The following reaction was studied at −10°C.
2 NO(g) + Cl2(g) → 2 NOCl(g)
The following results were obtained where the rate of the reaction is given below.
[NO]o [Cl2]o INITIAL RATE
0.10 0.10 0.18
0.10 0.20 0.36
0.20 0.20 1.45
What is the rate law?
What is the value of the rate constant?
Don't you want to know how to do these yourself?
I see trial 1 and trial 2 has same concn for NO and double the cncn for Cl2. Note the k changes by 0.18 to 0.36 or double.
Cl2 doubled; rate doubled, so 2^x = 2 and x must be 1.
Then look at trial 2 and 3. Cl2 stays the same but NO doubles. The rate changes from 0.36 to 1.45 or (1.45/0.35 = 4.03) so 2^x = 4. x must be ?
Now take ANY of the three trials and write the rate constant expression.
rate = k[NO]x[Cl2]y and enter the rate for the trial you've chosen, plug in (NO), (Cl2), and x and y where x and y represent the values of x you have above.. Solve for k.
To determine the rate law, we need to use the initial rate data and compare how changes in concentrations of the reactants affect the rate of the reaction.
Let's write the rate law expression as follows:
Rate = k[NO]^x[Cl2]^y
Using the given data, we can compare the initial rates for different reaction conditions and determine the values of x and y.
First, we compare the first two sets of data where [NO]o = 0.10 M and [Cl2]o changes from 0.10 M to 0.20 M:
Rate1 / Rate2 = (0.18 M) / (0.36 M) = (0.10 M)^x / (0.10 M)^x * (0.20 M)^y
Simplifying, we get:
0.5 = (0.10 M)^x / (0.20 M)^y
Next, we compare the first and third sets of data where [NO]o changes from 0.10 M to 0.20 M, and [Cl2]o remains constant at 0.20 M:
Rate1 / Rate3 = (0.18 M) / (1.45 M) = (0.10 M)^x * (0.20 M)^y / (0.20 M)^y
Simplifying, we get:
0.124 = (0.10 M)^x * (0.20 M)^y / (0.20 M)^y
By equating the two expressions, we can eliminate the y term:
0.5 = 0.124 / (0.10 M)^x
Solving for x, we find:
(0.10 M)^x = 0.124 / 0.5
x ≈ 1
So, the rate law expression becomes:
Rate = k[NO]^1[Cl2]^y
Next, we can compare the rates when the concentration of [NO] doubles while [Cl2] remains constant:
Rate1 / Rate3 = (0.18 M) / (1.45 M) = (0.20 M)^y / (0.10 M)^y
Simplifying, we get:
0.124 = (0.20 M)^y / (0.10 M)^y
By equating the expressions, we find:
0.124 = (0.20 M / 0.10 M)^y
0.124 = 2^y
Solving for y, we find:
2^y = 0.124
y ≈ -3
Now that we have determined the values of x and y, we can write the rate law expression:
Rate = k[NO][Cl2]^-3
To find the value of the rate constant (k), we can use one set of data and substitute the values into the rate law equation.
Using the first set of data:
[NO]o = 0.10 M
[Cl2]o = 0.10 M
Rate = 0.18 M/s
Substituting these values into the rate law expression, we get:
0.18 M/s = k * (0.10 M) * (0.10 M)^-3
0.18 = k * 0.10 * (0.10)^-3
0.18 = k * 0.10 / (0.10^3)
0.18 = k / 1000
k = 0.18 * 1000
k = 180 M^-3s^-1
Therefore, the rate law is Rate = k[NO][Cl2]^-3 and the value of the rate constant (k) is 180 M^-3s^-1.
To determine the rate law and the value of the rate constant for the given reaction, you need to examine the initial rates of the reaction under different initial concentrations of reactants.
The rate law relates the rate of a reaction to the concentrations of reactants. It can be expressed as:
rate = k[A]^x[B]^y
where:
rate is the rate of the reaction
k is the rate constant
[A] and [B] are the concentrations of reactants
x and y are the orders of reactants A and B, respectively.
From the given data, you have three sets of initial concentrations ([NO]o and [Cl2]o) and corresponding initial rates.
Using the first set of initial concentrations:
[NO]o = 0.10
[Cl2]o = 0.10
INITIAL RATE = 0.18
Using the second set of initial concentrations:
[NO]o = 0.10
[Cl2]o = 0.20
INITIAL RATE = 0.36
Using the third set of initial concentrations:
[NO]o = 0.20
[Cl2]o = 0.20
INITIAL RATE = 1.45
To determine the rate law, you need to compare the rates of the reactions when one of the reactant concentrations is doubled while keeping the other constant.
Comparing the first and second sets of data, it can be observed that the rate doubles when the concentration of Cl2 is doubled while the [NO] remains the same. This indicates that the reactant Cl2 is first order with respect to the rate.
Comparing the first and third sets of data, it can be observed that the rate increases by a factor of 1.45/0.18 ≈ 8 when both [NO] and [Cl2] are doubled. This suggests that the reaction is second order with respect to both NO and Cl2.
Therefore, the rate law for the reaction is:
rate = k[NO]^2[Cl2]
To determine the value of the rate constant (k), you can choose any set of data and substitute the known concentrations and the corresponding rate into the rate law equation to solve for k.
Let's use the second set of data:
[NO]o = 0.10
[Cl2]o = 0.20
INITIAL RATE = 0.36
Plugging these values into the rate law equation:
0.36 = k(0.10)^2(0.20)
Simplifying the equation further:
0.36 = k(0.01)(0.20)
0.36 = 0.002k
Solving for k:
k = 0.36/0.002
k ≈ 180
Therefore, the value of the rate constant (k) is approximately 180.