2CIO2 (aq) + 2OH (aq) <---> CIO3 (aq) + CIO2 (aq) + H2O (l)
[CIO2]0 [OH]0 initial rate
(mol/L) (mol/L) (mol/L * s)
0.0500 0.100 5.75 * 10^-2
0.100 0.100 2.30 * 10^-1
0.100 0.050 1.15 * 10^-1
determine the rate law and the value of the rate constant. what would be the initial rate for an experiment with [CIO2]0 = 0.175 mol/L and [OH]0 = 0.0844 mol/l?
how can I approach the problem?
[CIO2]0 [OH]0 initial rate
(mol/L) (mol/L) (mol/L * s)
0.0500 0.100 5.75 * 10^-2
0.100 0.100 2.30 * 10^-1
0.100 0.050 1.15 * 10^-1
determine the rate law and the value of the rate constant. what would be the initial rate for an experiment with [CIO2]0 = 0.175 mol/L and [OH]0 = 0.0844 mol/l?
how can I approach the problem?
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I hate rate laws lol. They're not hard, but they're hard to explain. Try to stay with me.
Look at Trial 2 and Trial 3. From this we see that when the [OH] is cut in half so is the rate.
(1/2^n) = 1/2
n = 1
Therefore this reaction must be First order with respect to [OH]
Now look at Trial 1 and Trial 2. From this we see that when [ClO2] is cut in half the rate is only 1/4 of what it was.
(1/2^n) = 1/4
n=2
From this we can conclude that the reaction is Second order with respect to [ClO2]
So our rate law looks like this;
Rate = k[OH][ClO2]^2
Knowing that our Rate is in M/sec we know that the units of our rate constant, k, must be (M^2 *sec)^-1
To find the value of the rate law we just need to plug in the concentrations of OH and ClO2 from one of our trials into our rate law. Lets do Trial 2
2.30*10^-1 (M/s) = k* (0.100M) *(0.100M)^2
k = 230(M^2 *sec)^-1
Now for the final part of this problem.
Rate = 230* (0.0844M) * (0.175M)^2
Rate = 0.594 M/s
I Hope you were able to understand that!
Look at Trial 2 and Trial 3. From this we see that when the [OH] is cut in half so is the rate.
(1/2^n) = 1/2
n = 1
Therefore this reaction must be First order with respect to [OH]
Now look at Trial 1 and Trial 2. From this we see that when [ClO2] is cut in half the rate is only 1/4 of what it was.
(1/2^n) = 1/4
n=2
From this we can conclude that the reaction is Second order with respect to [ClO2]
So our rate law looks like this;
Rate = k[OH][ClO2]^2
Knowing that our Rate is in M/sec we know that the units of our rate constant, k, must be (M^2 *sec)^-1
To find the value of the rate law we just need to plug in the concentrations of OH and ClO2 from one of our trials into our rate law. Lets do Trial 2
2.30*10^-1 (M/s) = k* (0.100M) *(0.100M)^2
k = 230(M^2 *sec)^-1
Now for the final part of this problem.
Rate = 230* (0.0844M) * (0.175M)^2
Rate = 0.594 M/s
I Hope you were able to understand that!