First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed.

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.

∆Stotal = ∆Ssys + ∆Ssurr > 0Spontaneous process:

∆Stotal = ∆Ssys + ∆Ssurr = 0Equilibrium process:

How far? Entropy

Entropy“ spontaneous” reaction as time elapses

Organizes effort requiring energy input

A chemical system and its surroundings.

the system

the surroundings

Entropy (S) is a measure of the randomness or disorder of a system.

order SdisorderS

If the change from initial to final results in an increase in randomness

∆S > 0For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state

Ssolid < Sliquid << Sgas

H2O (s) H2O (l) ∆S > 0

The equation below shows the equilibrium existing between nitrogen, oxygen and nitrogen monoxide.

N2(g) + O2(g) 2NO(g)The equilibrium constant, Kp, at 298 K is 5.0 ×10–31(a)(i) Write an expression for the equilibrium constant, Kp, in terms of the partial pressures of the three gases.

(ii) Why does the value for Kp have no units?

(b) An equilibrium mixture of these three gases was found to contain nitrogen, at a partial pressure of 0.87 atm, and oxygen, at a partial pressure of 0.23 atm.(i) Calculate the partial pressure exerted by the nitrogen monoxide.

(ii) Deduce the value of the total pressure of the equilibrium mixture of gases.

(iii) Assuming that the total pressure on the mixture of gases is doubled, what, if any, would be the effect on the• partial pressure of nitrogen monoxide• equilibrium constant, Kp ?(c) Inside a car engine, air (a mixture of nitrogen and oxygen) is drawn in and, under the high temperatures operating, the value of Kp increases dramatically.This increase is also accompanied by an increase in the value of ΔStotal. Typical values of Kp and ΔStotal are shown in the table below.

Although the value of ΔSsystem is unlikely to alter very much, the value for ΔSsurroundings will change significantly.(i) At a temperature of 1500 K, ΔStotal is negative.Does this mean that the reaction between nitrogen and oxygen cannot occur at this temperature? Explain your reasoning.(ii) Why is the value for ΔSsystem for this equilibrium approximately constant when the temperature rises above 298 K?

Temperature / K Temperature / K Kp ΔStotal/ J mol–1K–1298 5.0 × 10–31 –5801500 1.0 × 10–5 – 96

(iii) What is the sign of ΔSsurroundings for an endothermic reaction? Justify your answer.(iv) Explain why an endothermic reaction results in an increase in the value of ΔStotal as the temperature increases.(d) A student used the value for Kp at 1500 K to calculate the partial pressure of nitrogen monoxide inside a working car engine.Why might the actual partial pressure be lower than the calculated answer?

17.When barium nitrate is heated it decomposes as follows:Ba(NO3)2(s) BaO(s) + 2NO2(g) + ½O2(g) ΔH = +505.0 kJ mol-1(a) Use the following data when answering this part of the question.

(i) Explain why:ΔSƟ [NO2(g)] is greater than ΔSƟ[BaO(s)]NO2 is a gas whereas BaO is a solidΔSƟ[Ba(NO3)2(s)] is greater than SƟ[BaO(s)].Ba(NO3)2 has a more complicated structure than BaO

Substance Standard entropy, SƟ /J mol-1 K-1

Ba(NO3)2(s) + 213.8BaO(s) + 70.4NO2(g) + 240.0O2(g) + 205.0

(ii) Calculate the entropy change of the system, ΔSƟsystem , for this reaction. Include a sign and units in your answer.ΔSƟsystem = 70.4 + (2×240.0) + (½×205.0) — 213.8 = +439.1 J mol-1K-1(b) Calculate the entropy change of the surroundings, ΔSƟsurroundings, for the reaction at 298 K. Include a sign and units in your answer.ΔSƟsurroundings = - ΔH = - 505 x 1000 J mol-1 = -1700 J mol-1K-1 T 298 K(c) Calculate ΔSƟtotal, and explain the significance of the sign for this value.ΔSƟtotal = +439.1 — 1695 = — 1260 J mol-1K-1(d) Calculate the minimum temperature at which the decomposition of barium nitrate should occur.You can assume that ΔH and ΔSsystem are not affected by a change in temperature.

ΔSƟsurroundings = -491. 1 J mol-1K-1 T =505 x 1000 J mol-1 = 1150K

493.1 J mol-1 K-1

16.Ammonia can be oxidised to form nitrogen(II) oxide and water according to the equation

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) ΔH ○ = –905.6 kJ mol-1.In industry, the reaction is carried out at 1123 K with a platinum/rhodium catalyst.The standard entropy of one mole of each substance in the equation, measured at 298 K, is shown in the table below.

