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General Chemistry II 2302102

General Chemistry II 2302102. Chemical Equilibrium for Gases and for Sparingly-Soluble Ionic Solids. Part 1. i.fraser@rmit.edu.au Ian.Fraser@sci.monash.edu.au. Chemical Equilibrium - 2 Lectures. Outline - 4 Subtopics. Equilibrium and Le Chatelier’s Principle

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General Chemistry II 2302102

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  1. General Chemistry II2302102 Chemical Equilibrium for Gases and for Sparingly-Soluble Ionic Solids Part 1 i.fraser@rmit.edu.au Ian.Fraser@sci.monash.edu.au

  2. Chemical Equilibrium- 2 Lectures Outline - 4 Subtopics • Equilibrium and Le Chatelier’s Principle • The Reaction Quotient and Equilibrium Constant • Temperature and Pressure Effects • Sparingly-Soluble Ionic Compounds in Aqueous Solution

  3. Chemical Equilibrium Objectives - Lecture 1 • By the end of this lecture AND completion of the set problems, you should be able to: • Understand the concepts of: the chemical equilibrium condition, dynamic equilibrium as the balance of forward and reverse reaction rates. • Know the definition of Le Chatelier’s Principle, and understand its application to the prediction of the direction of change in a chemical reaction at equilibrium, following changes in pressure, volume, temperature and amount of reactants and products. • Understand the definition of the reaction quotient Q and the equilibrium constant K (or Kp & Kc) for a chemical reaction. • Calculate the final equilibrium conditions for gas-phase reactions, and for heterogeneous reactions involving gases, from given non-equilibrium initial conditions.

  4. Equilibrium in Chemical Processes Equilibrium constants • Take the simple process (at a given temperature) H2O (l) H2O (g) • An equilibrium partial pressure of H2O known as the “vapour pressure” po(H2O) is reached; at this point the rates of evaporation and condensation are equal. • The equilibrium condition is very simply stated as pH2O = po(H2O) = Kp where Kp is a constant whose value characterises the equilibrium.

  5. A more complex equilibrium process is the chemical reaction 2NO2 (g) N2O4 (g) • Here, colour changes • (NO2 is brown, N2O4 is colorless) • and total pressure changes (P = pN2O4 + pNO2 = n RT/V ) • (the number of molecules changes) P can be monitored to measure progress and find the position of equilibrium.

  6. For equilibrium in the reaction 2 NO2 (g) N2O4 (g) + heat • At any given temperature, one finds that pN2O4 ------- has ~the same equilibrium value, Kp, whatever the pressures pNO22 • And, as the temperature rises, the value of Kpfalls (this finding is what tells us that the reaction is exothermic).

  7. Le Chatelier’s Principle The Tendency Towards Equilibrium • Any system which is not at equilibrium will tend to change spontaneously toward a state of equilibrium (i.e. without a need for us to adjust external variables such as temperature and pressure). • It follows that if a chemical system at equilibrium is disturbed or “stressed” away from equilibrium then the system will tend to react so as to remove the “stress” and return to equilibrium. • This response is known as Le Chatelier’s Principle (The approach to equilibrium may be fast, gradual or even undetectable).

  8. Ideal Gas Reaction Equilibrium Law “Law of Mass Action” aA + bB + … cC + dD + …(all gaseous) pCcpDd… at equilibrium = constant, Kp pAapBb… This “ideal” equilibrium relationship between partial pressures of products and reactants in any gas reaction is closely followed at normal temperatures and total gas pressures up to about 10 atmospheres .

  9. Ideal Gas Reaction - Concentrations & Kc“Law of Mass Action” Remember pBV ~ nBRT => pB~ (nB/V) RT = [B]RT aA + bB + … cC + dD + …(all gaseous) [C]c [D]d… at equilibrium = constant, Kc [A]a [B]b… This “ideal” equilibrium relationship between concentrations of products and reactants is closely followed in any gas reaction at normal temperatures and total gas pressures up to about 10 atmospheres • deviations at higher pressures arise from intermolecular forces

  10. Relationship between Kc and Kp for gas phase reactions Kp = Kc(RT)  = gaseous products – gaseous reactants (= c + d + …) - (a + b + ….. ) gases only!

