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Levi Howard Jordan Leach. In a Reaction When a reaction occurs, eventually the molarity (concentration) of the reactants and products will become constant,

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Presentation on theme: "Levi Howard Jordan Leach. In a Reaction When a reaction occurs, eventually the molarity (concentration) of the reactants and products will become constant,"— Presentation transcript:

1 Levi Howard Jordan Leach

2 In a Reaction When a reaction occurs, eventually the molarity (concentration) of the reactants and products will become constant, not necessarily equal. This determines when the reaction is “over”, however the reaction never truly stops. This is demonstrated below.

3 In the Reaction Which of the following on the graph shows when the reaction has reached Equilibrium? C A B

4 Equilibrium Cont. The correct answer is C, because it is the only point at a straight horizontal line. B shows where the two concentrations are equal, but that has nothing to do with equilibrium. A is just… wrong.

5 Equilibrium Constants (K) There are two very important constants that go with equilibrium, and these two are Kc and Kp. Kc is for using concentrations, which is associated with most problem, but some will have the partial pressures for compounds, where you would use the pressure constant, Kp. If these K values are large, that means that the reaction is spontaneous. If it is small, the reverse reaction would be spontaneous. K also only depends on temperature, T. Knowing which type of K to use helps to organize information and use the correct equations. Since reactions can be revered, the constant can be also. If a reaction is reversed, you would use the reciprocal of the normal K value. If it is multiplied by a number, the original K will have an exponent of that number.

6 Equilibrium Expression (K) Writing an equilibrium expression is simple. The important thing to remember is to write Products over Reactants equals the constant, K. Phases are important, because for equilibrium expressions, the elements must be in either the gas phase or aqueous (aq). Solids and liquids are removed from the expression completely. If you are using Kc, then you would put in the equilibrium concentrations for the compounds in their respective spot. If you are using Kp, then write products over reactants, and plug in the partial pressures at equilibrium that are given.

7 Equilibrium Expressions Cont. After taking note of the phases, and concentrations (or partial pressures) of the reactants and products, be sure to look at the coefficients (the big number before the chemical formula). This coefficient decides the exponent of the compound. For example, in the reaction below, N2 + 3H2 2NH3 You notice the 3 before hydrogen, and 2 before ammonia. This just means that in an expression, H2 is cubed, and NH3 is squared. Since N2 technically has a 1, it is to the first power which means it isn’t changed.

8 Writing Expressions When 0.40 mole of SO2(g) and 0.60 mole of O2(g) are placed in an evacuated 1.00-liter flask, the reaction represented below occurs. After the reactants and the products reach equilibrium and the initial temperature is restored, the flask is found to contain 0.30 mole of SO3. Based on these results, the equilibrium constant, Kc, for the reaction is… 2 SO2(g) + O2(g) 2 SO3(g) A (o.30)^2 / [(o.45)(0.10)^2] B (0.30)^2 / [(o.60)(0.40)^2] C(2 x 0.30) / [(0.60)(0.40) ^2] D(0.30) / [(0.45)(0.10)]

9 Expressions Cont. The answer is A. This is because in the reaction, 0.30 mol SO3 was found. SO3 also had a 2 as a subscript, so this amount would be squared in an expression. This means that the answer is going to have [0.30^2] on top. This means C and D cant be the answer. Also, when a reaction happens, reactants will be used up. In B, the concentrations of the reactants are the exact same as before the reaction happened, which is impossible. This leaves A as your answer.

10 Expressions Cont. Hydrofluoric acid, HF, dissociates in water as represented by the equation below. Write the equilibrium constant expression for the dissociation of HF in water. HF(aq) + H2O(l) H3O+(aq) + F-(aq)

11 Expressions Cont. The first thing to look for is the phases of all the reactants and products. You will notice H2O as a liquid, which means it will not be used in our expression. This leaves H3O and F- on top, over HF on the bottom. [H3O+] [F-] ___________ [HF]

12 Is it done yet? You now remember that Equilibrium is when the concentrations are remaining constant. But how exactly can you tell when it is when you aren’t looking at a graph? Well, this is where Q comes in. Q is calculated just like K, by putting products over your reactants. However, with Q, you use whatever concentrations you have, which aren’t necessarily at equilibrium. To know if it is at equilibrium, Q would be equal to you K value, because the numbers you used for Q in fact were the equilibrium concentrations. To put it simply… Q = Products over Reactants for any concentration QK means it is not at EQ, need to make more reactants.

13 Organizing with ICE Tables ICE is an acronym. These tables are great for organizing information in an equilibrium reaction by showing the change in concentration as the reaction goes on to equilibrium. Like equilibrium expressions, liquids and solids are omitted from ICE tables. It stands for… I – Initial concentration. This is before the reaction starts. C – The change in concentrations. This is usually where an unknown variable (x) is put into. All changes in concentrations are proportional to the coefficients. E – The concentration of the atom (or compound) at Equilibrium.

14 Ice Tables in Action Hydrofluoric acid dissociates in water according to the reaction below. If the Ka of hydrofluoric acid is 7.2 x 10^-4, calculate the molar concentration of H3O in a.40M HF solution. HF(aq) + H2O(l) H3O+(aq) + F-(aq)

15 Ice Tables Cont. The ICE table shown below is a great way to organize information. Hydrofluoric acid, HF, was given to be 0.4M, so that will be put in the box for initial concentrations. It shows that the products start out at 0, because they havent been made yet. After the equation a certain amount, x, has been added to the product’s size, while being taken away from the reactant side. To find the molar concentration, we need to find the value for x. This can be done by putting all of this information into an equilibrium expression. [x] [x] _______ = 7.2 x 10^-4 0.4M 0M 0M [0.4] -x +x +x x^2 = 2.88 x 10^4 x = 0.0170M 0.4-x x x

16 Le Chatelier’s Principle The scientist Henry Louis Le Chatelier discovered how to change equilibrium in a reaction. This is possible by doing a number of things, such as changing the concentration, temperature, volume, or partial pressure. Changing one of these will cause the reaction to shift left or right, having a larger concentration of either products or reactants to counteract the change.

17 Le Chatelier In the exothermic equation below, HCO3(ap) + OH(aq) H2O(l) + CO3(aq) ΔH=-41.4kJ Raising the temperature would shift the reaction how? A. Left B. Right C. It would not shift

18 Why A? If a reaction is exothermic, it means that as it goes on, heat is produced. In other words, heat is a product. Raising the temperature would mean adding a product, which would make the equation want to make more products, and have the equation shift left.


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