thermochemistry

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Thermochemistry and Equilibrium:

Thermochemistry and Equilibrium NS 2012 Note this has parts of both P&T and T&E in it.

Collision Theory:

Collision Theory Particles must collide with enough energy and in the correct orientation in order to react. Rate of reaction depends on the number of effective collisions per unit time .

PowerPoint Presentation:

Concentration (pressure) More particles exist in a given volume so more effective collisions occur per unit time . Temperature Increasing temperature makes particles move faster therefore increasing the frequency of effective collisions per unit time. Changing Reaction Rate Altering any of four factors will change the rate of a chemical reaction:

PowerPoint Presentation:

Changing Reaction Rate (cont.) Per unit time and change in activation energy Altering any of four factors will change the rate of a chemical reaction: Surface area Smaller particles expose more reactant to collisions per unit time. Catalyst Allows a reaction to follow a different path which requires less activation energy . This means more particles have sufficient energy to overcome the activation barrier and therefore there is a higher frequency of effective collisions.

Catalysis:

Catalysis Increasing the rate of reaction involves a higher frequency of effective collisions What about Endothermic reactions?

Enthalpy:

Enthalpy Enthalpy is the energy in a substance It is made of: Kinetic energy of the particles Potential energy in the chemical bonds Enthalpy of products and reactants cannot be measured, but… enthalpy change during a reaction can be measured as energy is released or absorbed from the surroundings

Enthalpy Change ∆H:

Enthalpy Change ∆H Chemical reactions involve breaking and re-making chemical bonds. Energy is needed to break chemical bonds so… Energy is given out when bonds are made so… ∆H is the difference between the energy needed to break the bonds in the reactants, and the energy given out when new bonds are made in the products. It is measured in Kj/mol . ∆H = H P - H R heat energy is absorbed ( Endothermic ) heat energy is released ( Exothermic )

Energy Level Diagrams:

Energy Level Diagrams Exothermic reactions energy

Energy Level Diagrams:

Energy Level Diagrams Exothermic reactions energy course of reaction

Energy Level Diagrams:

Energy Level Diagrams Exothermic reactions energy course of reaction reactants

Energy Level Diagrams:

Energy Level Diagrams Exothermic reactions energy course of reaction reactants products

Energy Level Diagrams:

Energy Level Diagrams Exothermic reactions energy course of reaction reactants products energy given out ∆H is negative

Exothermic Reactions:

Exothermic Reactions Energy is given out (exits) The products have less energy than the reactants Combustion and neutralisation reactions are exothermic Surroundings gain heat energy Enthalpy Change is negative ( ∆H - )

Endothermic Reactions:

Endothermic Reactions Energy is taken in (enters) The products have more energy than the reactants Surroundings lose heat energy Enthalpy Change is positive ( ∆H + ) Photosynthesis or ethanoic acid with sodium carbonate

Energy Level Diagrams:

Energy Level Diagrams Endothermic reactions energy

Energy Level Diagrams:

Energy Level Diagrams Endothermic reactions energy course of reaction

Energy Level Diagrams:

Energy Level Diagrams Endothermic reactions energy course of reaction reactants

Energy Level Diagrams:

Energy Level Diagrams Endothermic reactions energy course of reaction reactants products

Energy Level Diagrams:

Energy Level Diagrams Endothermic reactions energy course of reaction energy taken in ∆H is positive reactants products

Summary Table:

Summary Table Exothermic reactions Endothermic reactions

Summary Table:

Summary Table Exothermic reactions Endothermic reactions Energy is given out to the surroundings Energy is taken in from the surroundings

Summary Table:

Summary Table Exothermic reactions Endothermic reactions Energy is given out to the surroundings Energy is taken in from the surroundings ∆H is negative ∆H is positive

Summary Table:

Summary Table Exothermic reactions Endothermic reactions Energy is given out to the surroundings Energy is taken in from the surroundings ∆H is negative ∆H is positive Products have less energy than reactants Products have more energy than reactants

Calculating Energy:

Calculating Energy Energy temp change change ∆E = m x s x ∆T 1ml water weighs 1g Specific heat of water is the energy required to raise the temp of 1g water by 1 0 C = mass x specific heat x

