Chemical Equilibrium Unlike most of the reactions we have been looking at until this point, many reactions occur in both directions, rather than having a clear forward direction with obvious reactants and products. If a reaction can progress in both directions, there is usually nothing that stops the reverse reaction from happening at the same time. Instead of using an arrow ( → ) to indicate the → or reaction, we will typically write a double arrow (either ← ) to show a reversible reaction. CH 4( g ) + 2O 2( g ) → 2H 2 O( g ) + CO 2( g ) A non-reversible reaction → PCl3(g) + Cl2(g) PCl5(g) + heat ← A reversible reaction A chemical equilibrium happens when the forward rate of reaction (turning reactants into products) matches the rate of the reverse reaction (turning products into reactants). When this happens, we cannot see any further progress in the reaction on the large scale. In reality, the chemicals are still reacting, but since the products and reactants are converting at the same rate, nothing appears to be happening. A good way to visualize what is happening is to think of yourself as stacking bricks to make a wall. Suppose that as you stack bricks, someone else is unstacking them, but at a lower rate. As the wall grows taller, you slow down and the person taking down the wall speeds up. At some point, you will have the same rates. When that happens, the wall does not get any bigger or smaller. If you were far away, it would look like the wall stayed at a constant height, but up close you could see bricks being added and bricks being taken away. This is exactly what happens at equilibrium. The rate of products turning into reactants equals the rate of reactants turning into products. Nothing seems to be happening when we look at the reaction container, but in reality, at a molecular level, both reactions are still happening. If you could shrink down and get in with the molecules, you would see both reactions happening. For this reason, it is refered to as a dynamic equilibrium. Dynamic just means active or in motion, so it is a way of describing that the reaction is still occurring. The fact that the reaction is still happening is very important for a later concept (Le Chatlier’s Principle). Since the equilibria are based on the rates of reactions, they are also tied to temperature (recall that the rate of any reaction increases as temperature increases). This means that each reaction has a different constant of equilibrium at each temperature represented by Kx , where x represents the type of equilibrium. This also means that at a given temperature, the equilibrium expression has a constant value. That is the key to all of these calculations; when at a given temperature, the value for Kx is a constant. This value is in general generated by taking the concentration of the products to the power of their respective coefficients, and dividing by the concentration of the reactants raised to the power of their respective coefficients as in the example below. → 3 H3PO4(g) + PH3(g) 4 H3PO3(g) ← [ H PO ] [ PH3 ] Kc = 3 4 4 [ H3PO3 ] 3 Anything in brackets represents a concentration, and that means only gases or solutions can be expressed in this equation. Solids and pure liquids whose concentrations do not change are not represented in this expression because they have constant values. Also, even though the concentrations are molar (M or mol/L), the value of the equilibrium constant is always has no units (mainly to keep things simple). The reason for the use of concentrations in the expression is because when equilibrium is reached, the concentrations are no longer changing. Since the concentrations are not changing, the value of the K must be a constant. Even if we change one value, the others shift concentration in response to reach equilibrium again (we will discuss this process more later – it is called Le Chatlier’s Principle). Kc is the generic equilibrium constant for any equilibrium. Kp is the generic equilibrium constant for any gaseous equilibrium expressed in pressure rather than in concentrations (pressure is directly proportional to concentration). Ka is the equilibrium constant for acidic reactions. Kb is the equilibrium constant for basic reactions. Ksp is the solubility product constant used in largely insoluble ionic salt solutions. Kf is the constant of formation of coordination complexes in solution. Summary: • • • Equilibrium is where the forward rate and the backward rate are equal o This is referred to as dynamic equilibrium because the reaction is still happening, even though we can’t see it. [ products] Equilibrium expressions are written as follows: K c = [ reactants] o Brackets means concentration (in molarity) o Coefficients in the equation become powers in the equilibrium expression. Only gases and aqueous solutions are used in equilibrium expressions. Equilibrium Terminology Equilibria (which is the plural of equilibrium) can occur with substances in different phases, as we implied in the previous section. If an equilibrium reaction occurs all in one phase (all gases, for example) this is referred to as a homogeneous equilibrium. If the equilibrium occurs in multiple phases (gases and aqueous, for example) it is referred to as a heterogeneous equilibrium. Homogeneous Equilibria Examples: Heterogeneous Equilibria Examples: → CH4(g) + H2O(g) CO(g) + 3H2(g) ← → Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g) ← → H2(g) + I2(g) 2HI(g) ← → Ni(s) + 4CO(g) Ni(CO)4(g) ← → CH4(g) + H2O(g) CO(g) + 3H2(g) ← → NH3(g) + H2S(g) NH4HS(s) ← Equilibrium Calculations: Part 1 Calculating equilibrium constants is actually quite easy. If we want to know the equilibrium constant, all we have to do is write the equilibrium expression and plug in the concentration values for the reactants and products. We have to be