 # Using Rate Law Tutorial

## Key Concepts

• Reaction rate is the time rate of change of concentration of a reactant or product in a chemical reaction.
• The rate of consumption of a reactant is always negative.
• The rate of formation of a product is always positive.

 For the reaction: aA + bB → cC + dD If the reaction rate for the production of D = rate(1) Then during the experiment, the reaction rate for the consumption of A and B and for the production of C is as follows: (a) rate(A) = -a/d rate(1) (b) rate(B) = -b/d rate(1) (c) rate(C) = c/d rate(1)

• Rate law for a chemical reaction is the algebraic expression of the relationship between concentration and the rate of a reaction at a particular temperature.
example 1:
rate concentration

example 2:
rate concentration2

• The constant of proportionality in the algebraic expression is given the symbol k and is refered to as the specific rate constant for the reaction.
example 1:
rate = k1 × concentration

example 2:
rate = k2 × concentration2

• The rate law cannot be determined from the chemical equation for a reaction, it can only be determined by experimental measurements of a reaction rate at a particular temperature.

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## Examples of Writing the Rate Law for a Reaction Given the Value of the Specific Rate Constant, k

Rate Law can ONLY be determined experimentally.

The sign (+ or -) of a rate tells us whether reactants are being consumed, or, if products are being produced:

• If rate is positive (+) products are being produced.
• If rate is negative (-) reactants are being consumed.

Let's look at some experimentally determined rate laws:

1. Azomethane gas, C2H6N2(g), decomposes to form ethane gas, C2H6(g), and nitrogen gas, N2(g).
C2H6N2(g) → C2H6(g) + N2(g)

Experiments have shown that rate of consumption of azomethane is directly proportional to the concentration of azomethane, so,

rate ∝ [azomethane]

Because we are talking about the rate of consumption of a reactant, the rate will be negative (-) :

-rate ∝ [azomethane]

If we use the constant of proportionality, the specific rate constant, k, we can write the expression:

-rate = k[C2H6N2]

At a particular temperature, the value of k, the specific rate constant for this reaction, has been shown to be 1.60 × 10-2min-1.

k = 1.60 × 10-1min-1

So, the rate of decompostion of azomethane at this particular temperature can be calculated:

-rate = 1.60 × 10-2[C2H6N2]

The rate of this reaction will be negative since the reactant is being used to produce products.

2. Nitrogen dioxide gas, NO2(g), decomposes to form nitric oxide gas, NO(g), and oxygen gas, O2(g).
2NO2(g) → 2NO(g) + O2(g)

Experiments at 383°C have shown that the rate of decomposition of NO2(g) is proportional to the square of the concentration of nitrogen dioxide, so,

rate ∝ [nitrogen dioxide]2

Because this is the rate of decomposition, that is, reactants are being consumed, the rate will be negative (-):

-rate ∝ [nitrogen dioxide]2

If we use k to represent the constant of proportionality, the specific rate constant for this reaction, then we can write:

-rate = k[NO2]2

The value of k, the specific rate constant for this reaction has been shown to be 10.1 M-1s-1.

k = 10.1 M-1s-1

So the rate law for this reaction at 383°C is:

-rate = 10.1[NO2]2

Since the rate of this reaction refers to the disappearance of reactants its sign will be minus.

3. For the reaction:
C2H4Br2 + 3KI → C2H4 + 2KBr + KI3

it has been determined by experiment that the rate of production of KI3 is directly proportional to the concentrations of both C2H4Br2 and KI, so,

rate ∝ [C2H4Br2][KI]

Using the constant of proportionality, k, the specific rate constant for this reaction under these conditions, we can write:

rate = k[C2H4Br2][KI]

At a particular temperature, the value of k, the specific rate constant for the production of KI3 in this reaction, is equal to 0.299,

k = 0.299

so at this temperature:

rate = 0.299[C2H4Br2][KI]

Since the rate of this reaction refers to the production of products, the rate will be positive.

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## Worked Examples: Calculating Reaction Rate from Rate Law

### Determining Reaction Rate for the Production of a Product Given Reaction Rate for the Consumption of a Reactant

Question : For the reaction:

2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)

a student determined the rate law during an experiment and found it be k[H2S]2

If the reaction rate for the consumption of O2 during the experiment is -2.6 x 10-4 Ms-1, what is the reaction rate for production of H2O?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate rate of reaction for production of H2O
rate(H2O) = ?

2. What data (information) have you been given in the question?

Extract the data from the question:

rate(O2) = k[H2S]2

rate(O2) = -2.6 × 10-4Ms-1 (negative for consumption of O2 reactant)

For the consumption of O2(g):