  11. Pressures, Concentrations & Activities • For any gas in a reaction, the numerical value of partial pressure in atmospheres is a good approximation to its “activity” (or tendency to react, taken by convention, relative to ideal gas behaviour at 1 atm) Any gas: agas , = pgas / atm • For any solute in a reaction, the numerical value of concentration in moles per litre is a good approximation to its “activity” (or tendency to react, taken by convention, relative to ideal solute behaviour at 1 mol/L) Any solute, asolute , = [solute] / M

  12. For any pure solid or pure liquid in a reaction, its “activity” (or tendency to react) is taken (by convention) to be 1.000 [areas affect rates but not “activities”] • Any pure solid or liquid: apure solidor liquid = 1.000 • For any solid or liquidsolvent (in dilute solution) the “activity” is taken as 1.00 (by convention, relative to pure solvent behaviour, assuming solvent mole fraction > 98.5%, [otherwise, Raoult’s law => ideal asolvent~ xsolvent/1.000] • Any solvent (dilute solution): asolvent , = 1.00 • [The above “ideal” approximations are easy to use. • More accurate values can be obtained from measurements • of vapour pressures, osmotic pressures, m.pts, b.pts, etc.]

  13. Ideal Solution Reaction Equilibrium Law Many reactions occur between solutes in solutions aA + bB + … cC + dD + …(all in a liquid solvent) here we use concentrations/(mole per litre) to represent activities ( partial pressures are NOT relevant within solutions) - Thus “Reaction Quotient” Q = ([C]/M)c ([D]/M)d… --> constant, K, at equilm ([A]/M)a ([B]/M)b… This “ideal” relationship between the quotient of numerical values of concentrations/ (mol/L) at equilibrium, and the equilibrium constant, K, is closely followed in dilute solutions .

  14. 10.3 The Equilibrium Constant • Suppose we mix 0.6 atm of CO and 1.1 atm of Cl2. Say we can measure the equilibrium partial pressure of COCl2 and find that it is 0.1 atm. The temperature is 600oC. CO(g) + Cl2(g) COCl2(g) 0.6-0.1 atm 1.1-0.1 atm 0.1 atm at equilm • The equilibrium expression in terms of partial pressures is Q = (pCOCl2/atm)/ (pCO/atm) x (pCl2 /atm) --> K

  15. 0 . 1 K = = 0 . 2 (0 . 6 - 0 . 1 ) ( 1 . 1 - 0 . 1 ) • Using the known stoichiometry we now insert the activities (as pressure/atm data), as follows: (Take care if the stoichiometric coefficients are not all equal to one!!) [For reasons to be covered in later chapters, K should be dimensionless. This is ensured by using “activities” (approximated here by dividing all partial pressures by the reference pressure of 1 atm, or the equivalent pressures 760 torr, 101.3 kPa, etc, to obtain purely numerical values). A partial pressure of 1 atm is often referred to as the standard state for a gaseous substance.]

  16. Favoured Reactions • If K (for a gas reaction) is: >> 1 then equilibrium tends to favour “products” << 1 then equilibrium tends to favour “reactants” ~ 1 then substantial amounts of both products and reactants will tend to be present at equilibrium [this depends to some extent on the initial mixture that is used]

  17. Relationships among Equilibrium Expressions (i) forward reaction K1 A  B reversed equation K2 A  B K2 = 1/K1 (ii) Multiplying an equation by a factor n : • reaction (1) K1 A  B • reaction (x n) Kn = K1n nA  nB (iii) addition and subtraction of reactions • reaction 1 K1 A  B • reaction 2 K2 C  D 1 + 2 K1+2 =K1K2 A + C  B + D 1 - 2 K1-2 = K1 /K2 A + D  B + C

  18. Equilibrium Calculations for Gas-Phase Reactions • Sometimes equilibrium problems lead to complex equations which can be solved via simple assumptions. For example CH4(g) + H2O(g) CO(g) + 3H2(g) • for which K at 600 K is 1.8 x 10-7. • Suppose our initial conditions are as follows: Gas p/(atm) CH4 1.4 H2O 2.3 CO 1.6 H2 0.0

  19. 3 ( 1 . 6 + y ) ( 3 y ) - 7 K = = 1 . 8 x 10 ( 1 . 4 - y ) ( 2 . 3 - y ) • Let's assume the partial pressure of CH4 decreases by y atm to reach equilibrium. Again we write an equilibrium expression, • At equilibrium the following pressures will apply: Gas p/(atm) CH4 1.4-y H2O 2.3-y CO 1.6+y H2 3y • By substitution we obtain This looks like real trouble ( a quartic poynomial in y) !! DON'T TRY TO SOLVE THIS DIRECTLY. There is a much easier way.

  20. 3 ( 1 . 6 ) ( 27 y ) - 7 = 1 . 8 x 10 ( 1 . 4 ) ( 2 . 3 ) • We'll make the assumption that y<<1.4 and check at the end that everything is OK. i.e. Which transposes to y3 = 1.34 x 10-8 or y = 2.38 x 10-3 (cube root) Clearly our assumption of y<<1.4 is quite reasonable. • At equilibrium, then, the partial pressure will be pH2/atm = 3y = 7.1 x 10-3 .

