A Level Chemistry 3.1.9 Rate Equations PDF

A LEVEL CHEMISTRY 3.1.9 RATE EQUATIONS We should remember from Year 12 that the rate of a chemical reaction can depend on the concentrations of the reactants. Previously all we needed to do was describe how the rate is affected by changes in...

A LEVEL CHEMISTRY 3.1.9 RATE EQUATIONS We should remember from Year 12 that the rate of a chemical reaction can depend on the concentrations of the reactants. Previously all we needed to do was describe how the rate is affected by changes in concentration. We must now calculate the rate of a chemical reaction given the concentrations of the reactants. However, first we must understand “Orders” of reaction… ORDERS OF REACTION We learned in Year 12 that an increase in the concentration of a reactant led to an increase in the rate of a chemical reaction. Well, strictly speaking, that’s not true! Depending on the role that a reactants plays in the chemical reaction, changing the concentration of a reactant can have one of three effects, known as “Orders” of reaction. 2nd Order 1st Order Rate (mol.dm-3.s-1) Zero Order [Concentration] (mol.dm-3) What this graph shows is that as you increase the concentration of a reactant, it can have one of three effects on the rate of reaction: Zero Order - An increase in concentration has no effect on the rate 1st Order - An increase in concentration has a directly proportional effect on the rate of reaction. i.e. if you double the concentration the rate will also double 2nd Order- An increase in concentration has an exponential effect on the rate of reaction. i.e. if you double the concentration the rate will increase by a factor of 4 (22). AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.9 RATE EQUATIONS DEDUCING ORDERS OF REACTION You are expected to use data for a given reaction to deduce the order of a reaction with respect to a particular reactant. There are two ways to do this… 1. Using a Concentration Vs Time Graph Here, the only order of reaction you will be expected to work with is Zero Order. If you track the concentration of a reactant over time, its concentration will decrease as it gets used up. A Zero Order reactant will get used up at a constant rate. This is why we get a straight line. To calculate the rate of a reaction you need to find the gradient of the line. ΔY Rate of Reaction (mol.dm-3.s-1) = ΔX Concentration (mol.dm-3) Zero Order Time (s) You will not be expected to do this for 1st or 2nd Order reactants as they give a curve. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.9 RATE EQUATIONS 2. Using Reaction Data Here, you are given a table of experimental results that details how changing the concentrations of different reactants affects the rate of reaction. These are very common in exams. You must look at the data and deduce the order of reaction with respect to each reactant. A series of experiments is carried out for a reaction. Each experiment (or “run”) is carried out with different concentrations of reactants and the Initial Rate is measured. Example 1 A + B → C + D Experiment [A] [B] Initial Rate -3 -3 Number mol.dm mol.dm mol.dm-3.s-1 1 0.5 1.0 2.0 2 0.5 2.0 8.0 3 1.0 2.0 16.0 The trick here is to look for rows where the concentration of ONE reactant has changed, but the other stays the same. That way you can deduce the effect of that change of concentration has on the initial rate of reaction and deduce the Order. [A] In experiments 2 & 3, you can see that the [A] has been doubled from 0.5 to 1.0, and the [B] remains constant. The initial rate doubles from 8.0 to 16.0 So, if [A] has doubled and this causes the Initial rate to double, the Order with respect to [A] = 1st Order [B] In experiments 1 & 2, you can see that the [B] has been doubled from 1.0 to 2.0, and the [A] remains constant. The initial rate increases from 2.0 to 8.0. (x4) So, if [B] has doubled (x2) and this causes the Initial rate to increase by x4, the Order with respect to [B] = 2nd Order AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.9 RATE EQUATIONS Once you have deduced the orders with respect to each reactant, you can deuce the overall order of reaction. Overall Order = The Sum of the Orders of Reactants e.g. In the example on the previous page: Order w.r.t [A] = 1st Order w.r.t [B] = 2nd Therefore the overall order of reaction = 1+2 = 3rd