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Worked solutions: Chapter 6 Energetics

© Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008.

This page from the Chemistry: For use with the IB Diploma Programme SL Teacher’s Resource may be reproduced for classroom use.

Section 6.1 Exercises

1 a endothermic

b endothermic

c exothermic

d exothermic

e exothermic

f endothermic

2 Reaction e has the greatest difference between the enthalpy of the reactants and the products: 2C2H5OH(l) + 7O2(g) → 4CO2(g) + 6H2O(l) ∆H = –2734 kJ mol–1

3 a The spark supplies the energy required to overcome the activation energy of the reaction and so initiate the combustion process.

b Covalent bonds within the molecules of hydrogen and molecules of oxygen are broken.

c Covalent bonds in water molecules are generated in the products.

d This reaction is exothermic as the reaction was ‘explosive’, indicating that a considerable amount of energy was released.

4 a The total enthalpy of the products is greater than that of the reactants, as the temperature decreases (energy was absorbed from the surroundings).

b The reaction is endothermic.

c Ionic bonds between ammonium cations and nitrate anions were broken.

5 a The activation energy is +1100 kJ mol–1.

b The change in enthalpy, ∆H, is –900 kJ mol–1.

6

This reaction is endothermic.




7 a Hydrogen and oxygen would both be gases.

b The hydrogen and oxygen would both be at a pressure of 101.3 kPa (1 atm).

8 a

b The forward reaction is exothermic.

c ∆H = –90 – 80 = –170 kJ mol–1

d The amount of energy released as the products are formed = 200 + 90 = 290 kJ mol–1.

e Activation energy of the reverse reaction = 290 kJ mol–1.

9 a The total enthalpy of the solid NaOH will be greater than that of the dissolved solution, as energy was released during the dissolution process.

b The reaction is exothermic.

c Ionic bonds between sodium cations and hydroxide anions, and some hydrogen bonds between water molecules were broken.

10 Ea = +150 kJ mol–1

∆H = –100 – 0 = –100 kJ mol–1


1 q = m × c × ∆T

∴ ∆T = 46.2350

30022×

=× cmq = 25.9°C

2 q = m × c × ∆T = 1000 × 1000 × 0.448 × (550 – 24.0) = 2.36 × 108 J = 236 MJ

3 q = m × c × ∆T = 724 × 4.18 × (50.7 – 7.44) = 1.31 × 105 J = 131 kJ

From the equation, 2 mol of C4H10 yields 5748 kJ

∴ x mol of C4H10 yields 131 kJ




57482 =

131x

x = 5748

1312× = 0.0456 mol

m(C4H10) = n × M = 0.0456 × 58.14 = 2.65 g

4 n(NH4NO3) = 06.802.12

=Mm = 0.152 mol

From the equation, 1 mol of NH4NO3 absorbs 25 kJ

∴ 0.152 mol of NH4NO3 absorbs x kJ

251 =

x152.0

x = 25 × 0.152 = 3.8 kJ = 3800 J

q = m × c × ∆T

∴ ∆T = 18.40.85

3800×

=×cmq = 10.7°C

The water will reach a temperature of (28.3 – 10.7) = 17.6°C.

5 n(C3H8) = 11.44

100=

Mm = 2.27 mol

From the equation, 1 mol of C3H8 yields 2217 kJ

∴ 2.27 mol of C3H8 yields x kJ

22171 =

x27.2

x = 2217 × 2.27 = 5.03 × 103 kJ

The energy released by the combustion reaction is 5.03 × 103 kJ.

6 n(C4H9OH) = 14.74

806=

Mm = 10.9 mol

From the equation, 2 mol of C4H9OH yields 5354 kJ

∴ 10.9 mol of C4H9OH yields x kJ

53542 =

x9.10

x = 2

9.105354× = 2.91 × 104 kJ

The energy released by the combustion reaction is 2.91 × 104 kJ.




7 From the equation, 1 mol of C5H12 yields 3509 kJ

∴ x mol of C5H12 yields 500 kJ

35091 =

500x

x = 3509500 = 0.142 mol

m(C5H12) = n × M = 0.142 × 72.17 = 10.2 g

10.2 g of pentane would release 500 kJ of energy.

8 n(B2H6) = 68.271000.1 3×

=Mm = 36.1 mol

From the equation, 36.1 mol of B2H6 yields 73 700 kJ

∴ 1 mol of B2H6 yields ∆H

700731.36 =

H∆1

∆H = 1.36

70073 = 2.04 × 103 kJ

The thermochemical equation for the combustion of borane is:

B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g) ∆H = –2.04 × 103 kJ mol–1

9 a i Heat of combustion = –725 kJ mol–1

ii Heat of combustion = 05.32

725 = –22.6 kJ g–1

iii In 1.0 dm3 of methanol there is 787 g

∴ heat of combustion = 787 × 22.6 = –17.8 × 103 kJ dm–3 = –17.8 MJ dm–3

b m(CH3OH) = d × V

= 0.787 × 15.0 × 103

= 1.180 × 104 g

n(CH3OH) = 05.32

10180.1 4×=

Mm = 368.2 mol

n(CO2) = n(CH3OH) = 368.2 mol

V = P

nRT = 3.10198.0

)35273(31.82.368×

+×× = 9.49 × 104 dm3





1 Fe(s) + 21 O2(g) → FeO(s) ∆Hο = –272 kJ mol–1 ............................... (1)

3Fe(s) + 2O2(g) → Fe3O4(s) ∆Hο = –1118 kJ mol–1 ............................ (2)