  21. Example 2NO2 (g) N2O4 (g) + heat • A sample of atmospheric NO2is concentrated by freezing as N2O4 and then vaporizing. Immediately upon vaporizing the vessel would contains 200 kPa of N2O4(g) before any reaction. • What are the pressures of N2O4(g) and NO2(g) at equilibrium for T=298 K ?? Step 1: From tables of chemical data K298K = 11.3 Step 2: Rough prediction. • at time = 0, we have only product (N2O4 and noreactant). • so reaction must go backwards to reach equilibrium. Step 3: Write down equilibrium expression. (pN2O4/atm)eq (pNO2/atm)eq2= K = 11.3

  22. 2 ( 1 - ) . = 2  ( 4 ) Step 4: Work out expressions for the pressures. • We start with 202 kPa of N2O4 • Say that a fraction  reacts => 1-  unreacted pN2O4 = (1- ) 202 kPa = (1- ) x 2.00 atm. • Next consider NO2: pNO2 = (2 ) 202 kPa = 2  x 2.00 at = 4.00  atm Step 5: Substitute each p/atm in the equilibrium expression 2.00 (1 - ) / (4.00 )2 = 11.3  11.3 x (16.002) = 2.00 - 2.00  or 90.4 2 + 1.00  - 1.00 = 0, a quadratic, in which  = 0.0998 (discarding the negative root*) (i.e. about 10% of N2O4 has reacted) • *note: a negative a is not physically sensible => reject

  23. = = K 2 1.801 . 313 11 0.399 • So, at equilibrium: pNO2(eq) = 2  202 kPa = 39.9 kPa = 0.399 atm pN2O4(eq) = (1- ) 202 kPa = 180.1 kPa = 1.801 atm  Ptotal = 2.200 atm = 222.9 kPa I.e. 223 kPa • Notice that the total pressure has gone up during the reaction. • Finally, it's a good idea to check our calculations by back-substituting each calculated value of pressure/atm (not /kPa) into the equilibrium constant expression: • All is well!!

  24. The Reaction Quotient The "reaction quotient”, Q, contains “activities of products” / “activities of reactants” - activity of gas, a(gas) = p(gas)/atm - activity of solute, a(solute) = [solute]/M Pure liquid and solid substances need not appear in reaction quotients (their activities can be taken as 1.000 and normal pressures have negligible effect and areas ). Activity of a solvent (close to its mole fraction) is taken as 1.00. • By definition, the value of Q at equilibrium is the "equilibrium constant", K.

  25. Reactions always proceed in such a direction as to make Q = K. eg: • if Q < K it follows that the reaction can only go forward. (eg. when we start with only “reactants”) • if Q > K it follows that the reaction can only go backward. (eg. When we start with only “products”) • if Q = K it follows that the reaction is at equilibrium • Therefore by finding Q at any point in time compared with K we can predict the direction in which the reaction tends to proceed [The value of Q compared with K tells us nothing about the rate of reaction, but the minimum work required to drive reaction is G = RT ln Q/K (negative G => spontaneous reaction).]

  26. p p (0 ) (0 ) SO Cl 2 2 Q = = = 0 p 1.0 SO Cl 2 2 • Example: • The reaction: SO2Cl2(g) SO2(g) + Cl2(g) has an equilibrium constant K = 2.4 • Suppose the initial values of partial pressure/atm are: pSO2Cl2 /atm = 1.0 pSO2 /atm = 0 pCl2 /atm = 0 • We would proceed by computing the reaction quotient • Therefore since Q < 2.4 the reaction tends to proceed spontaneously (unaided, without being driven) to the right.

  27. Alternatively, if the conditions had been pSO2Cl2 = 0.01 atm pSO2 = 0.1 atm pCl2 = 0.4 atm Since Q > 2.4 the reaction now tends to proceed spontaneously to the left (this reaction approaches equilibrium quite rapidly and without need of catalyst). pSO2/atm x pCl2/atm 0.1 x 0.4 Q = ----------------------- = -------- = 4 pSO2Cl2/atm 0.01

  28. External Effects and Le Chatelier's “Principle” ‘When a system at equilibrium is subject to change it will respond in such a way as to reduce the change.’ • For example, in the reaction 2NO2 (g) N2O4 (g) initially at equilibrium • Some possible changes we can explore are • addition of NO2 • increase in total volume • addition of an inert gas

  29. pN2O4/atm Q = ---------------- (pNO2/atm)2 • First we need the reaction quotient • If we increase pNO2 , Q will decrease to less than K and the reaction will adjust by forming more N2O4 (g). • If we increase the volume of the system, then each partial pressure drops. Now Q increases, and the reaction must proceed in the reverse direction. • What happens if we increase Ptotal (by adding an inert gas)? Nothing! The partial pressures do not change!!

