Rates of Reaction and Chemical Equilibrium Chemistry Ms. Piela

Rates of Reaction A rate is any change that occurs within an interval of time Collision theory states that particles will react to form products when they collide This requires a certain amount of kinetic energy As an example, imagine throwing clay balls at one another Without enough force, they will ricochet off from one another. When they have enough kinetic energy, they will smash and fuse into something entirely new. This is the same in chemical reactions

Activation Energy Activation energy (Ea) is the minimum amount of energy required for a chemical reaction to occur Think of activation energy like a barrier

Activation Energy Diagram Diagram can be used to determine in reaction is exothermic (- ∆H) or endothermic (+∆H) Activated complex (or transition state) is the arrangement of atoms at the peak of the Ea barrier Unstable and only appears briefly

Activation Energy Diagram

Factors Affecting Rates of Reactions Four Major Factors Temperature Concentration Particle Size Catalysts

Factors Affecting Rates of Reaction Temperature Generally, an increase in temperature will increase the rate of a reaction, and vice versa Explanation: Increasing temperature speeds up particles, increasing kinetic energy This allows a reaction to overcome its activation energy Example Popcorn

Factors Affecting Rates of Reaction Concentration Increasing the number of particles (thereby increasing concentration) will increase the reaction rate Explanation: More particles will increase the collision frequency More collisions will mean more chances for a reaction to occur Example Wooden splint in pure oxygen

Factors Affecting Rates of Reaction Particle Size (aka Surface Area) The smaller the particle, the larger the surface area for a given mass of particles An increase in surface area will increase the rate of reaction Explanation: With more exposed area to react, this increases the rate of reaction Example Fine Dust Explosions (1902 Chicago Fire) Lycopodium dust explosion

Factors Affecting Rates of Reaction Catalysts Any substance that increases the rate of a reaction without being consumed Catalysts allow a reaction to occur at lower activation energy Enzymes are biological catalysts

Activation Energy Diagram

Factors Affecting Rates of Reaction

Review of Factors Affecting Rates Temperature Increase causes higher kinetic energy Concentration Higher concentration causes more collisions Particle Size More surface area leads to more exposed spaces to react Catalysts Substances that increase the rates of reaction without being spent

Chemical Equilibrium Reversible reactions are reactions that simultaneously occur in both directions Chemical equilibrium is the state in which the forward and reverse reactions take place at the same rate Sample Reaction: 2 SO2 (g) + O2 (g)  2 SO3 (g)

Chemical Equilibrium Differences between rates and concentration One doesn’t imply the other! Although rates of forward and reverse reactions are equal at chemical equilibrium, the concentrations are not necessarily the same Example A  B 1% 99% If A reacts to give B and the equilibrium mixture contains 99% B and only 1% A, the formation of B is said to be favored

Effect of Catalyst on Equilibrium Although catalysts speed up a reaction rate, they will not affect the concentration of reactants and products at equilibrium Catalysts can only decrease the time required to reach equilibrium Affects speed, not concentration! Meep meep!

Le Châtelier's Principle When a stress is applied to a system at equilibrium, the equilibrium will shift in order to relieve the stress

Factors Affecting Equilibrium Three Major Factors Concentration Temperature Pressure

Le Châtelier's Principle Concentration Changing the amount, or concentration, of any reactant or product in a system at equilibrium disturbs that equilibrium Example equilibrium: 2 H2CO3 (g)  2 H2O (l) + 2 CO2 (g) Adding CO2 Shifts equilibrium to the left Concentration of H2O will decrease Concentration of H2CO3 will increase

Le Châtelier's Principle Example equilibrium: 2 H2CO3 (g)  2 H2O (l) + 2 CO2 (g) Removing CO2 Shifts equilibrium to the right Concentration of H2O will increase Concentration of H2CO3 will decrease

Le Châtelier's Principle Temperature Increasing the temperature causes an equilibrium position of a reaction to shift in the direction that will absorb the heat. Consider heat as if it were a part of the reaction 2 SO2 (g) + O2 (g)  2 SO3 (g) + heat Adding heat Shifts equilibrium to the left. Concentration of SO2 and O2 will increase. Concentration of SO3 will decrease.

Le Châtelier’s Principle 2 SO2 (g) + O2 (g)  2 SO3 (g) + heat Removing heat Shifts equilibrium to the right. Concentration of SO2 and O2 will decrease. Concentration of SO3 will increase.

Le Châtelier's Principle Pressure Changing the pressure only affects an equilibrium with an uneven number of reactants and products Increasing pressure will shift an equilibrium in the direction of the least number of moles Be aware that a change in volume also affects pressure Reducing volume will increase the pressure, and vice versa

Le Châtelier's Principle Example Equilibrium: CO + 3 H2 (g)  CH4 (g) + H2O (g) Increasing pressure (reducing volume) will shift the equilibrium to the… right Decreasing pressure (increasing volume) will shift the equilibrium to the… left

Haber Bosch Process & Le Châtelier’s Principle

Equilibrium Constant Expressions A mathematical relationship exists between concentration of the reactants and products once equilibrium has been reached that is independent of the initial concentration of the participants Given the symbol Keq Can be used to determine whether products or reactants are favored in an equilibrium

Equilibrium Constant Expressions

Steps for Determining Keq Make sure the chemical reaction is balanced and at equilibrium Place the product concentrations in the numerator, and the reactant concentrations in the denominator [products] [reactants] Concentrations of any solid or liquid is left out because the concentrations never change Water is always omitted in aqueous solution. Water is almost constant during the reaction Keq =

Steps for Determining Keq To complete the expression, raise each substance’s concentration to the power equal to the substance’s coefficient in the balanced chemical equation For a general reaction: aA+ bB  cC + dD The equilibrium constant expression can be written as The equilibrium constant is unitless [C]c[D]d [A]a[B]b Keq =

H2CO3 (aq) + H2O (l)  HCO3- (aq) + H3O+ (aq) Example #1 An aqueous solution of carbonic acid reacts to reach equilibrium described below: H2CO3 (aq) + H2O (l)  HCO3- (aq) + H3O+ (aq) Write the equilibrium expression, Keq, for this reaction

N2O4 (colorless)  2 NO2 (brown) Example #2 The brown gas, NO2, is in equilibrium with the colorless gas N2O5. Write the equilibrium expression for the following reaction: N2O4 (colorless)  2 NO2 (brown)

Practice Problems Practice #1 Practice #2

H2CO3 (aq) + H2O (l)  HCO3- (aq) + H3O+ (aq) Calculating using Keq Example #1 H2CO3 (aq) + H2O (l)  HCO3- (aq) + H3O+ (aq) The solution contains the following solute concentrations: H2CO3; 3.3 x 10-2 M; HCO3- ion, 1.19 x 10-4 M; and H3O+, 1.19 x 10-4 M. Determine the Keq expression, and the value of Keq

Example #1

NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq) Example #2 Keq for the equilibrium below is 1.8 x 10-5 at a temperature of 25 °C. Calculate [NH4+] when [NH3] = 6.82 x 10-3 M and [OH-] = 3.45 x 10-3 M. NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)

Example #2

Practice Problems Problem #1 Problem #2

Magnitude of Keq The meaning of Keq If Keq is very large, products will be favored This implies the numerator is large in the Keq expression compared to denominator If Keq is very small, reactants will be favored This implies the denominator is large in the Keq expression compared to numerator If Keq is close to 1, then roughly equal amounts of reactants and products are present at equilibrium