Calorimeter Calculations:

Calorimeter Calculations A styrofoam cup is filled with 80ml H 2 0 at 18 0 C. 9.8g of sulfuric acid is added and the volume adjusted to 100ml. The temperature reaches 33 0 C. Calculate a) energy released given s=4.2j/C/ml b) energy released per mole of sulfuric acid. M (H 2 SO 4 ) = 98g/mol ∆E = m x s x ∆T n = m/M ∆H = ∆E /n = 6300j = 9.8/98 = 0.1mol =6300/0.1 = 63Kj/mol ∆H ∆E Exo or Endo? ∆H = -63Kj/mol Reacting substance

Calculate ∆H:

Calculate ∆H 0.01g Mg (s) reacts with 2mL HCl and the temperature rises by 12 0 C. Calculate the heat of reaction in kJmol -1 S= 4.2j/g/ 0 C ∆E = m x s x ∆T = 2 x 4.2 x 12 = 100.8J Moles of Mg n=m/M n= 0.01/24 n= 4.167 x 10 -4 moles ∆H = ∆E /n = 100.8/ 4.167 x 10 -4 = 242kJmol -1

Calculating ∆E :

Calculating ∆E Energy is measured in kJ or joules CH 4 + 2O 2 CO 2 + 2H 2 0 ∆H = -888kJmol -1 How much heat is released if 1.5mols of CH 4 burns? 1332kJ Enthalpy change for this reaction i.e. 888kJ of heat is released when 1 mol methane reacts with 2 mols of oxygen to produce 1 mol carbon dioxide and 2 mols of water

Calculating ∆E:

Calculating ∆E CH 4 + 2O 2 CO 2 + 2H 2 0 ∆H = -888kJmol -1 How much energy is released when 1kg of CO 2 is produced? M ( C ) = 12gmol -1 , M (O ) = 16gmol -1 m = 1000g, M (CO 2 ) = 44gmol -1 Moles of CO 2 produced n=m/M n= 1000/44 n= 22.72 moles 888kJ are produced to make 1 mole CO 2 therefore 888 x 22.72 =20181.81kJ

Calculating amount (moles):

Calculating amount (moles) CH 4 + 2O 2 CO 2 + 2H 2 0 ∆H = -888kJmol -1 How many moles of water are produced when 10000kJ of energy are released? 2 moles produced when 888kJ are released Therefore 1 mole produced for every 444kJ 10000/444 = 22.5moles

Dynamic Equilibrium:

Dynamic Equilibrium But most reactions go forward and backward! These reversible reactions come to a chemical dynamic equilibrium . When the rate of the forward reaction is equal to the rate of the backward reaction Reactions continue until completion i.e. when one of the reactants is used up.

Equilibrium Constant (K):

Equilibrium Constant (K) K eq is a measure of the position of the equilibrium i.e. how far a reaction proceds to the right [ ] = concentration in molL -1 At equilibrium the ratio of reactants and products are constant but the reactions keep going.

Keq > 1 = More products than reactants (Forward reaction is favoured):

K eq > 1 = More products than reactants (Forward reaction is favoured) Equilibrium Constant (K) K eq < 1 = More reactants than products (Reverse reaction is favoured) The only thing that can change K eq for a specified reaction is temperature

Le Chatelier’s Principle:

Le Chatelier ’ s Principle If changes are made to a system in equilibrium, the reaction will shift in the direction that will minimise the change. A system stays in equilibrium unless change is made. Then either the forward or reverse reaction will increase until equilibrium is reached again.

N2(g) + 3H2(g) 2NH3(g) ∆H = -92kJmol-1:

N 2 (g) + 3H 2 (g) 2NH 3 (g) ∆H = -92kJmol -1 What factors might affect this equilibrium? Factors Equilibrium shifts + [reactant] - [reactant] + [product] - [product] + temp. - temp. + pressure - pressure Catalyst increases the rate of both forward and reverse reactions but has no effect on the equilibrium.

