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
it is 6.1 second inverse (s-)
To find the rate law and the value of the rate constant, we can use the method of initial rates by comparing the initial concentrations and initial rates between different experiments. Here's how we can approach this problem step-by-step:
Step 1: Determine the rate law based on the initial concentrations.
In the given equation: I- + ClO- → IO- + Cl-
Since the rate is determined by the concentrations of I- and ClO-, we can write the rate law as:
Rate = k [I-]^a [ClO-]^b
Step 2: Determine the values of a and b in the rate law by comparing the initial rates in different experiments.
From the given data, we can compare the initial rates for different experiments and determine the exponent values a and b.
Experiment 1:
Rate = 12.2 x 10^-2 mol/(L*s)
[I-] = 0.010 M
[ClO-] = 0.020 M
Experiment 2:
Rate = 12.2 x 10^-2 mol/(L*s)
[I-] = 0.020 M
[ClO-] = 0.010 M
Comparing these two experiments, we can see that doubling the concentration of either I- or ClO- results in no change in the rate. Therefore, a and b are both 0.
Hence, the rate law becomes:
Rate = k [I-]^0 [ClO-]^0
Rate = k
Step 3: Determine the value of the rate constant.
From the given data, we have the initial rate for experiment 3:
Rate = 6.1 x 10^-2 mol/(L*s)
Since the rate law simplifies to Rate = k, we can plug in the value of the rate and solve for the constant:
6.1 x 10^-2 = k
Hence, the rate constant k is 6.1 x 10^-2 L/mol*s.
So, your value of 610 L/mol*s is correct, but it should be written as 6.1 x 10^-2 L/mol*s in scientific notation.
To determine the rate law and the value of the rate constant, you need to analyze the given concentration data and initial rates.
First, let's consider the rate law. The rate law defines the relationship between the rate of a chemical reaction and the concentrations of the reactants. In this case, we have:
Rate = k [I-]^a [ClO-]^b [OH-]^c
Where k is the rate constant, [I-], [ClO-], and [OH-] are the concentrations of Iodide Ion, Hypochlorite Ion, and Hydroxide Ion, respectively, and a, b, and c are the orders of the reaction with respect to I-, ClO-, and OH-, respectively.
To determine the orders of the reaction, you can compare the initial concentrations and initial rates of the four experiments.
From experiments 1 and 2, you can observe that doubling the concentration of [I-] does not change the initial rate. This suggests that the reaction is not dependent on the concentration of [I-]. Therefore, the order with respect to I- is zero (0).
From experiments 1 and 3, you can observe that doubling the concentration of [ClO-] doubles the initial rate. This indicates that the reaction is first order with respect to ClO-. So, the order with respect to ClO- is one (1).
From experiments 1 and 4, you can observe that doubling the concentration of [OH-] halves the initial rate. This suggests that the reaction is inversely proportional to the concentration of OH-. Therefore, the order with respect to OH- is negative one (-1).
Putting all the information together, the rate law for this reaction can be written as:
Rate = k [ClO-]^1 [OH-]^(-1)
Next, let's find the value of the rate constant (k). To determine the value of k, we can use any of the experiments, since the concentration of I- remains constant in all of them. Let's use experiment 1:
Rate = k [ClO-]^1 [OH-]^(-1)
12.2 x 10^-2 = k (0.020)^1 (0.010)^(-1)
k = 12.2 x 10^-2 / (0.020 x 0.010)
k = 610 L/mol*s
So, the rate constant (k) is indeed 610 L/mol*s.
Therefore, the rate law for the reaction is Rate = 610 [ClO-] [OH-]^(-1), and the value of the rate constant is 610 L/mol*s.