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Unifying Concepts Bringing back quite a bit of How Far, How Fast?, it might be good revision to have a quick look through that topic before moving to Unifying Concepts. You should already know that: We can calculate this using a special type of equation called a rate equation. Rate equations are related to chemical equations. If we had the chemical equation: mA + nB > products, the rate equation would look like this:
For example, if we had I_{2} + 2S_{2}O_{3}^{2} > 2I^{} + S_{4}O_{6}^{2}, the rate equation would be:
Because there is one mole of I_{2}, we call it first order with respect to I_{2} and because there are two moles of S_{2}O_{3}^{2}, we call it second order with respect to S_{2}O_{3}^{2}. The total order of reaction, over all, is just a case of adding together the individual orders. 1+2=3. This reaction is third order overall. So, if we have k=100 mol^{2}dm^{6}s^{1}, initial concentration of 0.03 mol dm^{3} of I_{2} and 0.05 mol dm dm^{3} of S_{4}O_{6}^{2}, the rate would be:
If we rearrange the rate equation, we can calculate k quite easily. The units are more interesting. Rate, of course, is mol dm^{3}s^{1} and each concentration is mol dm^{3}. However, if we have [A][B]^{2} for example, we'd have mol dm^{3} and another the mol dm^{3} is squared to get mol^{2}dm^{6}. The mol dm^{3}s^{1} is divided by the bottom bit: The two mol dm^{3} cancel out, leaving us: Using basic rules of indices, we can move the bottom part to the top, to get:
Not too hard, huh? A question: KI and K_{2}S_{2}O_{8} react in the following equation:
The initial concentration of KI is 1.0 x 10^{2} mol dm^{3} and the initial concentration of K_{2}S_{2}O_{8} is 5.0 x 10^{4} mol dm^{3}. The initial rate of disappearance of K_{2}S_{2}O_{8} is 1.02 x 10^{8 }mol dm^{3}. Construct a rate equation and find the rate constant. The rate equation uses the products, in the above equation, and so is:
In order to find k, we can rearrange this to make k the subject of the formula: Now it's just a case of putting in the numbers to calculate k:
Next, we figure out the units: The mol dm^{3} on the top and the bottom cancel, leaving: mol^{1}dm^{3}s^{1}. So, put together, we get the answer: k = 2.04 x 10^{4} mol^{1}dm^{3}s^{1}. We've encountered various mechanisms over time. A rate mechanism is a description of a chemical reaction as a series of onestep processes. Some of these steps might be slower than others. The slowest step is known as the rate determining step because, as the reaction cannot go faster than the slowest step, it determines the rate of the reaction. How do we know what the rate determining step is? Through something called the rate law. This law states that the order of reaction with respect to a reactant indicates how many molecules of that reactant are involved in the rate determining step. For example, if the reaction is first order with respect to a reactant, there will be one mole of that reactant in the rate determining step. If it's second order, there will be two moles. If we take the following:
In the stoichiometric equation (fancy name for the equations we usually see), we see one mole of H_{2} reacts with two moles of ICl_{(g)}. However, if we look at the rate equation, we see that one mole of H_{2} reacts with only one mole of ICl. This means there must be a slower step in the overall equation that reacts with a ratio of 1:1. If the number of moles in the overall equation don't correspond to those in the rate equation, you know that there is a RDS step somewhere. From the above example, it could be:
It's possible to measure the concentration of reactants as a reaction occurs. There are two different methods, measuring the rate against the concentration and the concentration against the time. They produce quite different graphs, however, so you need to be clear of them in your head. The graphs are different, too, depending on whether the reactions are zero, first or second order, so are useful for distinction.
Half life can greatly vary, depending on reactions, from seconds, to minutes, to hours, to years. In order to calculate it on an exponential, use the following method:
The gradient of any concentrationtime graph gives us the rate of reaction. If a question asks us to find the rate of reaction at a particular point, we need to draw a triangle from the point on the graph: If we divide the height of this triangle by its width, we get the rate of reaction... easy! Another way of figuring out whether reactants in an equation are first, second or zero order is to use the initial rate method. This means that the initial rate is found in several experiments, using different concentrations. It's usual to just change one concentration per equation, and see how this affects the rate of reaction.