Test Yourself:

Test Yourself 2N 2 O 5(g) 4NO 2(g) + O 2(g) ∆H negative colourless brown colourless Describe what you would observe when heated. Describe the observed effect of increasing [N 2 O 5 ]on the rate of reaction and on the equilibrium. As the forward reaction is exothermic the system moves in the reverse direction to minimise the effect of adding heat. The concentration of NO 2 is therefore decreased and the system turns lighter brown. Increasing the concentration of N 2 O 5 means there are more particles in the same volume so there will be more effective collisions in the same time and the rate of reaction will increase. The equilibrium would temporarily shift in favour of the forward direction to minimise the effect of the change which has been made. The concentration of NO 2 is therefore increased and the system turns darker brown.

Balancing Equations:

Balancing Equations Balance the following: HCl + NaOH NaCl + H 2 0 3 H 2 SO 4 + 2 Al(OH) 3 Al 2 (SO 4 ) 3 + 6 H 2 O 2 Fe 2 O 3 + 3 C 4 Fe + 3 CO 2 Ensure atoms on both sides of the equation are the same Try balancing H and O first

Brønsted – Lowry defined Acid/Base reactions:

Brønsted – Lowry defined Acid/Base reactions Involve the transfer of Hydrogen ions (H + ) H + is a proton Acids – donate H + in solution Water dissociates to form hydronium and hydroxide ions H 3 O + and OH - Bases – accept H + in solution HCl(g) + H 2 O(l) H 3 O + (aq) +Cl - (aq) NH 3 (g) + H 2 O(l) NH 4 + (aq) +OH - (aq) Acid Base ? or

Acid/Base reactions:

Acid/Base reactions Involve the transfer of Hydrogen ions (H + ) H + is a proton Acids – donate H + in solution Water dissociates to form hydronium and hydroxide ions H 3 O + and OH - Bases – accept H + in solution HCl (g) + H 2 O (l) H 3 O + (aq) +Cl - (aq) NH 3 (g) + H 2 O (l) NH 4 + (aq) +OH - (aq) Acid Base ? or Water can act as both it is amphiprotic

PowerPoint Presentation:

A Brønsted-Lowry acid is a proton donator A Brønsted-Lowry base is a proton acceptor acid conjugate base base conjugate acid By definition NH 4 + is an acid! NH 3 and NH 4 + are conjugate pairs

Conjugate Pairs:

Conjugate Pairs

Learning Check!:

Learning Check! Label the acid , base , conjugate acid , and conjugate base in each reaction: HCl + OH - Cl - + H 2 O H 2 O + H 2 SO 4 HSO 4 - + H 3 O + CH 3 COOH + H 2 O CH 3 COO - + H 3 O + CH 3 NH 2 + H 2 O CH 3 NH 3 + +OH -

Learning Check!:

Learning Check! Label the acid , base , conjugate acid , and conjugate base in each reaction: HCl + OH - Cl - + H 2 O H 2 O + H 2 SO 4 HSO 4 - + H 3 O + CH 3 COOH + H 2 O CH 3 COO - + H 3 O + CH 3 NH 2 + H 2 O CH 3 NH 3 + + OH -

Strong or Weak Acids:

Strong or Weak Acids Strong Acids e.g. HCl Donate all protons, full dissociation when dissolved Weak Acids e.g. Ethanoic Only donate a few protons, partial dissociation (K eq >1) Depends on the degree of dissociation HCl (g) + H 2 O (l) H 3 O + (aq) + Cl - (aq) CH 3 COOH (l) + H 2 O (l) H 3 O + (aq) + CH 3 COO - (aq)

Strong or Weak Bases:

Strong or Weak Bases Strong Base e.g. NaOH Readily accept protons Weak Base e.g. NH 3 Accept only a few protons Depends on the degree of dissociation NaOH (s) Na + (aq) + OH - (aq) NH 3(g) + H 2 O (l) NH 4 + (aq) + OH - (aq) Don ’ t confuse strong and weak with dilute and concentrate

Comparing Acid Strength:

Comparing Acid Strength HCl Ethanoic acid Concentration 0.1molL-1 0.1molL-1 Conductivity Excellent Poor pH 1 2.88 Rate of reaction with Mg Fast Slow Strength Strong Weak

