Collision Theory of Reactions A chemical reaction occurs when collisions between molecules have sufficient energy to break the bonds in the reactants molecules collide with the proper orientation bonds between atoms of the reactants (N2 and O2) are broken and new bonds (NO) form

Collision Theory of Reactions (continued) A chemical reaction does not take place if the collisions between molecules do not have sufficient energy to break the bonds in the reactants molecules are not properly aligned

Activation Energy The activation energy is the minimum energy needed for a reaction to take place upon proper collision of reactants

Reaction Rate and Temperature is the speed at which reactant is used up is the speed at which product forms increases when temperature rises because reacting molecules move faster, thereby providing more colliding molecules with energy of activation

Reaction Rate and Concentration Increasing the concentration of reactants increases the number of collisions increases the reaction rate

Reaction Rate and Catalysts A catalyst speeds up the rate of a reaction lowers the energy of activation is not used up during the reaction

Reversible Reactions In a reversible reaction, there are both forward and reverse reactions. Suppose SO2 and O2 are present initially. As they collide, the forward reaction begins. 2SO2(g) + O2(g) 2SO3(g) As SO3 molecules form, they also collide in the reverse reaction that forms reactants. This reversible reaction is written with a double arrow. forward reverse

Chemical Equilibrium At equilibrium, the rate of the forward reaction becomes equal to the rate of the reverse reaction the forward and reverse reactions continue at equal rates in both directions

Chemical Equilibrium (continued) When equilibrium is reached, there is no further change in the amounts of reactant and product

Equilibrium At equilibrium, the forward reaction of N2 and O2 forms NO the reverse reaction of 2NO forms N2 and O2 the amounts of N2, O2, and NO remain constant N2(g) + O2(g) 2NO(g)

Equilibrium Constants For the reaction aA bB The equilibrium constant expression, Kc, gives the concentrations of the reactants and products at equilibrium: Kc = [B]b = [Products] [A]a [Reactants] The square brackets indicate the moles/liter of each substance. The coefficients b and a are written as superscripts that raise the moles/liter to a specific power.

Guide to Writing the Kc Expression

Writing a Kc Expression Write the Kc expression for the following: STEP 1 Write the balanced equilibrium equation: 2CO(g) + O2(g) 2CO2(g) STEP 2 Write the product concentrations in the numerator and the reactant concentration in the denominator: Kc = [CO2] [products] [CO][O2] [reactants] STEP 3 Write the coefficients as superscripts: Kc = [CO2]2 [CO]2 [O2]

Heterogeneous Equilibrium In heterogeneous equilibrium, Gases and solid and/or liquid states are part of the reaction. 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) The concentration of solids and liquids is constant. The Kc expression is written with only the compounds that are gases. Kc = [CO2][H2O]

Guide to Calculating the Kc Value

Example of Calculating Equilibrium Constants What is the Kc for the following reaction? H2(g) + I2(g) 2HI(g) Equilibrium concentrations: [H2] = 1.2 moles/L [I2] = 1.2 moles/L [HI] = 0.35 mole/L STEP 1 Write the Kc expression: Kc = [HI]2 [H2][I2] STEP 2 Substitute the equilibrium concentrations in Kc: Kc = [0.35]2 = 8.5 x 10-2 [1.2][1.2]

Reaching Chemical Equilibrium A container initially filled with SO2(g) and O2(g) or only SO3(g) contains mostly SO2(g) and small amounts of O2(g) and SO3(g) at equilibrium reaches equilibrium in both situations

Equilibrium Can Favor Product If equilibrium is reached after most of the forward reaction has occurred, the system favors the products

Equilibrium with a Large Kc At equilibrium, a reaction with a large Kc produces a large amount of product; very little of the reactants remain Kc = [NCl3]2 = 3.2 x 1011 [N2][Cl2]3 a large Kc favors the products N2(g) + 3Cl2(g) 2NCl3(g) When this reaction reaches equilibrium, it will essentially consist of the product NCl3.