So, to take O_{2(g)} + 2ClO_{(g)} > 2ClO_{2(g)} you might be presented by a table like this:
Now, if you look at experiments 1 and 2, you'll see that ClO concentration is being increased and O_{2} is staying constant. ClO goes from 1.0 to 2.0, and so is doubled. If we look at the rate of disappearance, we see that it goes from 0.7 to 1.4, and so is also being doubled. The rate increases proportionally to the concentration, so the order of reaction for ClO is first order. If we consider experiment 2 and 3, we see that the O_{2} concentration is increased, while the ClO stays constant. The O_{2} goes from 1.0 to 2.0, and so is doubled. The rate goes from 1.4 to 5.6, and so is multiplied by four...or 2^{2}. This means that it is second order, because the rate change is the square of the concentration change. So, the overall rate equation will look like this:
We can find k by the usual process:
That's just about all you need to know for this section... get learning! K_{c} Now we've looked at the How Fast part of this topic, it's time to consider the How Far section. Everything in this section refers to equibriums and equilibrium positions. K_{c} is something called the equilibrium constant. It's easiest to explain, if we look at it in terms of an equilibrium's general equation: aA_{(aq)} + bB_{(aq)} cC_{(aq)} + dD_{(aq)}. From this, the equilibrium law states that: In other words, K_{c} are the products divided by the reactants.
The units for K_{c} vary, depending on the equilibrium in question. Each part is in mol dm^{3}, but if the reactant/product is squared due to two moles, for example, the part becomes mol^{2} dm^{6}. Take the following equilibrium, for example:
The K_{c} here would be: and the units would be: A question might give you some volumes of the gases and ask you to work out the value for K_{c}. For example: Calculate the K_{c} for
given that [H_{2(g)}]_{eq} = 0.53 mol dm^{3}, [CO_{2(g)}]_{eq} = 80.53 mol dm^{3}, [CO_{(g)}]_{eq} = 9.47 mol dm^{3}and [H_{2}O_{(g)}]_{eq} = 9.47 mol dm^{3}. The K_{c} equation is: Sticking in the numbers, we get: Times and divide these numbers and we get an answer of 2.1. This has no units because the mol dm^{3}s on the top are cancelled out by those on the bottom. If you've given initial concentrations rather than equilibrium concentrations, you need to figure out what the equilibrium concentrations are before you can work out the K_{c}. For example: 0.067 mol of SO_{3} is placed in a equilibrium vessel (volume of 1.52 dm^{3}) and allowed to reach the following equilibrium.
At equilibrium, the concentration of SO_{2} is 0.00142. Calculate K_{c} for the equilibrium. First off, draw up a table of what you have and fill in what you know:
Next, work out how many moles of SO_{3} have been used up. The reaction is 2mol>2mol, so 0.00142 must have been used. Take this away from the initial concentration to get 0.05668 mol. Then figure out how many moles of O_{2} have been produced. The reaction in this case is 2mol>1 mol, so divide 0.00142 by 2 to get 0.00071 mol.
