i need help with a rate law question:
The reaction is
2ClO2(aq) + 2OH -(aq)---> ClO3-(aq)+ ClO2-(aq) + H2O(l)
Experiment 1:
[ClO2]0= 0.0500
[OH-]0= 0.100
INITIAL RATE= 5.75*10^-2
Experiment 2:
[ClO2]0= 0.100
[OH-]0= 0.100
INITIAL RATE= 2.30*10^-1
Experiment 3:
[ClO2]0= 0.100
[OH-]0= 0.050
INITIAL RATE= 1.15*10^-1
a) determine the rate law and the value of the rate constant.
b) what would be the initial rate for an experiment with [ClO2]0= 0.175 mol/L and [OH-]0= 0.0844 mol/L?
Thanks for your answer.
I don't get that.
I have an order of 3 and k = 230
rate = k(ClO2^-)x(OH^-)y.
For x I have 2. For y I have 1. The order overall is 3. Check my work.
WHen i did it the answers i got was order of reaction m= 2 and k = 4.6
Can someone please double check
To determine the rate law and the value of the rate constant, we can use the method of initial rates. By comparing the initial rates of different experiments while varying the initial concentrations of reactants, we can determine the order of the reaction with respect to each reactant and overall.
a) Determine the rate law and the value of the rate constant:
Let's start by comparing the initial rates of the three experiments:
Experiment 1: [ClO2]0= 0.0500, [OH-]0= 0.100
Initial rate = 5.75*10^-2
Experiment 2: [ClO2]0= 0.100, [OH-]0= 0.100
Initial rate = 2.30*10^-1
Experiment 3: [ClO2]0= 0.100, [OH-]0= 0.050
Initial rate = 1.15*10^-1
Since we are given the initial concentrations of [ClO2] and [OH-], we can assume that the rate law is of the form:
Rate = k[ClO2]^a[OH-]^b
Now, let's compare the initial rates of Experiments 1 and 2 (where [OH-] is constant):
(Initial rate of Experiment 2) / (Initial rate of Experiment 1) = (0.23) / (0.0575)
= 4
This suggests that the reaction rate quadruples when [ClO2] doubles. This indicates that the reaction is first order with respect to [ClO2].
Now, let's compare the initial rates of Experiments 2 and 3 (where [ClO2] is constant):
(Initial rate of Experiment 2) / (Initial rate of Experiment 3) = (0.23) / (0.115)
= 2
This suggests that the reaction rate doubles when [OH-] doubles. This indicates that the reaction is first order with respect to [OH-].
Overall, the rate law for the reaction is:
Rate = k[ClO2][OH-]
Now, to find the value of the rate constant (k), we can use any of the given experiments. Let's use Experiment 2:
[ClO2]0= 0.100, [OH-]0= 0.100
Initial rate = 2.30*10^-1
Plugging these values into the rate law equation, we get:
2.30*10^-1 = k * (0.100) * (0.100)
Solving for k:
k = (2.30*10^-1) / (0.100 * 0.100)
k = 2.30
So, the value of the rate constant (k) is 2.30.
b) To find the initial rate for an experiment with [ClO2]0= 0.175 mol/L and [OH-]0= 0.0844 mol/L, we can use the rate law equation:
Rate = k[ClO2][OH-]
Plugging in the provided values, we get:
Rate = (2.30) * (0.175) * (0.0844)
Calculating the value:
Rate = 0.0346
Therefore, the initial rate for this experiment is 0.0346 mol/L•s.
To determine the rate law and rate constant for a given reaction, we need to use the initial rate data from several experiments. The rate law describes the relationship between the concentrations of reactants and the rate of reaction.
Let's start by comparing the initial rates of Experiments 1, 2, and 3. Remember that the rate is directly proportional to the concentrations of the reactants raised to some powers.
Experiment 1:
[ClO2]0 = 0.0500 mol/L
[OH-]0 = 0.100 mol/L
Initial Rate = 5.75 * 10^-2 mol/L·s
Experiment 2:
[ClO2]0 = 0.100 mol/L
[OH-]0 = 0.100 mol/L
Initial Rate = 2.30 * 10^-1 mol/L·s
Experiment 3:
[ClO2]0 = 0.100 mol/L
[OH-]0 = 0.050 mol/L
Initial Rate = 1.15 * 10^-1 mol/L·s
By comparing Experiments 1 and 2, we can see that the concentration of ClO2 doubles while all other factors remain the same, and the rate also doubles. This suggests a direct relationship between the concentration of ClO2 and the rate of reaction. Therefore, the rate law includes [ClO2] to the power of 1.
By comparing Experiments 1 and 3, we can see that the concentration of OH- doubles while all other factors remain the same, and the rate remains the same. This indicates that the concentration of OH- does not affect the rate of reaction. Therefore, the rate law does not include [OH-].
Based on these observations, we can write the rate law as:
Rate = k * [ClO2]^1
Now let's determine the value of the rate constant (k) using any of the given experiments. We'll use Experiment 1:
[ClO2]0 = 0.0500 mol/L
Initial Rate = 5.75 * 10^-2 mol/L·s
Using the rate law equation, we have:
5.75 * 10^-2 mol/L·s = k * (0.0500 mol/L)^1
Simplifying the equation:
5.75 * 10^-2 = k * 0.0500
k = (5.75 * 10^-2) / 0.0500 ≈ 1.15 s^-1
Therefore, the rate constant (k) is approximately 1.15 s^-1.
Now, let's move on to part b) where we need to calculate the initial rate for an experiment with [ClO2]0 = 0.175 mol/L and [OH-]0 = 0.0844 mol/L.
Using the rate law equation:
Rate = k * [ClO2]^1
Plugging in the values:
Rate = (1.15 s^-1) * (0.175 mol/L)^1
Rate = 0.20125 mol/L·s
Therefore, the initial rate for this experiment would be approximately 0.20125 mol/L·s.