order How To Deduce Orders Using Data More examples to work through … AQA www.chemistrycoach.co.uk © scidekick ltd 2024 AS CHEMISTRY 3.1.9 RATE EQUATIONS THE RATE EQUATION Now we know that different reactants have different impacts on the rate of the reaction we can use this information and build an equation quantify the rate. A + B → C + D The Rate Constant “k” (variable units) x y rate = k [A] [B] Initial rate of reaction Concentrations of each reactant (mol.dm-3.s-1) raised to the power of their order of reaction (1st or 2nd) e.g. Order w.r.t. [A] = x order Order w.r.t. [B] = y order Let’s say that we know that: The order with respect to [A] = 1st Order The order with respect to [B] = 2nd Order the rate equation becomes… 2 rate = k [A] [B] [A] is raised to the power “1”. The “1” isn’t shown in the equation, as anything to the power “1” is just the original value. [B] is raised to the power “2” as it has an exponential effect on the rate. Anything that has a zero order is not included in the equation. This makes sense, as if it doesn’t affect the rate of reaction, it isn’t used in the calculation. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.9 RATE EQUATIONS HINTS | TIPS | HACKS In exam questions it is likely that you will have to use data to deduce the orders of reaction first and then construct the rate equation for the reaction. Following on from this, you will likely be asked to calculate the rate constant “k” for the reaction. To do this… 1. Select any experiment (row of data) in the table where you have the concentrations for each reactant and the initial rate of reaction. 2. Rearrange the rate equation to make “k” the subject. rate k = 2 [A] [B] 3. Input the data from the row you selected and calculate k. You can use ANY row from the table of data and you should get the same answer for k as it is a “constant” for the reaction. You may also be asked to deduce the units for k. To do so, use the equation above and insert the units for rate (usually mol.dm-3.s-1) and concentration (mol.dm-3). From there cancel out the “mol.dm-3” where you can. How To Use the How To Deduce the Units Rate Equation of the Rate Constant k AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.9 RATE EQUATIONS THE ARRHENIUS EQUATION The rate equation shows how changes in concentration affect the rate of the reaction. However, we also know that there are two other major factors that can affect the rate of a reaction that are not included in the rate equation: Temperature Activation Energy (Ea) The Arrhenius Equation brings all of these things together: minus activation energy (J.mol-1) The Rate Constant “k” -Ea RT k= Ae Universal Gas Constant “R” (8.314 J.mol-1.K-1) The Arrhenius Constant “A” Temperature (Kelvin) AKA The pre-exponential factor Exponent “e” (J.mol-1) HINTS | TIPS | HACKS A is variable. It basically represents the frequency at which the reactants collide in a way that leads to a reaction. Don’t worry about the definition. You will either be given it or asked to calculate it. -Ea Essentially the part calculates the fraction of molecules that possess RT enough energy to react. A low Ea and a high RT means a faster reaction! AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.9 RATE EQUATIONS The exponential, e, in the Arrhenius equation makes it quite tricky to rearrange for calculations. So, taking the natural log (ln) of both sides, we can eliminate the exponential and make it easier to work with. -Ea lnk = + lnA RT You could be asked to use this equation to calculate k, A, Ea or T: -Ea lnA = lnk - RT Ea = RT (lnA - lnk) -Ea T= R (lnA + lnK) Once you have found lnk or lnA, enter “shift ln” on your calculator followed by the value you calculated for lnk or lnA. How To Manipulate the Arrhenius Equation AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.9 RATE EQUATIONS THE ARRHENIUS GRAPH One rearrangement of the Arrhenius equation gives a straight line graph. -Ea extract “T” -Ea 1 lnk = + lnA lnk = x + lnA RT R T y = m x 𝓍 +c This is the formula for a straight line graph Y intercept = lnA -Ea Gradient = R So, Ea = Gradient x R (J.mol-1) lnk 1/T Be sure to know your way around this How To Use the graph! Arrhenius Graph Be prepared to find: - lnA using the y intercept. - Ea by calculating the gradient and multiplying it by R. AQA www.chemistrycoach.co.uk © scidekick ltd 2024