In the original question we had a volume of 1.52 dm^{3}, so divide the moles by this to get the equilibrium concentrations:
These final figures are the ones we use in the K_{c} equation: Multiply these through and work out the units and you get the answer:
K_{c} only changes if the temperature changes. Although the equilibrium position may change with pressure, resulting in more or less yield, the K_{c} remains constant... (if you could explain this to me, I'd appreciate it!) So, what happens to K_{c} when the temperature changes? It depends on the equilibrium in question. If we think back to AS How Far, How Fast, we might recall Le Chatalier's Principal. Chatalier said that if there is a change to an equilibrium, the equilibrium will change position in order to oppose the change. If an equilibrium is exothermic in the forward direction and we increase the temperature, the reaction will move to the endothermic LHS, producing lots of reactants. Therefore, the K_{c} decreases. The reverse is true: if the equilibrium is exothermic in the forward direction and we decrease the temperature, the equilibrium will move to the RHS, with a high yield, and so K_{c} will increase. If, instead, an equilibrium is endothermic in the forward direction, an increase in temp will move it to the RHS with a higher yield, and so K_{c }will increase. If an equilibrium is endothermic in the forward direction, a decrease in temperature will move it to the LHS, decreasing the yield, and so decreasing K_{c}. Not too bad, hmm? Just remember that pressure doesn't affect it! K_{c} is a measure of equilibrium in aqueous solutions. However, for equilibriums that involve gases, it's often much more convenient to measure pressure rather than concentration. Instead of K_{c}, we use something called K_{p}, which means expressing the equilibrium in terms of partial pressures. Each gas has a partial pressure, which is the pressure exerted on that gas alone in the same volume as the mix of gases. The total pressure of the gas mixture (p_{total}) is the sum of all the partial pressures added together. p_{total} = p_{1} + p_{2} + p_{3} + .... To work out the partial pressure on a gas in the equilibrium, we use the following equation:
The mole fraction is the number of moles of that individual gas, divided by the total moles in the equation. As with K_{c}, we have the following equation for K_{p}: The partial pressures of the products (C and D) go on the top of the equation and the partial pressures of the reactants (A and B) on bottom. The number of moles of A, B, C, and D determine to what power they are raised. For example, the K_{p} of the equilibrium PCl_{3(g)} + Cl_{2(g)} PCl_{5(g)} would be: and the units would be kPa^{1}. So a question might look something like this: In the equilibrium N_{2}O_{4}_{(g)} 2NO_{2(g)}, there were 0.1 mol of N_{2}O_{4}_{} and 0.4 mol of NO_{2}. The total pressure was 100 kPa. Work out the partial pressures of each, and the K_{p}. Okay, so first off, we work out the mole fractions for both. There are a total of 0.1 + 0.4 mol of gas, so 0.5 mol. We divide each number of moles by this to get their mol fraction, and then times this by the total pressure:
Next, we add these into the kPa equation:
Finally, work out the units... (kPa)(kPa) / (kPa)... so one on the top and bottom will cancel, leaving kPa. Nothing to it! Acids and Bases This is quite an important topic, particularly for any biologists out there. If you think back to How Far, How Fast, you'll probably remember the BronstedLowry Theory of acids and bases:
For example:
Bases must always have at least one lone pair of electrons, in order to accept the H^{+} ion and form a dative covalent bond. In solutions, H^{+} can't really exist, because it is immediately attracted to the lone pairs on the oxygens in water, to form the H_{3}O^{+} ion. This is known as the oxonium ion or the hydroxonium ion. Water can act as an acid if a strong enough base is present, for example:
Because it can act as either acid or base, water is said to be amphoteric. Acid strength depends on how much an acid dissociates. For strong acids, we assume this is 100%. Weak acids only partially dissociate. The more the dissociation, the stronger the acid. In a reaction between two 'acids', the stronger will force the other to act as a base, such as in a nitricsulphuric equilibrium:
