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, C₂H₆N_2(g), decomposes to form ethane gas, C₂H_6(g), and nitrogen gas, N_2(g).

   C₂H₆N_2(g) → C₂H_6(g) + N_2(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, NO_2(g), decomposes to form nitric oxide gas, NO_(g), and oxygen gas, O_2(g).

   2NO_2(g) → 2NO_(g) + O_2(g)

   Experiments at 383°C have shown that the rate of decomposition of NO_2(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.

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) + 3O_2(g) → 2SO_2(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 O_2(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

5. Is your answer plausible?

   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:

2NO_2(g) → NO_3(g) + NO_(g)

rate = k[NO₂]²

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

5. Is your answer plausible?

   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

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

Calculating the Specific Rate Constant

Question : A student investigating the reaction:

4NH_3(g) + 5O_2(g) → 4NO_(g) + 6H₂O_(g)

rate = k[NH₃]²[O₂]²

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

5. Is your answer plausible?

   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:

Cl_2(g) + 3F_2(g) → 2ClF_2(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
   [F_2(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

5. Is your answer plausible?

   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