# Unit two: Chemical Kinetics. 2.1.1 Introduction to Chemical Kinetics The rate at which a chemical reaction occurs is often very important to us. A common.

## Presentation on theme: "Unit two: Chemical Kinetics. 2.1.1 Introduction to Chemical Kinetics The rate at which a chemical reaction occurs is often very important to us. A common."— Presentation transcript:

Unit two: Chemical Kinetics

2.1.1 Introduction to Chemical Kinetics The rate at which a chemical reaction occurs is often very important to us. A common task of chemists is to find ways to change the rates of chemical reactions - we want our cars and buildings to rust more slowly and our food to spoil more slowly as well. But we need the reactions that produce economically important substances, such as ammonia and medical pharmaceuticals to occur quickly.

In this unit we are concerned with how fast the reaction occurs. We will be asking these key questions: 1. How do you measure the speed, or rate, of a reaction? 2. What factors influence how fast a reaction occurs?

2.1.2 Reaction Rates Reaction rate is a measure of how fast a reaction occurs, or how something changes during a given time period. Lab Complete the cooling of water lab

Consider the oxidation of glucose, C 6 H 12 O 6 : C 6 H 12 O 6 (s) + 6 O 2 (g) → 6 CO 2 (g) + 6 H 2 O(g) How could we measure how fast this occurs? Here are some other possible things that could be measured as they change: change in conductivity change in pH colour change change in pressure as gases are formed or used up

A common measure of reaction rate is how the concentration of a reaction participant changes over time (reactants decrease, products increase.) We also must consider the second key aspect of determining rate - time. Rate is a measure of how something changes over time. You measure your rate of speed when driving by determining how many kilometres you travel in an hour; you could measure your rate of speed during a race by dividing the distance of the course by the number of minutes it took to run the rate.

Rate Equation the rate of a chemical reaction: Rate =change in concentration change in time

Chemistry Notation In chemistry, we typically represent concentration by using square brackets around the chemical formula of the substance. For example to indicate the concentration of SO 2 (g) in the following reaction we would write it as [SO 2 ]. Also, the delta symbol, Δ is used to indicate a change. Δ T, for example, means "the change in temperature."

Reaction rate over time When the reaction first starts, the reactants are plentiful so things move along pretty fast - As the reaction continues, there will be less and less reactants available so the reaction will slow down. Eventually the reaction stops once all of the reactants are gone. A graph illustrating reaction rate is often useful graph So, if we measured the rate at the start of the reaction, it would be different than if we measured the rate near the end of the reaction. Rate is not a constant - it changes during the course of the reaction.

Average Rate For this reason, we will generally work with the average rate of reaction, and measure rate over a longer period of time. For example, if you were taking a long car trip, there might be times when you are driving fast (on the highway) and also times when you drive more slowly (through the cities). If you calculated your average rate for the entire trip, you would average out these fast and slow periods.

2.1.3 Calculating Reaction Rates Let's try an example of calculating a reaction rate. Consider the following reaction: A → B We could measure the rate of the reaction either by measuring how the concentration of reactant A changes or how the concentration of product B changes. Let's measure A's average rate of change first: Rate= Δ [A] = (0.250 - 1.00) = – 0.125 mol · L -1 · min -1 Δ time (6.00 - 0.00) Time (min) [A] (mol ·L -1 ) [B] (mol ·L -1 ) 0.01.0000.000 3.00.4000.600 6.00.2500.750

If we calculate the average rate based on the production of product B: Rate = Δ [B] = (0.750 - 0.00) = 0.125 mol · L -1 · min -1 Δ time (6.00 - 0.00) Notice that we must compare rates measured during the same time period. You will see that we got the same rate, except for the sign, when we calculated the overall average rate for the disappearance of A as for the formation of B. This is only true because our balanced equation shows us that one mole of A produces one mole of B.

Compare this rate to the rate of just the first three minutes of the reaction: Rate = Δ [A] = (0.400 - 1.00) = - 0.200 mol · L -1 · min -1 Δ time (3.000 - 0.000) We can see that the rate did slow down as the overall rate is slower than the rate of the first three minutes of the reaction.

We would not find the same rates if we did not have a 1:1 relationship between reaction participants. 2 H 2 O 2 → 2 H 2 O + O 2 we find that only one mole of oxygen forms for every two moles of hydrogen peroxide that decomposes. Therefore, we could make the following relationships: If 2 moles of H 2 O 2 decompose then 1 mole of O 2 is formed, if the rate of decomposition of H 2 O 2 is 4.00 mol · L -1 · min - 1 then the rate of formation of O 2 is ½ × 4.00 or 2.00 mol · L -1 · min -1

1. In the following decomposition reaction, 2 N 2 O 5 → 4 NO 2 + O 2 oxygen gas is produced at the average rate of 9.1 × 10 -4 mol · L -1 · s -1. Over the same period, what is the average rate of the following: a)the production of nitrogen dioxide b)the loss of nitrogen pentoxide

Solution #1: From the equation we see that for every 1 mole of oxygen formed, four moles of nitrogen dioxide are produced. Thus, the rate of production of nitrogen dioxide is four times that of oxygen: A) rate NO 2 production = 4 × (9.1 × 10 -4 mol · L -1 · s -1 ) = 3.6 × 10 -3 mol · L -1 · s -1 Nitrogen pentoxide is consumed at twice the rate that oxygen is produced: B) rate loss of N 2 O 5 = 2× (9.1 × 10 -4 mol · L -1 · s -1 ) = 1.8 × 10 -3 mol · L -1 · s -1

Example #2 2. Consider the following reaction: N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) If the rate of loss of hydrogen gas is 0.03 mol · L -1 · s -1, what is the rate of production of ammonia?

Solution #2: From the balanced equation we see that there are 2 moles NH 3 produced for every 3 moles H 2 used. Thus: rate NH 3 production = 2× (0.03 mol · L -1 · s -1 ) = 0.02 mol · L -1 · s -1 3

Example #3 In a reaction between C 4 H 9 Cl and water, the concentration of C 4 H 9 Cl is 0.220M at the beginning of the reaction. At 4.0s, the concentration of C 4 H 9 Cl is 0.100M. Calculate the average reaction rate.

Solution #3 [C 4 H 9 Cl] at 0s= 0.220M [C 4 H 9 Cl] at 4s= 0.100M t1=0s t2=4s Rate= ∆ [C 4 H 9 Cl]/ ∆ t = 0.100M-0.220M/ 4s- 0.0s = -0.120M/ 4s = 0.0300 mol/Ls

Assignment Assignment 2.1.3

Download ppt "Unit two: Chemical Kinetics. 2.1.1 Introduction to Chemical Kinetics The rate at which a chemical reaction occurs is often very important to us. A common."

Similar presentations