Iodide Ion is oxidized to hypoiodite ion IO- by hypochlorite ion ClO- in basic solution. The equation is
I- + ClO- --- IO- + Cl-
Initial concentraition
I- ClO- OH-
exp 1. 0.010 0.020 0.010
exp 2. 0.020 0.010 0.010
exp 3. 0.010 0.010 0.010
exp 4. 0.010 0.010 0.020
intial rate (mol/(L*s))
12.2 x 10^-2
12.2 x 10^-2
6.1 x 10^-2
3.0 x 10^-2
Find the rate law and the value of the rate constant.
I got 610 l/mol*s
6.1 s-
To determine the rate law and rate constant, we need to assess the relationship between the initial concentrations of the reactants (iodide ion, ClO-, and OH-) and the initial rates for each experiment.
By comparing experiments 1 and 3, we can see that doubling the initial concentration of I- leads to a doubling of the initial rate. Therefore, the rate is directly proportional to the concentration of I-.
Next, by comparing experiments 2 and 3, we can observe that doubling the initial concentration of ClO- leads to a doubling of the initial rate. Hence, the rate is directly proportional to the concentration of ClO- as well.
Experiment 4, with the same initial concentrations of I- and ClO- as experiment 3 but with elevated OH- concentration, shows that increasing the OH- concentration does not affect the initial rate. Thus, the rate does not depend on the OH- concentration.
Based on these results, we can conclude that the rate law for this reaction is:
Rate = k [I-] [ClO-]
Where k is the rate constant.
Since the rate law includes the concentrations of I- and ClO-, we can use any of the experiments to determine the rate constant. Let's use experiment 3:
Rate = k [I-] [ClO-]
6.1 x 10^-2 = k (0.010) (0.010)
Dividing both sides by (0.010) (0.010):
k = (6.1 x 10^-2) / (0.010) (0.010)
k = 610 L/(mol*s)
Therefore, the rate law for the reaction is rate = k [I-] [ClO-], and the value of the rate constant (k) is 610 L/(mol*s).
To find the rate law and the value of the rate constant, you need to analyze the initial rates of the reaction under different experimental conditions and determine how the concentrations of the reactants affect the rate.
From the given data, you have the initial concentrations of Iodide (I-), Hypochlorite (ClO-), and Hydroxide (OH-) for each experiment, as well as the corresponding initial rates.
Let's consider the general rate law equation: rate = k [I-]^x [ClO-]^y [OH-]^z
To determine the rate law, you need to compare the initial rates from different experiments while keeping the concentrations of two of the reactants constant and varying the concentration of the third reactant.
Comparing Exp 1 and Exp 2:
The concentration of ClO- is constant while the concentration of I- doubles. As the initial rates remain the same, we can conclude that the rate of the reaction does not depend on the concentration of Iodide. So, the rate law does not include [I-] and x = 0.
Comparing Exp 1 and Exp 3:
The concentration of OH- doubles while the concentrations of I- and ClO- remain constant. The initial rate is halved. From this comparison, we can conclude that the rate is directly proportional to [OH-], so z = 1.
Comparing Exp 3 and Exp 4:
The concentrations of I- and OH- remain constant while the concentration of ClO- doubles. The initial rate is doubled. So, the rate is directly proportional to [ClO-], and y = 1.
Therefore, the rate law equation becomes: rate = k [ClO-] [OH-]
Now, let's calculate the rate constant (k) using one set of experimental data. We can use Exp 1 since the initial rate is provided.
For Exp 1:
[I-] = 0.010 mol/L
[ClO-] = 0.020 mol/L
[OH-] = 0.010 mol/L
rate = 12.2 x 10^-2 mol/(L*s)
Substituting the values into the rate law equation:
12.2 x 10^-2 = k (0.020) (0.010)
k = (12.2 x 10^-2) / (0.020) (0.010)
k ≈ 610 L/(mol*s)
Therefore, the rate law for the reaction is rate = k [ClO-] [OH-], and the value of the rate constant (k) is approximately 610 L/(mol*s).