Reverse (1) and multiply × 3:

3FeO(s) → 3Fe(s) + 23 O2(g) ∆Hο = +816 kJ mol–1 .............................. (3)

Use (2) as is: 3Fe(s) + 2O2(g) → Fe3O4(s) ∆Hο = –1118 kJ mol–1 ............................ (2)

Add (2) and (3): 3FeO(s) + 21 O2(g) → Fe3O4(s) ∆Hο = –302 kJ mol–1

2 S(s) + 23 O2(g) → SO3(g) ∆Hο = –395.2 kJ mol–1 ............................(1)

2SO2(g) + O2(g) → 2SO3(g) ∆Hο = –198.2 kJ mol–1 ............................(2)

Reverse (2): 2SO3(g) → 2SO2(g) + O2(g) ∆Hο= +198.2 kJ mol–1 .............................(3)

Halve (3): SO3(g) → SO2(g) + 21 O2(g) ∆Hο = +99.1 kJ mol–1 .............................(4)

S(s) + 23 O2(g) → SO3(g) ∆Hο = –395.2 kJ mol–1 ...........................(1)

Add (1) and (4): S(s) + O2(g) → SO2(g) ∆Hο = –296.1 kJ mol–1

3 N2(g) + 2O2(g) → 2NO2(g) ∆Hο = +67.7 kJ mol–1 ..............................(1)

N2(g) + 2O2(g) → N2O4(g) ∆Hο = +9.7 kJ mol–1 ................................(2)

Reverse (1): 2NO2(g) → N2(g) + 2O2(g) ∆Hο = –67.7 kJ mol–1 ..............................(3)

N2(g) + 2O2(g) → N2O4(g) ∆Hο = +9.7 kJ mol–1 ................................(2)

Add (2) and (3): 2NO2(g) → N2O4(g) ∆Hο = –58.0 kJ mol–1

4 H2(g) + 21 O2(g) → H2O(l) ∆Hο = –285.8 kJ mol–1 ...........................(1)

N2O5(g) + H2O(l) → 2HNO3(l) ∆Hο = –76.6 kJ mol–1 ..............................(2)

21 N2(g) + 2

3 O2(g) + 21 H2(g) → HNO3(l) ∆Hο = –174.1 kJ mol–1 ............................(3)

Multiply (3) × 4: 2N2(g) + 6O2(g) + 2H2(g) → 4HNO3(l)

∆Hο = –696.4 kJ mol–1 ...........................(4)

Reverse and double (2): 4HNO3(l) → 2N2O5(g) + 2H2O(l)

∆Hο = +153.2 kJ mol–1 ...........................(5)




Reverse and double (1): 2H2O(l) → 2H2(g) + O2(g)

∆Hο = –571.6 kJ mol–1 .........(6)

Add (4), (5) & (6): 2N2(g) + 5O2(g) → 2N2O5(g) ∆Hº = –1114.8 kJ mol–1

5 C2H2(g) + 25 O2(g) → 2CO2(g) + H2O(l) ∆Hο = –1300 kJ mol–1 ..........(1)

C(s) + O2(g) → CO2(g) ∆Hο = –394 kJ mol–1 ............(2)

H2(g) + 21 O2(g) → H2O(l) ∆Hο = –286 kJ mol–1 .............(3)

Double (2): 2C(s) + 2O2(g) → 2CO2(g) ∆Hο = –788 kJ mol–1 ............(4)

Reverse (1): 2CO2(g) + H2O(l) → C2H2(g) + 25 O2(g) ∆Hο = +1300 kJ mol–1 ..........(5)

Use (3) as is: H2(g) + 21 O2(g) → H2O(l) ∆Hο = –286 kJ mol–1 ............(3)

Add (3), (4) & (5): 2C(s) + H2(g) → C2H2(g) ∆Hο = +226 kJ mol–1

6 C(s) + O2(g) → CO2(g) ∆Hο = –390 kJ mol–1 ............(1)

Mn(s) + O2(g) → MnO2(s) ∆Hο = –520 kJ mol–1 .............(2)

Use (1) as is: C(s) + O2(g) → CO2(g) ∆Hο = –390 kJ mol–1 ............(1)

Reverse (2): MnO2(s) → Mn(s) + O2(g) ∆Hο = +520 kJ mol–1.............(3)

Add (1) & (3): MnO2(s) + C(s) → Mn(s) + CO2(g) ∆Hο = +130 kJ mol–1

7. O2(g) + H2(g) → 2OH(g) ∆Hο = +77.9 kJ mol–1 ...........(1)

O2(g) → 2O(g) ∆Hο = +495 kJ mol–1.............(2)

H2(g) → 2H(g) ∆Hο = +435.9 kJ mol–1 .........(3)

Reverse & halve (2): O(g) → 21 O2(g) ∆Hο = –247.5 kJ mol–1 ..........(4)

Reverse & halve (3): H(g) → 21 H2(g) ∆Hο = –217.95 kJ mol–1 .......(5)

Halve (1): 21 O2(g) + 2

1 H2(g) → OH(g) ∆Hο = +38.95 kJ mol–1 .........(6)

Add (4), (5) & (6): O(g) + H(g) → OH(g) ∆Hο = –426.5 kJ mol–1


1 Bond breaking of reactants is an endothermic process and bond making of products is exothermic. If the sum of the bond enthalpies of the reactants is greater than the sum of the bond enthalpies of the products, then the reaction will be endothermic.




