2.0 Equilibrium Constant

Transcription

1 2.0 Equilibrium Constant When reactions are reversible and chemical equilibrium is reached, it is important to recognize that not all of the reactants will be converted into products. There is a mathematical relationship that exists to explain this: Ratio = [products] [reactants] For any general reaction: aa + bb cd + dd an equilibrium constant expression can be written as: [C] c [D] d [A] a [B] b where: Keq is a constant ratio called the equilibrium constant. the products are placed in the numerator and their concentrations are raised to the power of the coefficients from the balanced equation. the reactants are placed in the denominator, with their concentrations raised to the power of their coefficients. 2.1 Calculating Keq For example, for the reaction: H2 (g) + I2 (g) 2 HI(g) the equilibrium constant expression will be: [HI] 2 [H2] [I2 ]

2 If we were given concentrations for each substance as measured at equilibrium: [H2] = M At equilibrium: [I2] = M [HI] = M We can substitute these values into our equilibrium expression and solve for Keq: [HI] 2 [H2] [I2 ] = (0.156) 2 (0.022)(0.022) = 50.3 The value of Keq, which has no units, is a constant for any particular reaction, and its value does not change unless the temperature of the system is changed. Key Points: The value of Keq is dependent upon temperature but independent of the initial concentration of the reactants. Keq relates the concentrations of products to reactants at equilibrium. Only gases and aqueous ions are included in the equilibrium expression. For aqueous solutions, concentration is often measured as mol L -1. For gases, concentration is often measured as partial pressure. Solids or liquids are not included in Keq because while their amounts change during the reaction their concentrations do not. They are instead assigned a value of 1 when present in a balanced reversible equation. CaO (s) + CO2 (g) CaCO3 (s) 1 [CO2]

3 2.2 The Meaning of Keq The equilibrium expression shows the extent to which the reactants are converted into products. If Keq >1 The forward reaction is favoured; meaning the reaction essentially "goes to completion" and all or most of the reactants are used up to form the products. If Keq <1 The reverse reaction is favoured. The reaction does not occur to any great extent - most of the reactants remain unchanged, and there are few products produced. When Keq 1 Neither the forward nor the backwards reaction is favoured. Note: Favoured means that side of the equation has the higher number of moles and higher concentrations than the other. Here are some examples to consider: the decomposition of ozone, O3 2 O3 (g) 3 O2 (g) Keq is very large, indicating that mostly O2 is present in an equilibrium system, with very little O3 N2 (g) + O2 (g) 2 NO(g) production of nitrogen monoxide Very little NO is produced by this reaction; N2 and O2 do not react readily to produce NO (lucky for us - otherwise we would have little oxygen to breath in our atmosphere!)

4 2.3 Putting it all together Example 1: For the equilibrium system described by: 2 SO2 (g) + O2 (g) 2 SO3 (g) at a particular temperature the equilibrium concentrations of SO2, O2 and SO3 were 0.75 M, 0.30 M, and 0.15 M, respectively. At the temperature of the equilibrium mixture, calculate the equilibrium constant, Keq, for the reaction and determine which reaction is favoured. Solution: 1. Begin by setting up the equilibrium constant expression for the balanced equation: [SO3] 2 [SO2] 2 [O2 ] 2. Next, substitute in the known values, and solve for the unknown, which is Keq for this question. Don't forget to use the exponents! [0.15] 2 [0.75] 2 [0.30] = Keq <1 so the backwards reaction is favoured. Example 2: The equilibrium equation for the formation of ammonia is: N2 (g) + 3 H2 (g) 2 NH3 (g) At 200 C the concentrations of nitrogen, hydrogen, and ammonia at equilibrium are measured and found to be [N2] = 2.12; [H2] = 1.75, and [NH3] = Calculate Keq at this temperature. Solution: [NH3] 2 [N2][H2] 3 (84.3) 2 (2.12)(1.75) 3 ( ) (2.12)(5.36) = 626

5 2.0 Equilibrium Constant When reactions are reversible and chemical equilibrium is reached, it is important to recognize that not all of the reactants will be converted into products. There is a mathematical relationship that exists to explain this: Ratio = [products] [reactants] For any general reaction: aa + bb cd + dd an equilibrium constant expression can be written as: [C] c [D] d [A] a [B] b where: Keq is a constant called the. the products are placed in the and their concentrations are raised to the of the coefficients from the balanced equation. the reactants are placed in the, with their concentrations raised to the of their coefficients. 2.1 Calculating Keq For example, for the reaction: H2 (g) + I2 (g) 2 HI(g) the equilibrium constant expression will be: [HI] 2 [H2] [I2 ]

6 If we were given concentrations for each substance as measured at equilibrium: [H2] = M At equilibrium: [I2] = M [HI] = M We can substitute these values into our equilibrium expression and solve for Keq: [HI] 2 [H2] [I2 ] = (0.156) 2 (0.022)(0.022) = 50.3 The value of Keq, which has no units, is a for any particular reaction, and its value does not change unless the temperature of the system is changed. Key Points: The value of Keq is of the reactants. Keq relates the concentrations of products to reactants at. Only and are included in the expression. For aqueous solutions, concentration is often measured as mol L -1. For gases, concentration is often measured as partial pressure. or are not included in Keq because while their amounts change during the reaction their concentrations do not. They are instead assigned a value of 1 when present in a balanced reversible equation. CaO (s) + CO2 (g) CaCO3 (s) 1 [CO2]

7 2.2 The Meaning of Keq The equilibrium expression shows the extent to which the. If Keq >1 The reaction is favoured; meaning the reaction essentially "goes to completion" and all or most of the reactants are used up to form the products. If Keq <1 The reaction is favoured. The reaction does not occur to any great extent - most of the reactants remain unchanged, and there are few products produced. When Keq 1 the forward nor the backwards reaction is favoured. Note: Favoured means that side of the equation has the higher number of moles and higher concentrations than the other. Here are some examples to consider: the decomposition of ozone, O3 2 O3 (g) 3 O2 (g) Keq is very large, indicating that mostly O2 is present in an equilibrium system, with very little O3 N2 (g) + O2 (g) 2 NO(g) production of nitrogen monoxide Very little NO is produced by this reaction; N2 and O2 do not react readily to produce NO (lucky for us - otherwise we would have little oxygen to breath in our atmosphere!)

8 2.3 Putting it all together Example 1: For the equilibrium system described by: 2 SO2 (g) + O2 (g) 2 SO3 (g) at a particular temperature the equilibrium concentrations of SO2, O2 and SO3 were 0.75 M, 0.30 M, and 0.15 M, respectively. At the temperature of the equilibrium mixture, calculate the equilibrium constant, Keq, for the reaction and determine which reaction is favoured. Solution: 1. Begin by setting up the equilibrium constant expression for the balanced equation: 2. Next, substitute in the known values, and solve for the unknown, which is Keq for this question. Don't forget to use the exponents! 3. Keq <1 so the backwards reaction is favoured. Example 2: The equilibrium equation for the formation of ammonia is: N2 (g) + 3 H2 (g) 2 NH3 (g) At 200 C the concentrations of nitrogen, hydrogen, and ammonia at equilibrium are measured and found to be [N2] = 2.12; [H2] = 1.75, and [NH3] = Calculate Keq at this temperature. Solution: