphically+order+reactions&client=firefox-a&hs=LiV&rls=org.mozilla:en-US:official&channel=sb&tbm=isch&tbo=u&source=univ&sa=X&ei=2aypU7H0G7S2sASSwIHYCw&ved=0CDsQsAQ&biw=1143&bih=688&dpr=0.9
To determine the rate law for the reaction and the rate constant, we will start by using the method of initial rates. In this method, we compare the initial rates of the reaction at different concentrations to determine the order of reaction with respect to each reactant.
Step 1: Find the order of reaction with respect to N2O5.
We can do this by comparing the initial rates at different concentrations of N2O5 while keeping the concentrations of other reactants constant.
- Comparing the first and second sets of data points, we can see that when [N2O5] doubles from 0.92 M to 1.23 M, the rate approximately doubles from 0.95x10^-5 to 1.20x10^-5.
This suggests that the rate is directly proportional to the concentration of N2O5.
So, the order of reaction with respect to N2O5 is 1.
Step 2: Find the order of reaction with respect to NO2.
To do this, we need to compare the initial rates at different concentrations of NO2 while keeping the concentration of N2O5 constant.
- Comparing the second and third sets of data points, we can see that when [N2O5] remains constant at 1.23 M, the rate increases from 1.20x10^-5 to 1.93x10^-5 as [NO2] increases from 1.23 M to 1.79 M.
This suggests that the rate is directly proportional to the concentration of NO2.
So, the order of reaction with respect to NO2 is also 1.
Step 3: Write the rate law expression.
Based on the above results, we can write the rate law expression as follows:
Rate = k[N2O5]^a[NO2]^b
Since the order of reaction is 1 with respect to both N2O5 and NO2, we have:
Rate = k[N2O5]^1[NO2]^1 = k[N2O5][NO2]
Step 4: Calculate the rate constant.
We can use any set of data points to calculate the rate constant (k) once we know the rate and the concentrations of N2O5 and NO2.
Let's use the first set of data points: [N2O5] = 0.92 M and Rate = 0.95x10^-5 M/s.
0.95x10^-5 M/s = k(0.92 M)(0.92 M)
0.95x10^-5 M/s = k(0.92 M)^2
Solving for k:
k = (0.95x10^-5 M/s) / (0.92 M)^2
Using a calculator, we find k ≈ 1.2x10^-5 M^-1s^-1.
Therefore, the rate law for the reaction is Rate = k[N2O5][NO2], and the rate constant (k) is approximately 1.2x10^-5 M^-1s^-1.
To determine the rate law for the given reaction, we need to analyze the relationship between the initial concentration of N2O5 and the initial rate of the reaction (denoted by [N2O5] Initial rate). The rate law expression can be written as follows:
Rate = k[A]^m[B]^n
Where:
- Rate is the initial rate of the reaction
- k is the rate constant
- [A] and [B] are the concentrations of the reactants
- m and n are the reaction orders with respect to A and B, respectively.
In this case, we only have [N2O5] Initial rate and [N2O5], but we can assume that the reaction is only dependent on the concentration of N2O5. Therefore, we can simplify the rate law expression to:
Rate = k[N2O5]^m
To determine the reaction order (m) and the rate constant (k), we need to linearize the rate law expression. Taking the natural logarithm (ln) of both sides of the equation would help us achieve linearity:
ln(Rate) = ln(k) + m ln([N2O5])
By analyzing the given data, we can plot a graph of ln(Rate) versus ln([N2O5]) and determine the slope of the best-fit line. The slope of the line corresponds to the reaction order (m), and the y-intercept corresponds to ln(k).
Now let's calculate the values:
[N2O5] (M) [N2O5] Initial rate (M/s)
0.92 0.95x10^-5
1.23 1.20x10^-5
1.79 1.93x10^-5
2.00 2.10x10^-5
2.21 2.26x10^-5
Step 1: Calculate ln(Rate) for each data point.
ln(Rate) = ln([N2O5] Initial rate)
[N2O5] (M) [N2O5] Initial rate (M/s) ln(Rate)
0.92 0.95x10^-5 -11.40775
1.23 1.20x10^-5 -11.62867
1.79 1.93x10^-5 -11.57182
2.00 2.10x10^-5 -11.56076
2.21 2.26x10^-5 -11.52642
Step 2: Calculate ln([N2O5]) for each data point.
ln([N2O5]) = ln([N2O5])
[N2O5] (M) [N2O5] Initial rate (M/s) ln(Rate) ln([N2O5])
0.92 0.95x10^-5 -11.40775 -0.08329
1.23 1.20x10^-5 -11.62867 0.20412
1.79 1.93x10^-5 -11.57182 0.58320
2.00 2.10x10^-5 -11.56076 0.69315
2.21 2.26x10^-5 -11.52642 0.79299
Step 3: Plot ln(Rate) versus ln([N2O5]) and find the best-fit line.
By plotting the ln(Rate) values on the y-axis and ln([N2O5]) values on the x-axis, we obtain a scatter plot. Fit a straight line through the data points and determine the slope of the line.
Step 4: Determine the reaction order (m) and the rate constant (k).
The slope of the best-fit line corresponds to the reaction order (m), and the y-intercept corresponds to ln(k). Calculate these values from the graph.
Using the slope-intercept form (y = mx + b) of a straight line and substituting the corresponding values:
Slope (m) = (ln(Rate)₂ - ln(Rate)₁) / (ln([N2O5])₂ - ln([N2O5])₁)
Using any two points from the graph:
Slope (m) = (-11.62867 - (-11.40775)) / (0.20412 - (-0.08329))
Similarly, determine the value of ln(k) from the graph by noting the y-intercept.
Finally, calculate the rate constant (k) by taking the antilog of the ln(k) obtained.
By following these steps and analyzing the given data, you should be able to determine the rate law for the reaction and calculate the rate constant.