rate laws. rate law the concentration of reactants can affect the rate of a chemical reaction. even...

Rate Laws

Rate Law• The concentration of reactants can

affect the rate of a chemical reaction. • Even if the reaction has multiple

reactants, changing the concentration of just one reactant can affect the rate.

2 H2 + O2 2 H2O• In a container, if there

is only a small amount of hydrogen, and a small amount of oxygen, then the rate of water production will be slow.

• It takes time for the reactants to find each other.

2 H2 + O2 2 H2O• This time, there is a high

concentration of hydrogen, and a still a small amount of oxygen.

• Just changing one reactant will cause the rate of reaction to be faster because the reactants find each other more quickly.

Rate Laws• As concentrations of reactants change at

constant temperature, the rate of reaction changes!

• We write rate-law expressions or simply Rate Laws to describe how the rate of a reaction depends on the concentrations.

• These Rate Laws can only be determined through experimentation!

Rate Law Blueprint• The Rate law expression has the same

format for all equations. • Using this template for an equation:

aA + bB ------> cC + dD– lower case letters are coefficients– upper case letters are compounds

• Rate of the reaction = k[A]x[B]y

• Notice the coefficients are not part of the Rate Law.

Rate of reaction = k[A]x[B]y

• k = Rate Constant – this will be either given, or we will solve for it. Determined by experiment. • [A] = Concentration (Molarity) of reactant A.• X = (exponent) the “order” of the reaction in

regards to compound A.• [B] = Concentration (Molarity) of reactant B. • Y = (exponent) the “order” of the reaction in

regards to compound B.

Rate of reaction = k[A]x[B]y

• Usually X and Y will be small numbers. Generally 0, 1, or 2. • They tell us the “order” of the reaction in

regards to their respective compound or…• They can tell us the “order” for the whole

equation when added together!• X + Y = “Order” of the overall reaction.

Example: k[A]x[B]y

• Let’s analyze this Rate Law = k[Ag+][NO2-]

• What is the order of reaction in regards to the silver ion?

• What is the order of the reaction in regards to the nitrite ion?

• So we say the reaction is 1st order in Ag+ and 1st order in NO2

-

• What is the order of the reaction overall? • X = 1 and Y = 1 so overall 1 + 1 = 2nd order!

Practice Problems• Determine the order of the reaction in

regards to each reactant, and the overall order of the reaction.

A) Rate = k[NO3-] B) Rate = k[CO3

2-][Ag+]

C) Rate = k[NOBr]2 D) Rate = k[Na+]2[O2-]

Method of Initial Rates• We’ve mentioned earlier that rate law

constants can only be determined experimentally.

• We can also experimentally determine the rate law. (1st, 2nd order)

• By observing the initial concentrations and initial rate of the reaction, we can determine the order of the reaction.

A C• Let’s consider the above reaction. Our

goal is to write the rate law. • This expression will have the form:

rate = k[A]x

• Remember that the brackets [A] represent concentrations of the reactants.• The X represent the order of the

reaction. That’s what we’re trying to find.

A C

• To determine the order of the reaction we need to see how the concentration changes, and then how the rate of reaction changes.

• The ratio of the initial concentration changes by a factor of 2. (2.0 M / 1.0 M)

• The rate of the reaction changes by a factor of 2. (6.0 M/sec / 3.0 M/sec)

Experiment Initial Concentration [A]

Initial Rate of the Reaction

1 1.0 M 3.0 M/sec2 2.0 M 6.0 M/sec

A C• The rate law is = k[A]x

• We can solve for X (order of reaction) by: (Concentration change)X = Rate of the reaction change

• (2)X = 2 *Determined from previous slide*• X must equal one, indicating that this is a

first order reaction. • We would write the rate law as: Rate = k[A]

A + 2B C• Let’s consider the above reaction. Our goal

is to write the rate law for the above reaction.

• This expression will have the form: k[A]x[B]y

• Remember that the brackets[A] and [B] represent concentrations of the reactants.

• The X and Y represent the order of the reaction. That’s what we’re trying to find.

A + 2B C

• This reaction is trickier because there are two reactants. • We need to focus on each reactant individually to see

how it changes the rate. • We need to find experiments that keep one of the

reactants constant.

Experiment Initial [A] Initial [B] Initial Rate of the Reaction

1 1.0 M 1.0 M 2.0 M/sec2 1.0 M 2.0 M 4.0 M/sec3 2.0 M 2.0 M 16.0 M/sec

A + 2B C

• Experiment 1 and 2 have the same [A], that lets us see how [B] affects the reaction rate.

• When [B] doubles, the rate doubles. • 2Y = 2, so Y must be 1. The reaction is 1st order in

regards to reactant B.



A + 2B C

• Experiment 2 and 3 have the same [B], and this time will let us see how [A] affects the rate.

• When [A] doubles, the rate quadruples!!! • 2X = 4, so X must be 2. This time the reaction is 2nd

order in regards to reactant A.



A + 2B C• The rate law requires us to put it all

together.• Rate = k[A]x[B]y

• The reaction was second order in regards to A and first order in regards to B.• Rate = k[A]2[B]• What is the overall order of the reaction?

A + 2B C Rate = k[A]2[B]

• Now that we know the rate law, we can solve for the rate constant, K.

• Use any experiment, plug in the data and solve for k. • 2.0 M/sec = k[1.0 M]2[1.0 M]• K = 2.0 M-2 sec -1



• Lets determine the rate law for the above reaction. Look for experiments where the concentrations are constant.

• In experiment 1 and 2, the Nitrite ion is constant. When the [NH4

+] doubles, the rate doubles.

• [2]X = 2, The rate is first order in regards to [NH4+]

• In experiment 5 and 6, the [NH4+] ion is constant.

When the [NO2-] doubles, the rate doubles.

• [2]Y = 2, The rate is first order in regards to [NO2-]

• So the rate law = k [NH4+][NO2

-]• Determine the rate constant!• 5.4 x 10-7 M/sec = k [0.0100 M][0.200 M]• K = 2.7 x 10-4 M-1 sec-1

rate laws. rate law the concentration of reactants can affect the rate of a chemical reaction. even...

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