Why are some acids stronger than others? It's not enough to say that one is a stronger proton donor, and depending on the acid in question, there are two possible answers. When comparing phenol's acidity to ethanol's, there are two reasons why phenol is more acidic. First, the bond between the proton and the rest of the molecule is weakened. The electrons in the bond between OH are delocalised into the benzene ring, thus weakening the bond. Also, the anion is stable, as the charge delocalised, meaning it is more likely that it forms. Between HF and HCl, the strongest acid is HCl. The bond between H and F is very strong, meaning it is less likely to dissociate into its ions. Because the [H^{+}] concentration can vary widely, it's impractical to measure acid strength on concentration. Instead, we use the concentration in a log calculation, to result in an answer in a much smaller range (usually, pH=114, but be aware that negative pHs can be found, and alkalis stronger than pH=14.) In order to measure the pH of a strong acid, we must assume that the acid fully dissociates. We use the concentration of the acid in the following calculation (which you must learn!):
If you're ever asked to find the [H^{+}_{(aq)}] concentration from the pH, you need to use the reverse calculation:
So, if you have 0.01 mol dm^{3} HCl, simply put it into the log calculation:
Pretty straightforward, hmm? To make things more tricky, the exam question might say something like this: 25cm^{3} of 0.1 mol dm^{3} HCl was made up to 100 cm^{3} with water. What was the pH of the resulting solution? Here, we need to use the mol calculation for concentrations... So if we think back to AS Foundation, we might recall: First, we use this to work out the number of moles of HCl: (25 x 0.1) / 1000 = 0.0025 mol. Then we use this with the new volume in the mol equation, in order to calculate the new concentration. 0.0025 mol = ( M x 100cm^{3}) / 1000 2.5 = M x 100 M = 0.025 mol dm^{3} Now we have our concentration of [H^{+}] ions, we can put them into the log calculation:
There are shortcut methods to finding a new concentration, but if you're unsure of those, just stick to this old method. Although the log calculation for the pH is exactly the same for strong bases as acids, there's an extra step when you're faced with an aqueous solution of a base. The water in the solution ionises:
At 25^{o}C, the equilibrium lies to the left, and the ions on the right have equal concentrations of 10^{7} mol dm^{3}. As the concentration of H_{2}O is virtually unchanged, we can treat it as a constant. This means we can replace it with a new equilibrium constant, called K_{w}. So, just what is K_{w}?
and at room temp, K_{w} is 1 x 10^{14} mol^{2} dm^{6}. So how does this help us with a base equation? Quite simply, if we put the concentration of the base into the K_{w} equation, we can work out the concentration of [H^{+}] ions. Like the strong acids, though, we need to assume that the base fully dissociates. For example, if we have 0.01 mol dm^{3} of NaOH, we can use the K_{w} equation:
We can put in values for K_{w} and NaOH:
Rearranging, we get:
Now, stick this into the log formula:
And that's pretty much it for strong bases! As we know, weak acids are only partially dissociated. This means that the concentration won't equal the [H^{+}] concentration and we can't simply use the log equation to find the pH. Instead, we use something called K_{a}, the acid dissociation constant. K_{a} measures how much of the acid dissociates by dividing the concentration of undissociated, by the concentration of ions. If HA_{(aq)} H^{+}_{(aq)} + A^{}_{(aq)} represents a general equilibrium, K_{a} will look like this: The higher K_{a}, the higher the [H^{+}], so the stronger the acid. In order to calculate the pH using K_{a}, we need to make two assumptions:
From this, we can calculate pH. For example, if we have 0.1 mol dm^{3} of CH_{3}COOH, and the K_{a} = 4.8 x 10^{5}, we use the following method: Draw up a table of changes from the start to equilibrium. Call the unknown amount of ions x:
Next, we use the K_{a} equation and fill in what we know:
Now we've found x, the H^{+} concentration, we can use it in the log calculation:
Maybe weak acids take a little more thought, but that's all there is to it! Because K_{a} is really small, it's often easier to compare weak acids by using pK_{a}. This means sticking the K_{a} into another equation:
The values that come out work the opposite of K_{a}, meaning the smaller the pK_{a} value, the stronger the acid. If we're given a pK_{a} value, and we need to find the pH of the acid, we can convert pK_{a} back into K_{a} by using the following equation:
So, if we're given a K_{a} of 2.9 x 10^{3} for HA, the pK_{a} would be:
If we know the pK_{a} = 9.9, the K_{a} would be:
Nothing to it! Buffers Learn the proper definition of a buffer, not the one you might learn in biology: buffers are solutions that resist changes in pH. They are very important for anything biological, and we wouldn't be alive if we didn't have them inside of us! There are two main types of buffers: ones that use weak acids and ones that use weak bases. For both, you need equal concentrations of the acid/base and its salt. An example with CH_{3}COOH and CH_{3}COONa:
