# Chemical Equilibrium Equilibrium.

## Presentation on theme: "Chemical Equilibrium Equilibrium."— Presentation transcript:

Chemical Equilibrium Equilibrium

General Info on Equilibrium
Concerned with how far a reaction goes. Why does it have a low or high % yield? Why do some of the reactants never become products? Not concerned w/ how fast (not looking at rates) At equilibrium, a reaction: Macroscopically looks finished Microscopically the forward reaction rate = the reverse reaction rate Has both reactants and products present but not necessarily in equal concentrations or amounts. Is in a closed system

What determines where equilibrium is reached
What determines where equilibrium is reached? (Why do some reactions have low yields and others high yields?) The equilibrium state is a compromise between a loss in enthalpy (↓ in ∆H) and a gain in entropy (↑ in ∆S). ∆S = change in entropy. Entropy is a measurement of chaos. H20(l)  H2O(g) the forward reaction has a gain in ∆S and a gain in ∆H. = equilibrium Reactions which do not have this compromise such as combustion reactions, have no measurable equilibrium. The reaction only goes forward. Ex. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) this reaction loses enthalpy and no change in entropy when going in the forward direction. = no equilibrium

Quantitative Equilibrium
The equilibrium math expression: K = [Products]/[Reactants] K= the equilibrium constant Rules for writing the K expression: i) do not include solids or liquid water ii) Coefficients are written as exponents iii) Substitute into the expression only equilibrium concentrations Kc or equilibrium pressures Kp.

K constant The K constant for a reaction stays the same unless temperature is changed. (Changing temp. will change the K constant because it changes the thermodynamics of the reaction system. A new compromise between ∆S and ∆H is reached) Larger K constant = greater concentration of the products at equilibrium. Therefore, there is a higher % yield, or the forward reaction is more favored.

K constant cont. Categories of K constants and expressions: Ka = Acids
Kb =Bases Kp=gases using partial pressures Kc = using molar concentrations

Le Châteliers principle
Any change forced upon a reaction system at equilibrium will cause the reaction to respond in such a way as to counter act that change and restore equilibrium. Concentration’s effects: A change in concentration only changes the position of equilibrium (shifts equilibrium) but does not change the K constant.

Le Châteliers principle cont.
Concentrations effects (cont.) An increase of the concentration of a reactant or a decrease in the concentration of the product causes the reaction to shift in the forward direction to produce a product. A decrease in the concentration of a reactant or an increase in the concentration of the product causes the reaction to shift in the reverse direction.

Le châteliers principle cont.
Pressure (similar to concentration) does not effect the K constant. It only affects gases The side (reactants or products) which has the most moles of gas will experience the effect of the pressure change to the greatest extent. An increase in pressure = an increase in concentration. A decrease in pressure = a decrease in concentration. An increase in pressure causes the reaction to shift in the forward direction if there are more moles of gases as reactants A decrease in pressure causes the reaction to shift in the reverse direction with more moles of gases as reactants

Le Châteliers principle cont.
Temperature: alters both the position and the constant Endothermic: an increase in temperature causes a shift in the forward direction and an increase in the K constant, because there will be more products than reactants present Exothermic: an increase in temperature causes a shift in the reverse direction and a decrease in the K constant. Less products present at equilibrium.

Le châteliers principle cont.
Catalyst: has no effect on the position or the constant

Haber Process This reaction is used for the mass production of ammonia
The reaction is: N2 + 3H2  2NH3 all (g) ΔH = -92kJ K= [NH3]2/ [N2] [H2]3

The Contact Process

The industrial process of making ammonia

Yields of NH3 at various pressures and temperatures

Summary of the Haber process
Is a great example for the application of Le Chatelier’s principle Increases the pressure (300atm) to create a forward shift. Constantly supplies the reactants H2 & N2 Removes the NH3 through cooling (NH3 has hydrogen bonding)

Also applies principles of kinetics.
Uses a Fe catalyst which lowers the Ea Uses high temperatures to speed up the reaction rate. (The reaction will not reverse since there is no ammonia available)

The Contact Process This process consists of a series of reactions which produce sulfuric acid. S(s) + O2(g)  SO2(g) 2SO2(g) + O2(g) 2SO3(g) ΔH -196 kJ/mol SO3(g) + H2O (l)  H2SO4(l)

Similar presentations