Equilibrium Constant for Water Kw:

Equilibrium Constant for Water Kw H 2 O can function as both an ACID and a BASE. Equilibrium constant for water = K w = 10 -14 at 25 o C K w = [H 3 O + ] [OH - ] = 10 -14 at 25 o C 2H 2 O (l) H 3 O + (aq) + OH - (aq)

Measuring acidity or alkalinity:

Measuring acidity or alkalinity Acidic solutions [H 3 O + ] > [OH - ] Neutral solutions [H 3 O + ] = [OH - ] Basic (alkaline) solutions [H 3 O + ] < [OH - ] Aqueous solutions have quantities of H 3 O + and OH - ions

The pH scale is a way of expressing the strength of acids and bases. pH = - log [H+] Under 7 = acid 7 = neutral Over 7 = base Example: If [H+] = 10-10 pH = - log 10-10 pH = 10 :

The pH scale is a way of expressing the strength of acids and bases. pH = - log [H+] Under 7 = acid 7 = neutral Over 7 = base Example: If [H + ] = 10 -10 pH = - log 10 -10 pH = 10 pH Scale

Calculating the pH:

Calculating the pH Example: If [H + ] = 1.8 X 10 -5 pH = - log 1.8 X 10 -5 pH = 4.74 Find the pH of these: 1) A 0.15 M solution of Hydrochloric acid 2) A 3.00 X 10 -7 M solution of Nitric acid 3) A 0.025 M solution of Sulfuric acid

Calculate [H+] given pH:

Calculate [H + ] given pH If the pH of Coke is 3.12, [H + ] = ??? Because pH = - log [H + ] then - pH = log [H + ] Take antilog (10 x ) of both sides and get 10 -pH = [H + ] [H + ] = 10 -3.12 = 7.60 x 10 -4 molL -1 *** to find antilog on your calculator, look for “ Shift ” or “ 2 nd function ” and then the log button

Test Yourself:

Test Yourself A solution has a pH of 8.5. What is the concentration of hydrogen ions in the solution? [H+] = 10 - pH = 10 -8.5 = 3.16 x 10 -9 molL -1

pOH:

pOH Since acids and bases are opposites, pH and pOH are opposites! pOH does not really exist, but it is useful for changing bases to pH. pOH looks at the perspective of a base pOH = - log [OH - ] Since pH and pOH are on opposite ends, pH + pOH = 14

[H3O+], [OH-] and pH:

[H 3 O + ], [OH - ] and pH What is the pH of the 0.0010 molL -1 NaOH solution? [OH-] = 0.0010 (or 1.0 X 10 -3 molL -1 ) pOH = - log 0.0010 pOH = 3 pH = 14 – 3 = 11.0 OR K w = [H 3 O + ] [OH - ] = 10 -14 [OH-] = 1.0 X 10 -3 molL -1 so… [H 3 O + ] = 1.0 x 10 -11 molL -1 pH = - log (1.0 x 10 -11 ) = 11.0

PowerPoint Presentation:

The pH of rainwater collected in Auckland on a particular day was 4.82. What is the H + ion concentration of the rainwater? The OH - ion concentration of a blood sample is 2.5 x 10 -7 molL -1 . What is the pH of the blood? [H + ] = 10 -pH =10 -4.82 = 1.51 x 10 -5 molL -1 [OH - ] = 2.5x10 -7 pOH = -log[OH - ] = 6.6 So pH = 14 – 6.6 = 7.40 Test Yourself

PowerPoint Presentation:

[OH - ] [H + ] pOH pH 10 -pOH 10 -pH -Log[H + ] - Log[OH - ] 14 - pOH 14 - pH 1.0 x 10 -14 [OH - ] 1.0 x 10 -14 [H + ]

PowerPoint Presentation:

pH paper

PowerPoint Presentation:

Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water

Lewis Acids & Bases:

Formation of hydronium ion is also an excellent example. Lewis Acids & Bases Electron pair of the new O-H bond originates on the Lewis base.

Lewis Acid/Base Reaction:

Lewis Acid/Base Reaction

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