careful with our calculators when performing these calculations – make sure that you put the entire denominator in parentheses when you enter it into your calculator, or your calculations will never come out right. Example: Find the constant of equilibrium for the equation below, given that the equilibrium concentrations are as follows: [HI] = 0.250 M, [I2] = 0.175 M, [H2] = 0.175 M → H2(g) + I2(g) 2HI(g) ← Kc = [ I2 ][ H 2 ] 2 [ HI] All chemicals involved are gasses, so they are all in our equilibrium expression, and the coefficients are powers. Kc = ( 0.175)( 0.175) 2 ( 0.250 ) Substitute in the molarities. K c = 0.490 Note that the constant has no units. Of course, if we are given the equilibrium constant and some of the molarities, we can solve for the remaining one. Example: Given the Kc for the equation below at 298 K is 0.116. If the equilibrium concentration of NO2 is 0.020 M, what is the equilibrium concentration of N2O4? → 2NO2(g) N2O4(g) ← [ NO2 ] Kc = [ N 2O4 ] 2 0.020 ) ( 0.116 = [ N 2O4 ] 2 0.020 ) ( [ N 2O4 ] = 2 0.116 [ N 2O4 ] = 3.4 ×10−3 M There are a lot of different ways these problems can be presented, and various algebraic techniques might have to be used to solve them at the college level, but these nuances are not typically tested on the AP Exam. Just be aware that in a college class, you may be asked to go into much greater depth with the algebraic techniques. I.C.E. Table Review: Equilibrium calculations can get quite messy, so we need an organizational strategy to help us. The problems get difficult if we are given an initial value before equilibrium is reached. To set up those problems, we use what is called an I.C.E. table. That stands for Initial, Change, and Equilibrium concentrations. These are most easily explained by example. Remember, all K calculations must be done in terms of Molarity (mol/Liters, M), unless we are doing a Kp calculation, which must use pressure in atm, instead. Occasionally, we will be given amounts of moles and the size of the reaction vessel. We need to remember that Molarity moles = and use that to find the concentrations L If the K is very small compared to the concentrations, the problems get much easier as we can ignore the variable that is subtracted (because it is insignificant compared to the concentration). Example: → H+(aq) + CN-(aq) HCN(aq) ← The Ka for the reaction above is 4.9 ×10−10 . The initial concentration of HCN = 1.50 M. Find the [H+]. We first notice the Ka is very small, and so the substance will be mostly reactants at equilibrium. This makes the problem much easier. → H+(aq) + Concentrations (M) HCN(aq) CN-(aq) ← Initial 1.50 0.0 0.0 Change –x +x +x . Equilibrium 1.50 – x x x H + CN − Ka = [ HCN ] 4.9 ×10−10 = x2 (1.50 − x ) 4.9 ×10−10 = x2 (1.50 − x ) x2 1.50 = x2 4.9 ×10−10 = 7.35 ×10 −10 Since the x is very, very small, we cross it out because it is insignificant compared to the 1.50. We must write the following justification: x << 1.50 Solve for x, and we get that H + = 2.71× 10−5 M x = 2.71×10−5 Notice that our answer, H + = 2.71×10−5 M , is so small that if we subtract it from our initial value of 1.50 M, we would still get 1.50 M when we use significant figures. This is why we could ignore the subtracted x. Important Things to Remember: Several things that you must remember: • You will know it is an equilibrium by the double arrows in the chemical equation (and you must include this if you are asked to write a chemical equation involving an equilibrium) • Brackets means concentration (in Molarity, or moles/Liter of solution) • Only gases and aqueous substances are represented in equilibria (as these are the only things that can be expressed in concentrations). • Coefficients in the chemical equation become exponents in the equilibrium expression, and it is always products over reactants. • Gases are commonly put into equilibria expressions as pressures rather than as concentrations (the ideal gas law relates concentration directly to pressure, and many subsequent calculations only work with gaseous equilibria expressed in pressure). Note: This is not an absolute rule, gases can be expressed as concentrations as well, but it should be indicated in the problem, often by the label, Kp (the p representing pressure). Common Issues on the AP Test Rules for Equilibria: • Temperature changes will also change the value of K, the equilibrium constant. o This is because the temperature affects the rate of the reaction as well as the amount of products and reactants at equilibrium (depending on whether the reaction is exothermic or endothermic) o This is why every AP exam will list a temperature value for the equilibrium. You will not need this temperature for any part of the equilibrium calculations (though you might need it in a multi-part problem where they ask a non-equilibrium question that needs temperature). • Pressure changes will cause a gaseous equilibrium to shift towards the left or right, precisely because the K value is constant (so the pressure values shift to keep the K value the same) • Concentration changes will cause an aqueous equilibrium to shift towards the left or right, precisely because the K value is constant (so the concentration values shift to keep the K value the same) • If you write a chemical equation in the reverse, the value for the K for the new reaction is simply the reciprocal (1/K) of the original. • The value for the equilibrium constant for two or more chemical equations that are added together is the product of the K values of the individual equations. • Equilibria that have all products and reactants in one state are said to be homogeneous. If you have an equilibrium with substances in differing states, it is a heterogeneous equilibrium (remember, solids and liquids do not show up in the equilibrium expression, K).