  30. 2 p HI Q = p p H I 2 2 • Another example: • H2(g) + I2(g) 2HI(g) • Stage 1: reactants only => Q = 0 (Q < K )  react.  right • Equilibrium reached -> Q = K • Stage 2: inject iodine -> Q < K  react.  right • System always adjusts to counteract the perturbation which takes it out of balance.

  31. Heterogeneous Equilibrium • So far we have concentrated on homogeneous equilibria involving gases. We now consider heterogeneous reactions in which at least two different physical states of matter are present. • Pure solids and liquids appear in Q as 1.000 : H2(g) + I2(s) 2HI(g) Q = (pHI/atm)2 / (pH2/atm) x 1.000 C(s) + H2O(l) + Cl2(g) COCl2(g) + H2(g) Q =(pCOCl2/atm) x (pH2/atm) / 1.000 x 1.00 x (pCl2/atm)

  32. For a liquid solvent in a reaction equation its mole fraction should be used in Q, this can normally be taken as 1.00 I2(s) = I2(aq) Q = [I2(aq)] / 1.00 ---> solubility/(mol/L) at equilm CaCO3(s) = CaO(s) + CO2(g) Q = 1.000 x pCO2/atm / 1.000 ---> equilm pressure/atm Q < K ? = K ? > K ?

  33. The Lime Kiln CaCO3(s) CaO(s) + CO2(g) An industrially important example of heterogeneousequilibria involves the production of calcium oxide (CaO - builders’ lime, quicklime) from limestone (CaCO3 - garden lime, agricultural lime).

  34. The Lime Kiln At any given temperature the equilibrium pressure of CO2 (/atm) is equal to the equilibrium constant K and is independent of the relative amounts of CaO and CaCO3 present. K increases with T (=> endothermic reaction) and exceeds 1 at above 1000oC (--> equilm pCO2 > 1 atm). It is then simply necessary to allow the CO2 to escape to obtain more CaO from the CaCO3. CaCO3(s) CaO(s) + CO2(g)

  35. In all equilibria involving pure solids with gases, the partial pressures of gases are independent of the amounts of solid present. (n.b.as long as we have some solid present) • Similar behaviour is found for solid-liquid equilibria. For example, a sparingly soluble salt like PbSO4 dissolves in water to form ions: PbSO4(s) Pb2+(aq) + SO42-(aq) Q = ([Pb2+]/M) x ([SO42-]/M) / 1.000 ---> K (at equilm) • The concentrations of Pb2+(aq) and SO42-(aq) are independent of the amount of PbSO4(s) that is present.

  36. Some more general rules for the Law of Mass Action follow from these observations. 1. Each gas enters a reaction quotient as a numerical activity (partial pressure in atmospheres, p(gas)/atm) 2. Each dissolved species enters as a numerical activity (concentration in moles per litre, [solute]/M) 3. Activity of apure solid or pure liquidneed not appear in a reaction quotient (except asa =1.000), neither need a solvent taking part in the reaction, provided the solution is dilute (takea(solvent) as 1.00). 4. The reaction quotient Q contains the numerical activity of each “product” in the numerator and of each “reactant” in the denominator (each raisedto the power of its coefficient in the balanced chemical equation).

  37. 0 . 345 pH2/atm . • Example 1: • Consider the equilibrium • H2(g) + I2(s) 2HI(g) • Given K = 0.345 at 25°C • If pH2 = 1.00 atm and I2(s) is present, determine pHI • Solution • For the reaction given we have Q = • so that pHI/atm = = 0.5874 x 1.00 thus pHI = 0.587 atm 2 p HI = K p x 1.000 H 2

  38. Example 2: Find the equilibrium expression for the extraction of gold from ore with cyanide via the reaction: 4 Au(s) + 8 CN–(aq) + O2(g) + 2H2O(l) 4 Au(CN)2–(aq) + 4 OH–(aq)

  39. 4 4 - - [Au (CN) ] [OH ] 2 Q = 8 - [ CN ] . p O 2 4 Au(s) + 8 CN–(aq) + O2(g) + 2 H2O(l) 4 Au(CN)2–(aq) + 4 OH–(aq) Answer: = K 1.0004 1.002

  40. Chemical Equilibrium - End of Lecture 1 Objectives Covered in Lecture 1 • After studying this lecture you should be able to: • Understand the concepts of: the chemical equilibrium condition, dynamic equilibrium as the balance of forward and reverse reaction rates. • Know the definition of Le Chatelier’s Principle, and understand its application to the prediction of the direction of change in a chemical reaction at equilibrium, following changes in pressure, volume, temperature and amount of reactants and products. • Understand the definition of the reaction quotient Q and the equilibrium constant K (or Kp and Kc) for a chemical reaction. • Calculate the final equilibrium conditions for gas-phase reactions, and for heterogeneous reactions involving gases, from given non-equilibrium initial conditions.

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