Equilibrium
equilibrium • Many chemical reactions are reversible: • Reactants → Products or Reactants ← Products Reactants form ProductsProducts form Reactants • For example, under certain conditions, one mole of the colourlessgas N2O4 will decompose to form two moles of brown NO2 gas: • N2O42 NO2 • Under other conditions, you can take 2 moles of brown NO2 gas and change it into one mole of N2O4 gas: N2O42 NO2
equilibrium • In other words, this reaction is as written can go forward or in reverse depending on the conditions If we were to put some N2O4 in a flask, the N2O4 molecules would collide with each other some of them would break apart to form NO2 This is called a FORWARD reaction N2O4 2 NO2
equilibrium • Once this has happened for awhile, there is a build up of NO2 molecules in the same flask Once in awhile, two NO2 molecules will collide with each other and join to form a molecule of N2O4 ! This is called a REVERSE reaction: N2O4 2 NO2
Equilibrium • 2 things to notice: • 1. as long as N2O4 present, the forward reaction will keep on happening • 2. as long as there is NO2 present, the reverse reaction will keep on happening! • At one particular time a molecule of N2O4 might be breaking up, and at the same time two molecules of NO2 might be joining to form another molecule of N2O4
Equilibrium • In any reversible reaction, the forward reaction and the reverse reaction are going on at the same time N2O4 2 NO2 The double arrow means that both the forward and reverse reaction are happening at the same time.
Rate of reaction • Rate of the forward reaction Rate will slow down Rate of forward reaction will be fast at first
Rate of reaction • Rate of reverse reaction
Rate of reaction
Rate of reaction
Rate of reaction • The rate of the forward reaction = the rate of the reverse reaction • N02 is being used up at the same rate that it is being formed • N2O4 2 NO2 • Rate of reaction does not change because both N202 and NO2 are constant
Equilibrium • When this happens we say the system is at EQUILIBRIUM • We have reached a state of DYNAMIC EQUILIBRIUM
Key points • The reaction has not stopped • The forward and reverse reactions are still taking place but their rates are equal so there are no changes in concentrations of the reactants and products • It is happening at a microscopic level so we don’t see the individual molecules reacting (macroscopic)
Key points • All observable properties are constant • include the reactants, products, total pressure, temperature • If no changes were made to the condition a system would stay at equilibrium forever
Key points • Temperature must remain constant • A system at equilibrium is a closed system • Equilibrium can be approaches form the left (reactants) or from the right (products) https://www.youtube.com/watch?v=dUMmoPdwBy4 https://www.youtube.com/watch?v=wlD_ImYQAgQ
Calculations • Equilibrium calculations • We can figure out what the concentrations are for the products and reactants when they are at equilibrium • We use: • Keq • It tells us the ratio of products: reactants when the reaction is at equilibrium
Calculations • to figure out the concentration of the products and reactants at equilibrium we use: • Keq = [products] • [reactants] • A + B C + D • Keq = [C] [D] • [A] [B]
Calculations • Given the equilibrium system: • PCl5(g) PCl3(g) + Cl2(g) • The system is analyzed at a certain temperature and the equilibrium concentrations are as follows: • [PCl5] = 0.32 M, [PCl3] = 0.40 M and the [Cl2] = 0.40 M. • Calculate the Keq for this reaction at the temperature this was carried out.
Calculations • SOLUTION: • Step 1 - Use the balanced equation to write the Keq expression: • PCl5(g) PCl3(g) + Cl2(g) • Keq = [PCl3] [Cl2] • [PCl5] • Step 2 - "Plug in" the values for the equilibrium • Keq = [0.4] [0.4] • [0.32] • Step 3 - Calculate the value of Keq : • Keq = 0.5
Calculations • Given the equilibrium system: • 3 PCl5(g) 2PCl3(g) + Cl2(g) • [PCl5] = 0.32 M, [PCl3] = 0.40 M and the [Cl2] = 0.40 M. • We write it as: • Keq = [PCl3]2 [Cl2] • [PCl5]3 • Raise to the power of the coefficient • Keq = [0.4]2 [0.4] • [0.32]3
Calculations • At 200°C, the Keq for the reaction: N2(g) + 3H2(g) 2NH3(g) is 625 • If the [N2] = 0.030 M, and the [NH3] = 0.12 M, at equilibrium, calculate the equilibrium [H2]. • Solution: All concentrations given are at equilibrium so: Write out the Keq expression: • Keq = [NH3]2 • [N2] [H2]3 • 625 = [0.12]2 • [0.30] [H2]3 • (625) [H2] 3 (0.030) = (0.12) 2