Equilibrium Can Favor Reactant If equilibrium is reached when very little of the forward reaction has occurred, the reaction favors the reactants

Equilibrium with a Small Kc At equilibrium, a reaction that produces only a small amount of product has a small Kc Kc = [NO]2 = 2.3 x 10-9 [N2][O2] a small Kc favors the reactants N2(g) + O2(g) 2NO(g) When this reaction reaches equilibrium, it will essentially consist of the reactants N2 and O2.

Summary of Kc Values A reaction that favors products has a large Kc with about equal concentrations of products and reactants has a Kc close to 1 that favors reactants has a small Kc

Guide to Using the Kc Value

Using Kc to Solve for an Equilibrium Concentration At equilibrium, the reaction PCl5(g) PCl3(g) + Cl2(g) has a Kc of 4.2 x 10–2 and contains [PCl3] = [Cl2] = 0.10 M. What is the equilibrium concentration of PCl5?

Using Kc to Solve for an Equilibrium Concentration (continued) STEP 1 Write the Kc expression: Kc = [PCl3][Cl2] [PCl5] STEP 2 Solve for the unknown concentration: [PCl5] = [PCl3][Cl2] Kc STEP 3 Substitute the known values and solve: [PCl5] = [0.10][0.10] = 0.24M 4.2 x 10–2 STEP 4 Check answer by placing concentrations in Kc: Kc = [0.10][0.10] = 4.2 x 10–2 [0.24]

Le Châtelier’s Principle Le Châtelier’s principle states that any change in equilibrium conditions upsets the equilibrium of the system a system at equilibrium under stress will shift to relieve the stress there will be a change in the rate of the forward or reverse reaction to return the system to equilibrium

What happens when a system is at equilibrium and you upset the balance?

Le Chatelier’s Principle If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.

Le Chatelier’s Principle Used to predict how a equilibrium system will react to changes in concentration, pressure (volume) and temperature.

Effects of Changes on Equilibrium

Heat and Endothermic Reactions For an endothermic reaction at equilibrium, a decrease in temperature (T) removes heat, and the equilibrium shifts toward the reactants an increase in temperature adds heat, and the equilibrium shifts toward the products. CaCO3(s) + 133 kcal CaO(s) + CO2(g) Decrease T Increase T

Heat and Exothermic Reactions For an exothermic reaction at equilibrium, a decrease in temperature removes heat, and the equilibrium shifts toward the products an increase in temperature adds heat, and the equilibrium shifts toward the reactants. N2(g) + 3H2(g) 2NH3(g) + 22 kcal Decrease T Increase T

Saturated Solution A saturated solution contains the maximum amount of dissolved solute contains solid solute is an equilibrium system: solid ions in solution

Solubility Product Constant The solubility product constant for a saturated solution gives the ion concentrations at constant temperature is expressed as Ksp does not include the solid, which is constant Fe(OH)2(s) Fe2+(aq) + 2OH−(aq) Ksp = [Fe2+] [OH−]2

Guide to Calculating Ksp

Example of Calculating Solubility Product Constant Calculate the Ksp of PbSO4 (solubility 1.4 x 10–4 M). STEP 1 Write the equilibrium equation for dissociation: PbSO4(s) Pb2+(aq) + SO42−(aq) STEP 2 Write the Ksp expression: Ksp = [Pb2+][SO42−] STEP 3 Substitue molarity values and calculate: Ksp = (1.4 x 10–4) x (1.4 x 10–4) = 2.0 x 10–8

Molar Solubility (S) The molar solubility (S) is the number of moles of solute that dissolve in 1 L of solution determined from the formula of the salt calculated from the Ksp

Calculating Molar Solubility (S)

Solubility Calculation Determine the solubility (S)2 of SrCO3 (Ksp = 5.4 x 10–10). STEP 1 Write the equilibrium equation for dissociation: SrCO3(s) Sr2+(aq) + CO32−(aq) STEP 2 Write the Ksp expression: Ksp = [Sr2+][CO32−] STEP 3 Substitute S for the molarity of each ion into Ksp: Ksp = [Sr2+][CO32−] = [S][S] = S2 = 5.4 x 10−10 STEP 4 Calculate the solubility, S: S = = 2.3 x 10−5 M