1. The initial reaction rate for 0.02 moles produced in a 5-L vessel in the 1st 2 seconds is
a. 0.2 M/s
b. 0.04 M/s ***
c. 0.002 M/s
d. 0.05 M/s
2. Use the table below to calculate the initial reaction rate of the following reaction:
2A→B+C
[A]/M | t/s
0.10 | 0
0.08 | 2
a. 0.005 M/s
b. -0.005 M/s ***?
c. 0.01 M/s
d. -0.01 M/s
3. Use the data table below to determine the value of the rate law constant for the first order reaction:
Experiment Initial [A]/M Initial rate/M s-1
1 0.010 0.012
2 0.020 0.022
3 0.040 0.040
a. 1.3 s-1
b. 1.1 s-1
c. 0.9 s-1 ***
d. 0.7 s-1
4. For the rate law, rate=k[A]2, determine k using the data table below:
Experiment Initial [A]/M Initial rate/M s-1
1 0.40 | 0.016
2 0.20 | 0.0040
3 0.10 | 0.0010
a. 0.000020 M-1s-1
b. 0.00020 M-1s-1
c. 0.20 M-1s-1 ***
d. 0.10 M-1s-1
How did you arrive at that answer for #1? Does that answer give you the correct units of M/s?
1. To calculate the initial reaction rate, we divide the change in moles by the change in time. The change in moles is 0.02 moles and the change in time is 2 seconds.
Initial reaction rate = (Change in moles) / (Change in time)
= 0.02 moles / 2 seconds
= 0.01 moles/second
Since the volume of the vessel is given as 5 L, we need to convert the moles/second to molarity/second.
Molarity = (moles of solute) / (volume of solution in liters)
= 0.01 moles / 5 L
= 0.002 M/s
Therefore, the correct answer is c. 0.002 M/s.
2. To calculate the initial reaction rate, we need to determine the rate of change in the concentration of A with respect to time.
The rate of change of concentration can be determined by taking the slope of the plot of [A] versus time. In this case, from the given data table, we have:
[A]/M | t/s
0.10 | 0
0.08 | 2
Using the formula for slope (change in y / change in x), we can calculate the initial reaction rate.
Change in [A] = (0.08 M - 0.10 M) = -0.02 M
Change in time = (2 s - 0 s) = 2 s
Initial reaction rate = (Change in [A]) / (Change in time)
= (-0.02 M) / (2 s)
= -0.01 M/s
Therefore, the correct answer is b. -0.01 M/s.
3. To determine the value of the rate law constant, we can use the data given in the table.
Experiment | Initial [A]/M | Initial rate/M s-1
1 | 0.010 | 0.012
2 | 0.020 | 0.022
3 | 0.040 | 0.040
Since the reaction is first order, the rate law can be written as follows:
rate = k[A]
For the first experiment, we have:
0.012 M/s = k * 0.010 M
Solving for k, we get:
k = (0.012 M/s) / (0.010 M)
= 1.2 s-1
Therefore, the correct answer is a. 1.2 s-1.
4. For the rate law, rate = k[A]^2, we need to determine the value of k using the data table given.
Experiment | Initial [A]/M | Initial rate/M s-1
1 | 0.40 | 0.016
2 | 0.20 | 0.0040
3 | 0.10 | 0.0010
Using the rate law expression, we can calculate k for each experiment:
For the first experiment:
0.016 M/s = k * (0.40 M)^2
Solving for k, we get:
k = (0.016 M/s) / (0.40 M)^2
≈ 0.001 M-1s-1
Similarly, for the second experiment:
0.0040 M/s = k * (0.20 M)^2
Solving for k, we get:
k = (0.0040 M/s) / (0.20 M)^2
≈ 0.010 M-1s-1
For the third experiment:
0.0010 M/s = k * (0.10 M)^2
Solving for k, we get:
k = (0.0010 M/s) / (0.10 M)^2
≈ 0.010 M-1s-1
Therefore, the correct answer is c. 0.010 M-1s-1.