In the two equilibriums, we ignore the Na^{+} because it's just a spectator ion. We assume that the salt fully dissociates and that the acid doesn't really. So, there will be lots of CH_{3}COOH, from the first equilibrium and lots of CH_{3}COO^{} from the second equilibrium. If H^{+} ions are added (an acid is added), they react with the CH_{3}COO^{} to form CH_{3}COOH. Because the acid is only weak, it doesn't dissociate much, so the H^{+} concentration pretty much stays the same and the pH stays much the same...change is resisted. If OH^{} ions are added (an alkali), they react with the H^{+} to form H_{2}O. The weak acid equilibrium then moves to the right hand side to resist change (through Le Chatalier's principal) and restores the lost H^{+} ions, maintaining pH. For weak base reactions, let's take the following example:
As with Na^{+}, the Cl^{} is just a spectator ion, so we can ignore it. We assume the weak base doesn't really dissociate, but the salt dissociates completely meaning we have lots of NH_{3} from the first equilibrium and lots of NH_{4}^{+} from the second equilibrium. If H^{+} is added, this reacts with the OH^{} to form H_{2}O. The equilibrium is disrupted and thus moves to the RHS through Le Chatalier, to restore the balance. Thus, the OH^{} concentration is restored, and pH change is resisted. If OH^{} is added, it reacts with the NH_{4}^{+} to form NH_{3} + H_{2}O. Because NH_{3} is a weak base, it doesn't dissociate into many ions, so the OH^{} concentration stays roughly the same and pH change is resisted. To work out the pH of a buffer solution, we use the K_{a} equation; where the [HA] was before, we use the acid concentration, and where [A^{}] was before, we can replace it with the concentration of the salt. So: We can rearrange this: From this, we just use the old pH =  log_{10}[H^{+}] to work out the pH of the solution. For example: What is the pH of a buffer solution containing 0.02 mol dm^{3} of CH_{3}COOH and 0.05 mol dm^{3} of CH_{3}COONa, where the K_{a} = 1.7 x 10^{5} mol dm^{3}? We use the buffer equation:
Next, stick the [H^{+}] into the log equation:
Indicators We've been using indicators of various types for years, to find the pH or the acidity... but just what are they? They can be either weak acids or weak bases, that undergo a colour change when ions or undissociated. In represents the indicator and the equilibrium for a weak acid indicator may be written as:
The pH range over which an indicator changes colour is equal to pK_{a}=1. If the pH of the solution is more than one below the the pK_{a}, the colour of the indicator is colour I. If pH is at least one greater than the pK_{a}, then the colour is II. The colour change occurs, because if we flood with H^{+}, the equilibrium with move to the LHS to decrease the increase, appearing colour I. If we flood with OH^{}, these will react with the H^{+} to form water, and the equilibrium will move to the RHS to replenish the loss, appearing colour II. Two common indicators are phenolphthalein and methyl orange (fortunately, you don't need to know the names!). Phenolphthalein has a pK_{a} of 9.3. At less than pH=8.3, it will be colourless and at more than pH=10.3, it will be red. In the middle section, it will appear as a pale pink. Methyl orange has a pK_{a} of 3.7. At less than pH=2.7, it will appear red, and at more than 4.7, it will appear yellow. Between these two figures, it will be orange. In a titration, indicators can be used to show pH change, if the rapid change in pH coincides with the colour changes of the indicator. With a strong acid and strong base, the titration curve would look like this:
A weak acid and strong base might look like this:
A strong acid a weak base:
A weak acid and a weak base:
In order to work out when the sudden change of pH occurs, we do a simple moles equation. For example: When does the pH change occur if 25 cm^{3} of 0.01 mol dm^{3} HCl is titrated with an excess of 0.005 mol dm^{3} NaOH? We think about the equation: 1 mol of HCl reacts with 1 mol of NaOH. Next, we work out how many moles we have of HCl, using the moles in solution equation:
So, if we have 0.00025 mol of HCl, we will need 0.00025 of NaOH to neutralise. The NaOH has concentration of 0.005 mol dm^{3}, so we use the moles equation again:
The rapid change of pH will occur when 50 cm^{3} of NaOH is added to the HCl. The exam may throw any weird and wonderful indicators at you, but you don't need to learn any! All you need to know is how to use them, based on their pK_{a}s, and how to draw a titration curve for strong and weak acids and bases. That's it for Unifying Concepts... email if you think something's missing. All that's left to say is prepare for lots of moles equations, and good luck! 