OCR A Chemistry A-Level Module 5.1 Rates, Equilibrium and pH PDF

OCR A Chemistry A-level Module 5.1: Rates, Equilibrium and pH Detailed Notes This work by PMT Education is licensed under https://bit.ly/pmt-cc https://bit.ly/pmt-edu-cc CC BY-NC-ND 4.0...

OCR A Chemistry A-level Module 5.1: Rates, Equilibrium and pH Detailed Notes This work by PMT Education is licensed under https://bit.ly/pmt-cc https://bit.ly/pmt-edu-cc CC BY-NC-ND 4.0 https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc 5.1.1 How Fast? Orders, Rate Equations and Rate Constants Rates and Rate Equations The rate of a reaction shows how fast reactants are converted into products. It depends on the concentrations of the reactants and the rate constant. The rate of reaction is given by the rate equation: A+B→C+D Rate = k[A]m[B]n The constants m and n show the order of the reaction with respect to that species. This means that different species can have more of an effect on the rate of reaction than others. The values m and n can be 0, 1 or 2 - corresponding to zero order, first order or second order. The total order of reaction for this chemical reaction can be found as the sum of the separate orders. Total order = m + n The units for rate of reaction are mol dm⁻³s⁻¹. Rate Constant (k) The rate constant for a reaction is constant when the reaction temperature is constant. The rate constant relates the concentrations of the species that affect the rate of a reaction to the overall rate of reaction. The rate constant, k, can be calculated by rearranging the rate equation for that reaction. It has varying units depending on the number of species and their orders of reaction. The units of k can be found by substituting the relevant units into the rearranged equation and performing cancellations. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Rate Graphs and Orders The orders of reaction (that you need to know about at A-level) range from zero to second order. This means that changing the concentration of reactants can have different effects on the whole reaction: Zero Order - The concentration of this species has no impact on rate. - Shown on a rate-concentration graph as a horizontal line. - Rate = k First Order - The concentration of the species and the rate are directly proportional. - Doubling the concentration doubles the rate. - Rate = k[A] Second Order - The rate is proportional to the concentration of the species squared. - Doubling the concentration will increase the rate by four times. - Rate = k[A]2 https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Concentration-Time Graphs Reaction orders can be worked out by using rate-concentration graphs as shown above, but they can also be determined from the shapes of concentration-time graphs. These graphs can be generated by continuously monitoring the concentration of reactants during an experiment. Calculating the gradient of these graphs gives the rate. The concentration-time graph for a zero order reaction is linear: The concentration-time graphs for first order and second order reactions are curved: https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Initial Rates Using the initial rate of reactions is one way the order of a reaction can be determined. This involves varying the concentrations of reactants and measuring the initial rate of the reaction. Doubling the concentrations of zero, first and second order reactants would have the following effects: Zero order - No change to the initial rate. First order - Initial rate doubles. Second order - Initial rate quadruples (22). Example: Trial Initial [A] Initial [B] Initial [C] Initial rate (mol dm-3) (mol dm-3) (mol dm-3) (mol dm-3 s-1) 1 10 10 10 40 2 20 10 10 80 3 10 20 10 40 4 10 10 20 160 From this data, you can deduce that: A is a first order reactant B is a zero order reactant C is a second order reactant This would give the rate equation: Rate = k[A][C]2 Half-life Half-life (t1/2): The time taken for the initial concentration of the reactants to decrease by half. The half-life can be found from a concentration-time graph. The overall order of a reaction affects how the length of the half-life changes over the course of a reaction. First Order Reaction In a first order reaction, the half-life of a reaction is constant throughout the reaction. So the time taken for the reactant concentration to go from 100% to 50% is the same as the time taken for the reactant concentration to go from 50% to 25%, and so on. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Example: Half life of a first order reactant Experimental Techniques There are various experimental techniques that can be used to obtain rate data for reactions. This allows for the calculation of the overall order of reaction and the rate of reaction at given times. The two general ways this can be investigated is by: Measuring the change in a reactant mass or concentration over time. Measuring the change in a product mass or concentration over time. Collecting this raw data allows you to generate a concentration-time graph, mass-time graph or volume-time graph, which can then be used to calculate the rate of reaction. To find reactant orders and the overall order of reaction the concentration of reactants can be varied and their effects on the rate of reaction can be analysed. Mass Change If a gas is produced by a reaction, then the mass of the reaction mixture will decrease as the reaction proceeds. Plotting a mass-time graph and drawing a tangent to the curve can be used to find the rate of reaction. Volume of Gas Evolved If a gas is produced by a reaction, the rate of reaction can be found by measuring the volume of gas produced over the course of the reaction, and plotting a graph of volume evolved against time. A gas syringe or an underwater upside down measuring cylinder can be used to collect the gas. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Experiment setup: Titration Small samples of a reaction mixture can be removed at regular intervals throughout a reaction. These samples can then be titrated to determine the concentration of a given reactant or product at that time. A concentration-time graph can then be plotted. Colorimetry Colorimetry can be used to determine the rate of reaction for a reaction that involves the formation or depletion of a coloured species. A colorimeter is a device that measures the amount of light that is absorbed by a solution. The amount of light absorbed by the solution is proportional to the concentration of the coloured species. In a colorimetry experiment, a calibration curve is often generated. This involves using a colorimeter to measure the absorbance of solutions of known concentrations, from which a calibration curve is plotted. Then, throughout the experiment, the absorbance of samples from the reaction mixture can be measured and the calibration curve used to convert the absorbance readings into concentration values. A concentration-time graph can then be plotted. Example: Iodination of propanone The acid catalysed reaction of propanone and iodine can be monitored using colorimetry. CH3COCH3 + I2 → CH3COCH2I + H+ + I- https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc The initial solution is brown in colour due to the iodine present. As the iodine is used up in the reaction, the colour of the solution changes from brown to orange, to yellow and finally to colourless. The concentration of iodine can be found by continually taking samples of the reaction mixture and measuring the absorbance using a colorimeter. Rate Determining Step Not all stages of a reaction occur at the same rate, but the overall rate is determined by the slowest step of the reaction, also known as the rate determining step. Therefore, the rate equation contains all the species involved in the stages up to and including the rate determining step. The rate determining step can be identified from a reaction sequence by looking at which steps include the species in the rate equation. The rate determining step can also be used to predict the mechanism for the reaction. Example: In this question, step 2 would be the rate determining step as all the reactants of this step are in the rate equation given at the start. When constructing a reaction mechanism, the powers in the rate equation indicate the number of molecules of each substance involved in the slowest step and any steps before this. Any intermediates generated in the slowest step must be reactants in another step as they are not present in the balanced overall equation. Example mechanism: Nitrogen dioxide and carbon monoxide react to form nitrogen monoxide and carbon dioxide: NO2(g) + CO(g) → NO(g) + CO2(g) The rate equation for this reaction is: rate = k[NO2]2 - From the rate equation, the reaction is zero order with respect to CO(g) and second order with respect to NO2(g). - 2 molecules of NO2 are in the rate-determining step 1st step 2NO2(g) → NO(g) + NO3(g) (slow) nd 2 step NO3(g) + CO(g) → NO2(g) + CO2(g) (fast) Overall NO2(g) + CO(g) → NO(g) + CO2(g) https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Effect of Temperature on Rate Constants As the temperature increases, the rate constant increases and the rate of reaction increases. The Arrhenius Equation The Arrhenius equation shows how the rate constant, k, and temperature, T, are related exponentially: It is a very useful equation and the logged form can be used in the form ‘y = mx + c’ to show the relationship graphically. On a graph of ln(k) against 1/T, the gradient is -Ea/R which is negative and constant, and the y-intercept is ln(A): The above relationship shows how the activation energy for a reaction (the minimum energy required for two particles to react) can be found graphically using experimental methods and data. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc 5.1.2 How Far? Equilibrium Mole Fractions and Partial Pressure A mole fraction shows the proportion that a molecule accounts for of the total moles present. It is calculated by dividing moles of substance A by the total moles present. Within a gaseous system, each gas has a partial pressure. The partial pressures add up to give the total system pressure. The partial pressure of a substance is found using the mole fraction of that substance and the total pressure of the system. Partial pressure of A would be shown as (PA). Example: Total moles in the system = 0.51 + 0.28 + 0.52 = 1.31 moles Mole fraction of H2 = 0.28 / 1.31 = 0.21 (PH2) = 0.21 x 2.35 = 0.50 atm Partial pressures are commonly measured in Pascals but are occasionally measured in atmospheres. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Calculating Kc Kc is the equilibrium constant of a reversible reaction. Kc is equal to the concentration of the products divided by the concentration of the reactants at equilibrium. The concentration terms are raised to a power of the same value as the number of moles of that substance. Example: The Kc expression The equilibrium constant has varying units, depending on the chemical reaction. The units can be calculated by substituting the concentration units into the Kc expression. Some of these units will then cancel, giving the overall units of Kc for that reaction. Example: Finding the units of Kc Homogeneous and Heterogeneous The equilibrium constant, Kc, can be found for both homogeneous and heterogeneous reactions. Homogeneous reactions are reactions in which the reactants and products are in the same phase, whereas heterogeneous reactions are reactions in which some of the reactants and/or products are in different phases to each other. For homogeneous reactions, Kc is calculated as shown above. The difference when calculating Kc for heterogeneous reactions is that any terms representing a solid are not included in the calculation. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Example: H2O(g) + C(s) → H2(g) + CO(g) The solid carbon is not included in the equation for Kc: Gaseous Equilibrium Constant (Kp) Kp is the equilibrium constant used for gaseous equilibria. Kp is calculated from gaseous reactants and products. If all reactants and products are in the gaseous state, the system is said to be homogeneous. Calculating Kp Partial pressures allow the value of Kp for a gaseous equilibrium to be found. Kp is equal to the product of the partial pressures of products over the product of the partial pressure of reactants. It is similar to Kc in that any variation in moles raises the partial pressure to a power of equal quantity to the number of moles. The gaseous equilibrium constant has varying units, depending on the chemical reaction. The units can be calculated by substituting the partial pressure units into the Kp expression. Some of these units will then cancel, giving the overall units of Kp for that reaction. Example: Finding the units for Kp https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Homogeneous and Heterogeneous Similarly to Kc, the gaseous equilibrium constant, Kp, can be found for both homogeneous and heterogeneous reactions. For homogeneous reactions, Kp is calculated as shown above. The difference when calculating Kp for heterogeneous reactions is that any terms representing a solid are not included in the calculation. Example: CaCO3(s) ⇌ CaO(s) + CO2(g) The solid CaCO3 and CaO are not included in the equation for Kp: Factors Affecting Kc and Kp The values of Kc and Kp are not affected by concentration or pressure change or by the use of a catalyst. However, they are affected by changing the reaction temperature. Concentration and pressure changes and the addition of a catalyst affect the rate of the reaction (the kinetics) but not the position of the equilibrium. They only affect how fast the system reaches equilibrium, hence they have no impact on the equilibrium constant. Temperature, on the other hand, does change the position of the equilibrium, resulting in different concentrations of reactants and products. If the forward reaction is exothermic, an increase in temperature will decrease the rate of the forward reaction because the equilibrium shifts to the left to oppose the change and favour the reverse endothermic reaction. This will decrease the concentrations of products and increase the concentrations of reactants, and therefore the equilibrium constant (Kc or Kp) decreases. If the forward reaction is endothermic, an increase in temperature will increase the rate of the forward reaction because the equilibrium shifts to the right to oppose the change. This will increase the concentrations of products and decrease the concentrations of reactants, and therefore the equilibrium constant (Kc or Kp) increases. Similar arguments can be made for the effect of decreasing the temperature. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc 5.1.3 Acids, Bases and Buffers Brønsted–Lowry Acids and Bases Acid-base equilibria involve the transfer of protons between substances. Therefore, substances can be classified as acids or bases depending on their interaction with protons. A Brønsted-Lowry acid is a proton donor. For example, ammonium ions (NH4+). A Brønsted-Lowry base is a proton acceptor. For example, hydroxide ions (OH-). Brønsted–Lowry conjugate acid-base pairs A conjugate acid is the species formed when a base accepts a proton. A conjugate base is the species formed when an acid donates a proton. Conjugate acids and conjugate bases form conjugate acid-base pairs. Example: Ionic Equations An ionic equation shows the reacting ions in a chemical equation. Spectator ions are ions that do not change in the reaction and are left out of the ionic equation. It is important to balance elements and charge in ionic equations. In the reactions of acids with carbonates, metal oxides, and alkalis, H⁺ is the reacting ion. Example: 2H⁺(aq) + CO₃²⁻(aq) → H₂O(l) + CO₂(g) https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Acid and Base Strength Acid strength doesn’t refer to the concentration of a solution. A strong acid is defined as being: An acid that completely dissociates into its ions when in solution. Example: In comparison, a weak acid is defined as being: An acid that only slightly dissociates into its ions when in solution. Example: The same definitions are true for strong and weak bases. Strong acids have pH between 0-1 and weak acids have pH between 3-7. Strong bases have pH between 12-14 and weak bases have pH between 7-11. The Acid Dissociation Constant Weak acids and bases only slightly dissociate in solution to form an equilibrium mixture. Therefore, the reaction has an acid dissociation constant, Ka. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc The constant Ka can be found using pKa: The value pKa is a logarithmic acid dissociation constant, representing how acidic something is. A low value of pKₐ, or equivalently a large Kₐ, indicates a strong acid. pH and [H⁺(aq)] Determining pH pH is a measure of acidity and alkalinity. It is a logarithmic scale from 0 to 14 that gives the concentration of H+ ions in a solution. 0 is an acidic solution with a high concentration of H+ ions whereas 14 is a basic solution with a low concentration of H+ ions. pH is a specific example of pKₐ for when H⁺ ions are present. It can be calculated using the concentration of hydrogen ions, [H+], as follows: This equation also allows the concentration of H+ ions to be determined if the pH is known. When using these equations above, the concentration of H+ ions is given in mol dm-3. This concentration of H+ ions is equivalent to the concentration of a strong acid as it completely dissociates to ions in solution. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Diluting Acids If you dilute a strong acid 10 times its pH will increase by one unit, because pH is a logarithmic scale. Diluting it 100 times and 1000 times would, therefore, increase the pH by two units and three units, respectively. Weak acids do not behave in the same way. Weak acids are not fully dissociated in solution, so diluting them causes the equilibrium to shift to oppose the change. This means a 10x dilution of a weak acid would increase the pH by less than one unit. Ionic Product of Water Water slightly dissociates to form hydroxide and hydrogen ions in an equilibrium with its own equilibrium constant, Kw. At 25oC, room temperature, Kw has a constant value of 1.0 x 10-14. However, as temperature changes, this value changes. The forward reaction in the equilibrium of water is endothermic and is therefore favoured when the temperature of the water is increased. As a result, as temperature increases, more H+ ions are produced meaning the water becomes more acidic. H2O ⇌ H+ + OH- In the same way that pKa can be calculated from Ka, pKw can be calculated from Kw. The pH of a strong base can be calculated using pKw or Kw. For a strong base, the concentration of OH- will be the same as the concentration of the base. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Calculations The relationships of Ka, pKa and [H+] can be used to find the pH of weak acids and bases. Depending on the reaction and the relative concentrations, a different method may have to be used: HA in excess - Use [HA] and [A-] along with Ka to find [H+], then pH. A- in excess - Use Kw to find [H+], then pH. HA = A- - In this case, pKa is equal to pH, therefore find pKa. Calculating the Ka of Weak Acids Weak acids only partially dissociate in solution. Therefore, the equilibrium of a weak acid has to be taken into account. 1. The initial concentrations, change in concentrations and equilibrium concentrations of the reactants and products have to be found. 2. The concentration of H+ ions can then be found using the pH given. 3. This value can be used to find the actual equilibrium concentrations. 4. Finally, these values can be substituted into the expression for Ka. Example: A weak acid, HA, with a concentration of 0.25 M has a pH of 3.5. What is its Ka? HA + H2O ⇌ A- + H3O+ We assume that ‘x’ mol of HA dissociates to A- and H3O+. HA A- H3O+ Initial 0.25 0 0 Change -x +x +x End 0.25 - x x x Using the pH value give, the [H+] (= x) can be calculated: https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Substituting into the expression for Ka and using the equilibrium concentrations: We can now substitute in the value of x calculated earlier: x = 3.16 x 10-4 M Approximations For weak acid calculations, the following approximations are made: [HA]equilibrium ~ [HA]undissociated i.e. [HA] ≫ [H + ] [HA]equilibrium ~ [A− ]equilibrium i.e. negligible dissociation of H₂O These approximations are limited to calculations for weak acids. For stronger weak acids, the approximations breakdown because the first assumption may no longer be valid. Buffers: Action, Uses and Calculations Buffer Action A buffer solution is a system that minimises pH changes on addition of small amounts of an acid or a base. It is formed from a weak acid and its salt or an excess of a weak acid and a strong alkali. This produces a mixture containing H+ ions and a large reservoir of OH- ions which helps to resist any change in pH. Therefore, a buffer solution is defined as: A solution which is able to resist changes in pH when small volumes of acid or base are added. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc The large reservoir of OH- ions allows the ratio of acid to base in the mixture to be kept almost constant. Consider the following buffer solution: NH3 + H2O ⇌ NH4+ + OH- The OH- concentration will increase if a small amount of base is added, making the solution more basic. The extra OH- ions will react with the NH4+ ions, to form the original reactants. Therefore, the equilibrium will shift to the left to remove the OH- ions and stop the pH from changing largely. Buffer Calculations Buffer calculations are long calculations that use acid-base calculations. There are two types: Acid + Base - Find the number of moles of each species. - Calculate their concentration when at equilibrium using the total volume. - Use Ka to find [H+] and therefore pH. Acid + Salt - Find the moles of the salt. - Use Ka to find pH. Example: A buffer solution contains 0.35 mol dm-3 methanoic acid and 0.67 mol dm-3 sodium methanoate. For methanoic acid, Ka = 1.6 x 10-4 mol dm-3. Find the pH of this buffer. We assume that the sodium methanoate completely dissociates so that the equilibrium concentration of HCOO- is the same as the initial concentration of HCOO-Na+. Similarly, HCOOH only slightly dissociates so we assume that the equilibrium concentration is equal to the initial concentration. 1. First, find the expression for Ka for methanoic acid HCOOH ⇌ H+ + HCOO- 2. Rearrange the expression to find [H+] [H+] = 1.6 x 10-4 x (0.35/0.67) = 8.4 x 10-5 3. Convert [H+] to pH pH = -log10(8.4 x 10-5) = 4.08 https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Adding Small Volumes The pH of a buffer solution doesn’t change much but will change in the order of 0.1 or 0.01 units of pH when a small volume of acid or base is added. Adding small amounts of acid (H+) increases the concentration of the acid in the buffer solution, meaning the overall solution will get slightly more acidic. Adding small amounts of base (OH-) decreases the concentration of acid in the buffer solution, meaning the overall solution will get slightly more basic. Uses of Buffers Buffer solutions are common in nature in order to keep systems regulated. This is important as enzymes or reactions in living organisms often require a specific pH, which can be maintained using a buffer solution. Another important buffer in nature is found in the human circulatory system. The pH of human blood is maintained in a buffer between carbonic acid and hydrogencarbonate ions. These ions neutralise any acidic substances that enter the bloodstream, converting them to carbonic acid and water. This buffer is present in blood plasma, maintaining a pH between 7.35 and 7.45. Neutralisation Enthalpy Change of Neutralisation When strong acids and bases react together they will produce very similar enthalpies of neutralisation. This is because the solutions completely dissociate, so the same, simple acid-base reaction occurs between H+ and OH- ions to produce water in each case. The other dissociated ions present are simply spectator ions and do not affect the reaction. However, in reactions of weak acids and bases, the ions only slightly dissociate so other enthalpy changes also occur within the solution. As a result, the enthalpies of neutralisation can vary quite a lot. Titration Curves A pH titration curve shows how the pH of a solution changes during an acid-base reaction. When an acid and base react, a neutralisation point is reached, identified as a large vertical section of the pH curve, through the neutralisation/equivalence point. To obtain a pH titration curve, alkali is slowly added to an acid (or vice versa) and the pH is regularly measured with a pH probe. The smaller the added volumes, the more accurate the curve produced. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Example: The red dot indicates the equivalence point For a strong acid - strong base reaction, the neutralisation point occurs around pH 7. Other combinations of strong and weak acids and bases results in a different neutralisation point: Strong Acid + Strong Base = pH 7 Strong Acid + Weak Base = < pH 7 (more acidic) Weak Acid + Strong Base = > pH 7 (more basic) Weak Acid + Weak Base = normally pH 7 but hard to determine https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc Calculating Ka from Titration Curves The vertical region of a titration curve is the equivalence point. At half the equivalence point the pH is equal to pKa, by definition. Therefore, by reading the pH at half the equivalence point, Ka can be easily calculated from pKa. Indicators Specific chemical indicators have to be used for specific reactions as they can only indicate a pH change within a certain range. The two most common indicators used at A-level are methyl orange and phenolphthalein: Methyl Orange - used for reactions with a more acidic neutralisation point. - red in acids and turns yellow at the neutralisation point. Phenolphthalein - used for reaction with a more basic neutralisation point. - pink in alkalis and turns colourless at the neutralisation point. Indicator pH at colour change Colour in acid Colour in base Methyl orange 3-5 Red Yellow Phenolphthalein 8-10 Colourless Pink Litmus 5-8 Red Blue It is important that, depending on the strength of the titration reactants, the correct indicator is selected so that the colour change occurs at neutralisation. The pH at colour change must fall within the vertical section on the titration curve. Indicators are weak acids themselves, so only a few drops are used otherwise they could affect the overall pH. They should be considered as HA. The colour change occurs due to an equilibrium shift between the HA and A⁻ forms of the indicator. https://bit.ly/pmt-cc https://bit.ly/pmt-edu https://bit.ly/pmt-cc

OCR A Chemistry A-Level Module 5.1 Rates, Equilibrium and pH PDF

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