1. To determine the initial reaction rate, we need to divide the moles produced by the time taken. In this case, the moles produced is 0.02 moles and the time taken is 2 seconds. We can use the formula: initial reaction rate = moles produced / time taken. Plugging in the values, we get: initial reaction rate = 0.02 moles / 2 seconds = 0.01 moles/second. However, we need to convert this to M/s (Molarity per second) since we are given a 5-L vessel. Molarity is defined as moles of solute divided by liters of solution. In this case, the solution is 5 L, so we divide the moles produced by the volume of the vessel to get the Molarity. So, the initial reaction rate is: 0.01 moles / 5 L = 0.002 M/s.
Therefore, the correct answer is c. 0.002 M/s.
2. To determine the initial reaction rate, we need to calculate the change in concentration of the reactant A over time. We can use the formula: initial reaction rate = (change in [A]) / (change in time). In this case, the change in [A] is 0.10 M - 0.08 M = 0.02 M, and the change in time is 2 seconds. Plugging in the values, we get: initial reaction rate = 0.02 M / 2 seconds = 0.01 M/s.
However, based on the given equation 2A -> B + C, we need to consider the stoichiometry of the reaction. For every 2 moles of A that react, we get 1 mole of B and 1 mole of C. Therefore, the rate of formation of B or C is half of the rate of disappearance of A. So, the correct answer is b. -0.005 M/s.
3. To determine the rate law constant for a first-order reaction, we need to use the data provided and the rate law equation. The rate law equation for a first-order reaction is: rate = k[A], where [A] is the concentration of the reactant A.
From the data table, we can see that for Experiment 1, the initial [A] is 0.010 M and the initial rate is 0.012 M/s. Using the rate law equation, we can write: 0.012 M/s = k * 0.010 M. Solving for k, we get: k = 0.012 M/s / 0.010 M = 1.2 s-1.
For Experiment 2, the initial [A] is 0.020 M and the initial rate is 0.022 M/s. Using the rate law equation, we can write: 0.022 M/s = k * 0.020 M. Solving for k, we get: k = 0.022 M/s / 0.020 M = 1.1 s-1.
For Experiment 3, the initial [A] is 0.040 M and the initial rate is 0.040 M/s. Using the rate law equation, we can write: 0.040 M/s = k * 0.040 M. Solving for k, we get: k = 0.040 M/s / 0.040 M = 1.0 s-1.
Since the values for k are consistent across the experiments, we can take the average of the three values to get the rate law constant: (1.2 s-1 + 1.1 s-1 + 1.0 s-1) / 3 = 1.1 s-1.
Therefore, the correct answer is b. 1.1 s-1.
4. To determine the rate constant (k) for a reaction with a rate law of rate=k[A]^2, we need to use the data provided and the rate law equation.
From the data table, we can see that for Experiment 1, the initial [A] is 0.40 M and the initial rate is 0.016 M/s. Using the rate law equation, we can write: 0.016 M/s = k * (0.40 M)^2. Solving for k, we get: k = 0.016 M/s / (0.40 M)^2 = 0.10 M-1s-1.
For Experiment 2, the initial [A] is 0.20 M and the initial rate is 0.0040 M/s. Using the rate law equation, we can write: 0.0040 M/s = k * (0.20 M)^2. Solving for k, we get: k = 0.0040 M/s / (0.20 M)^2 = 0.10 M-1s-1.
For Experiment 3, the initial [A] is 0.10 M and the initial rate is 0.0010 M/s. Using the rate law equation, we can write: 0.0010 M/s = k * (0.10 M)^2. Solving for k, we get: k = 0.0010 M/s / (0.10 M)^2 = 0.10 M-1s-1.
Since the values for k are consistent across the experiments, we can take the average of the three values to get the rate constant: (0.10 M-1s-1 + 0.10 M-1s-1 + 0.10 M-1s-1) / 3 = 0.10 M-1s-1.
Therefore, the correct answer is c. 0.10 M-1s-1.