-2.6 × 10-4Ms-1 = k[H2S]2

3. What is the relationship between what you know and what you need to find out?
rate(H2O) = 2/3 × -rate(O2)
(i) Use stoichiometric ratio (mole ratio) to determine how much slower rate of formation of water will be.
(ii) Rate of production of H2O will be positive (opposite sign to rate of consumption of O2)
4. Substitute the values into the equation and solve for rate of production of H2O:
rate(H2O) = 2/3 × -rate(O2)
rate(H2O) = 2/3 × -(-2.6 × 10-4 Ms-1)
rate(H2O) = 1.73 × 10-4 Ms-1
Consider: 1 mole H2S requires 1.5 mol O2 to produce 1 mol H2O
Therefore O2 must disappear at a rate 1.5 times faster than the rate at which H2O appears.
Work backwards to check that this is true using our calculated value for the rate of production of H2O:
rate(O2 disappears) = 1.5 × rate(H2O appears) = 1.5 × 1.73 × 10-4 = 2.6 × 10-4
Sign of rate(O2 disappears) must be opposite to that for rate(H2O) appears, ie, + becomes -
rate(O2 disappears) = -2.6 × 10-4
which is the same as the value given in the question we are reasonably confident that our answer is plausible.
6. State your solution to the problem "rate of production of H2O":

rate(H2O) = 1.73 × 10-4 Ms-1

### Calculating the Reaction Rate Given the Specific Rate Constant and Concentration(s)

Question : A student investigating the reaction:

2NO2(g) → NO3(g) + NO(g)

found the reaction followed the rate law:
rate = k[NO2]2

Given that the concentration of NO2 was 0.594 M and the specific rate constant was 1.25 × 102, calculate the rate of the reaction.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate the rate of the reaction
rate = ?

2. What data (information) have you been given in the question?

Extract the data from the question:

rate = k[NO2]2

k = 1.25 × 102

[NO2] = 0.594 M

3. What is the relationship between what you know and what you need to find out?
rate = k[NO2]2
4. Substitute the values into the rate equation and solve:
rate = k[NO2]2

rate = 1.25 × 102[0.594 M]2

rate = 1.25 × 102 × 0.353

rate = 44.1

Round off the numbers and perform a "rough calculation" to make sure your answer is in the right "ball park" (that is, that your answer is the right order of magnitude).
Let [NO2] ≈ 0.6 M
and k ≈ 100
then rate ≈ 100[0.6]2 = 100 × 0.36 = 36
Our carefully calculated value of 44.1 is of the same order of magnitude as our "rough" calculation so we are reasonably confident our answer is plausible.
6. State your solution to the problem "rate of reaction":

rate = 44.1

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## Worked Examples of Using Rate Law to Calculate Specific Rate Constant and Concentrations

### Calculating the Specific Rate Constant

Question : A student investigating the reaction:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

found that the reaction followed the rate law:
rate = k[NH3]2[O2]2

If the concentrations of NH3 and O2 were 0.188 M and 0.789 M respectively and the reaction rate was 1.43 × 10-4 Ms-1, calculate the value of the specific rate constant for this reaction.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate specific rate constant, k
k = ?

2. What data (information) have you been given in the question?

Extract the data from the question:

rate law:
rate = k[NH3]2[O2]2

[NH3] = 0.188 M

[O2] = 0.789 M

rate = 1.43 × 10-4 Ms-1

3. What is the relationship between what you know and what you need to find out?

 We have been given the rate law equation: rate = k[NH3]2[O2]2 divide both sides by [NH3]2[O2]2 rate       [NH3]2[O2]2 = k[NH3]2[O2]2 [NH3]2[O2]2 rate       [NH3]2[O2]2 = k

4. Substitute values into the equation and solve for k

 k = rate       [NH3]2[O2]2 k = 1.43 × 10-4     [0.188]2[0.789]2 k = 1.43 × 10-4     0.0353 × 0.623 k = 1.43 × 10-4 0.0220 k = 6.50 × 10-3

Work backwards: Use your calculated value for k, the rate law and concentrations provided in the question to calculate the rate and compare it to that given in the question.

k = 6.50 × 10-3
[NH3] = 0.188 M
[O2] = 0.789 M

rate = k[NH3]2[O2]2
rate = 6.50 × 10-3[0.188]2[0.789]2
rate = 1.43 × 10-4

Since the rate we calculated here agrees with the rate given in the question we are confident our value for k is plausible.

6. State your solution to the problem "value of specific rate constant":
k = 6.50 × 10-3

### Calculating Concentration from Rate Law and Specific Rate Constant

Question : A student investigating the reaction:

Cl2(g) + 3F2(g) → 2ClF2(g)

found the rate law to be proportional to the square of the fluorine concentration.

If the rate and specific rate constant were found to be 9.34 × 10-5 Ms-1 and 1.35 × 10-3 respectively, calculate the fluorine concentration.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate concentration of fluorine
[F2(g)] = ? M

2. What data (information) have you been given in the question?

Extract the data from the question:

rate ∝ [F2]2

rate = 9.34 × 10-5 Ms-1

k = 1.35 × 10-3

3. What is the relationship between what you know and what you need to find out?

 rate ∝ [F2]2 Let k be the constant of proportionality: rate = k[F2]2 Divide both sides of equation by k rate k = k[F2]2 k rate k = [F2]2 Take the square root of both sides of equation √ rate   k = √[F2]2 √ rate   k = [F2]

4. Substitute the values into the equation and solve for [F2]

 [F2] = √ rate   k [F2] = √ 9.34 × 10-5 1.35 × 10-3 [F2] = √0.0692 [F2] = 0.263 M

Work backwards: use your calculated value for the concentration of F2 and the value for the specific rate constant (k) given in the question to calculate the rate and compare that to the value given in the question:

rate = k[F2]2 = 1.35 × 10-3[0.263]2 = 9.34 × 10-5
Since this value for rate is the same as that given in the question we are reasonably confident that our value for [F2] is plausible.

6. State your solution to the problem "concentration of fluorine":

[F2] = 0.263 M

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