Substance ΔSƟ /J mol-1K-1

NH3(g) +192.3O2(g) +205.0NO(g) +210.7H2O(g) +188.7

(a) (i) Use the values given to calculate the standard entropy change of the system, ΔS Ɵsystem, for this reaction. Include the sign and units in your final answer.[6 x 188.7 + 4 x 210.7] – [4 x 192.3 + 5 x 205] +180.8 J mol-1K-1

(ii) Is the sign for your value for ΔSƟsystem what you expected? Justify your answer.yes, as 9 molecules of gas are being changed to 10 molecules of gastherefore increase in disorder(iii) Calculate the entropy change of the surroundings, ΔSsurroundings, at 1123 K for this reaction. Include the sign and units in your final answer. ΔSsurroundings -(-905.6 x 1000 J mol-1) 1123 K

+ 806.4 J mol-1K-1 / 0.8064 k J mol-1K-1

(iv) Calculate the total entropy change, ΔStotal, for this reaction at 1123 K. Include the sign and units in your final answer. You may assume that ΔSsystem is unchanged at high temperatures. ΔStotal = ΔSsystem + Δssurroundings =+180.8 J mol-1K-1 + 806.4 J mol-1K-1

= +987.2 J mol-1K-1

(v) What does your answer to (iv) tell you about the extent of the reaction at 1123 K?Justify your answer.All products/reaction goes to completion because ΔStotal > 200 J mol-1K-1

6. Hydrogen can be manufactured by reacting methane with steam, as shown in the equation below.

CH4(g) + H2O(g) CO(g) + 3H2(g) ∆HƟ298 = + 206.1 k J mol Use these values:the standard entropy of 1 mol of H2(g) is (2 x65.3) = 130.6 J mol Kthe standard entropy of 1 mol of H2O(g) is 188.7 J mol KYou will also need to refer to the data booklet in the calculations which follow.(a) Calculate the standard entropy change of the system, ∆SƟsystemsystem, for this reaction at 298 K.∆SƟsystem = (3x2x65.3 +197.6) – (186.2 + 188.7)= (+) 214.5 / 215 J mol-1 K-1

(+) 0.2145 / 0.215 kJ mol-1 K-1

-1

-1-1

-1 -1

(b) Calculate the standard entropy change of the surroundings, ∆Sɵsurroundings , for this reaction at 298 K. Include a sign and units in your answer.(∆Sɵsurroundings ) = - ∆H T-206.1x(1000) J mol-1

298 K= -691.6 J mol-1 K-1 = -0.6916 kJ mol-1 K-1

(c) Calculate the total entropy change, ∆Sɵtotal, for this reaction at 298 K.∆Stotal = (214.5+(-691.6)) = -477.1 J mol-1 K-1

- 0.4771 kJ mol-1 K-1

Explain why this value shows that the reaction is not spontaneous at this temperature.Negative / less than zero (so not spontaneous) would be positive if spontaneous.

(d) The composition of an equilibrium mixture produced at 2.0 atmospheres pressure and at a much higher temperature is shown below.

CH4(g) + H2O(g) CO(g) + 3H2(g)Amount in equilibrium 0.80 0.80 1.20 3.60mixture / mol*(i) Write the expression for the equilibrium constant, Kp, of the reaction and calculate its value. Include units in your answer.Kp = (PH2) x (PCO) (PCH4)(PH2O) CH4 H2O CO H2 pp 0.25 0.25 0.375 1.125 atm Kp = (1.125 atm) x (0.375atm) = 8.54 atm (0.25atm)(0.25atm)

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(ii) The total entropy change in J mol-1 K-1 is related to the equilibrium constant by the equation∆Sɵtotal = RlnKp or ∆Sɵtotal = 2.3RlogKp

Calculate the total entropy change at the temperature of the reaction. [R = 8.31 J mol-1 K-1 ]∆Stotal = (8.31 J mol-1 K-1 x ln8.54) = (+)17.8 (J mol-1 K-1 )(iii) Calculate the temperature at which this equilibrium is reached using your answer to (ii) for ∆Sɵtotal. Assume that ∆H is still+206.1 kJ mol-1 and that ∆Sɵsystem = +225 J K-1mol-1. (This is not the same as the value for ∆Sɵsystem calculated in (a) which is at 298 K.)17.8 = 225 - 206.1 x 1000 T T = (206.1 x 1000) = 995 / 990 K 207.2

*(e) Use the magnitude and signs of the entropy changes to explain the effect of a temperature increase on the equilibrium constant of this endothermic reaction.∆Ssurroundings = -∆H T becomes less negative making ∆Stotal more positive (as T increases)Thus ∆Stotal increases equilibrium constant increases

7.(a) Crystals of hydrated cobalt(II) chloride, CoCl2.6H2O, lose water when they are heated, forming anhydrous cobalt(II) chloride, CoCl2.

CoCl2.6H2O(s) CoCl2(s) + 6H2O(l)(i)Calculate the entropy change of the system, ΔSɵsystem, at 298 K. Include a sign and units in your answer. You will need to refer to your data booklet.∆Sɵsystem = 109.2 + (6x 69.9) – 343 =(+)185.6 J mol-1 K-1 / (+)186 J mol-1 K-1

(ii) Explain whether the sign of your answer to (a)(i) is as expected from the equation for the reaction.Yes as (solid and) liquid forms (from solid) / number of moles increases (iii) The standard enthalpy change for the reaction, ΔHɵ, is +88.1 kJ mol-1

Calculate the entropy change in the surroundings, ΔSɵsurroundings, at 298 K for this reaction. Include a sign and units in your answer.∆Ssurroundings = –88.1 x (1000 ) 298 = -295.6375 J mol-1 K-1

= -295.6 J mol-1 K-1

(iv) Calculate the total entropy change, ΔSɵtotal, at 298 K for the reaction. (185.6–295.6) J mol-1 K-1

= -110 J mol-1 K-1

(v) Does your answer to (a)(iv) indicate whether hydrated cobalt(II) chloride can be stored at 298 K without decomposition? Explain your answer.Decomposition (at 298 K) will not occur as ∆Sɵtotal is negative / Reactions are only spontaneous if total entropy change is positive/ decomposition not thermodynamically feasible / (hydrated cobalt chloride) is thermodynamically stable(b) A student attempted to measure the enthalpy change of solution of anhydrous cobalt(II) chloride by adding 2.00 g of cobalt(II) chloride to 50.0 cm of water in a well-insulated container. A temperature rise of 1.5 °C was recorded.The student used a balance which reads to 0.01g, a 50.0 cm pipette, and a thermometer which can be read to 0.25 °C.(i) Which measuring instrument should be changed to give a result which is closer to the accepted value? Justify your answer.Thermometer , as temperature change is small (%) error in balance smaller than for temperature reading (%) error in pipette smaller than for temperature reading as scale with greater degree of precision needed / scale with more graduations needed.

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(ii) Suggest ONE other change the student could make to give a result which is closer to the accepted value. Justify your suggestion.Use more cobalt chloride / less water To increase temperature rise *(c) The lattice energies of magnesium chloride, MgCl2, calcium chloride, CaCl2, and strontium chloride, SrCl2 are shown in the table below.

(i) Use data on ionic radii, from your data booklet, to explain the trend in these values. Estimate a value for the lattice energy of cobalt(II) chloride, giving ONE piece of data to justify your estimate.Radius (of cation) increases (down group) OR any two values of radius: Mg 2+= 0.072, Ca 2+ = 0.100 / Sr 2+ = 0.113 (nm) data may be shown beside the table (1) Radius Co2+ = 0.065 nm (ii) Explain how lattice energy values, together with other data, can be used to predict the solubility of ionic compounds.*(d) Cobalt forms another chloride, CoCl3, but scientists predict that MgCl3 cannot be made. Suggest a reason for this.You should consider the enthalpy changes in the Born-Haber cycle, which provide evidence about why cobalt(III) chloride is known but magnesium(III) chloride is not.

Chloride Lattice energy/kJ mol-1

MgCl2 –2526CaCl2 –2258SrCl2 –2156

12.The oxidation of iron metal in the presence of oxygen is spontaneous.

4Fe(s) + 3O2(g) → 2Fe2O3(s)(a)Explain the meaning of spontaneous in a thermodynamic context.(b)(b) (i) Find the values of the standard molar entropies of iron and of iron(III) oxidefrom your data booklet.(ii) The standard molar entropy at 298 K for oxygen molecules O2 is +205 J mol–1 K–1.Calculate the standard entropy change of the system for the reaction betweeniron and oxygen. Include a sign and units in your answer.(iii) The standard enthalpy change for the reaction at 25 °C is –1648 kJ mol–1.Calculate ΔSsurroundings.(iv) Use your answers to (b)(ii) and (iii) to calculate the total standard entropychange for the reaction. Include a sign and units in your answer.*(v) The reaction is thermodynamically spontaneous.Use your answers to (b)(ii), (iii) and (iv) to explain, in terms of the physicalstates of the substances in the reaction and the movement of the molecules inthe surroundings, why this is so.