2 A C=C involves two shared pairs of electrons. This bond is stronger than C–C because there is a greater electrostatic attraction between the nuclei and the two bonding pairs of electrons of C=C than between the nuclei and one bonding pair of electrons in C–C. Consequently, more energy is required to break the C=C than C–C, so the average bond enthalpy is greater.

3 H–H + Cl–Cl → H–Cl + H–Cl

∆H = [DH–H + DCl–Cl] – [2DH–Cl]

= [436 + 242] – [2 × 431]

= 678 – 862

= –184 kJ mol–1

4

∆H = [6DC–H + 2DC–O + 2DO–H + 3DO=O] – [4DC=O + 8DH–O]

= [6 × 412 + 2 × 360 + 2 × 463 + 3 × 496] – [4 × 743 + 8 × 463]

= 5606 – 6676

= –1070 kJ mol–1

To find the heat of combustion, this value must be divided by 2, as the equation was balanced for 2 mole of methanol.

∴ Heat of combustion of methanol = –535 kJ mol–1

5 a 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

∆H = [2DC–C + 12DC–H + 7DO=O] – [8DC=O + 12DH–O]

= [2 × 348 + 12 × 412 + 7 × 496] – [8 × 743 + 12 × 463]

= 9112–11500

= –2388 kJ (for 2 mole ethane)

∴ Heat of combustion of ethane = –1194 kJ mol–1




b C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)

∆H = [DC=C + 4DC–H + 3DO=O] – [4DC=O + 4DH–O]

= [612 + 4 × 412 + 3 × 496] – [4 × 743 + 4 × 463]

= 3748 – 4824

= –1076 kJ mol–1

∴ Heat of combustion of ethene = –1076 kJ mol–1

c The heat of combustion of ethene is less exothermic than that of ethane because the C=C double bond has a higher bond enthalpy than C–C, so more energy is needed to break the bonds in the combustion of ethene, than for ethane, reducing the negativity of the final result.

6 Since ethyne has a triple bond between the two carbon atoms, more energy will be needed to break the bonds of ethyne than ethene (C=C), so ethene would have a more negative heat of combustion.

7 a The equation for the combustion of ethanoic acid is

CH3COOH + 2O2 → 2CO2 + 2H2O

∆H = [DC–C + 3DC–H + DC=O + DC–O + DO–H + 2DO=O] – [4DC=O + 4DH–O]

= [348 + 3 × 412 +743 + 360 + 463 + 2 × 496] – [4 × 743 + 4 × 463]

= 4142 – 4824

= –682 kJ mol–1

∴ Heat of combustion of ethanoic acid = –682 kJ mol–1




b The equation for the combustion of ethanal is

2CH3CHO + 5O2 → 4CO2 + 4H2O

∆H = [2DC–C + 8DC–H + 2DC=O + 5DO=O] – [8DC=O + 8DH–O]

= [2 × 348 + 8 × 412 + 2 × 743 + 5 × 496] – [8 × 743 + 8 × 463]

= 7958 – 9648

= –1690 kJ (2 mole of ethanal)

∴ Heat of combustion of ethanal = –845 kJ mol–1

c The heat of combustion of ethanal is greater than that of ethanoic acid because ethanoic acid has an extra C–O bond and a O–H (instead of C–H in ethanal), which have to be broken, creating a greater endothermic component to the calculation of heat of combustion, so reducing its value.

Chapter 6 Review questions

1 a The spark provides the activation energy required to initiate combustion of the methane gas.

b Covalent bonds within the methane and oxygen molecules are broken.

c Covalent bonds within the carbon dioxide and water molecule products are formed.

d It is exothermic, as all combustion reactions generate heat as a product of their reaction.

2 Endothermic: a, c and e

Exothermic: b and d

3 The required equation can be obtained by reversing the second equation and adding it to the first equation.

∆H = –393 + (+395) = +2.0 kJ mol–1

4 a ∆H = 2 × –272 = –544 kJ mol–1

b ∆H = 2 × (+824) = +1648 kJ mol–1

c ∆H = 3 × (+272) + (–1118) = –302 kJ mol–1




5 a

b The reaction is exothermic.

c Energy released as products formed = 250 kJ

d Activation energy of the reverse reaction = 330 kJ mol–1

6

7 n(Na2SO4.10H2O) = ==24.3220.25

Mm 0.0776 mol

From the equation, 1 mol of Na2SO4.10H2O absorbs 79.0 kJ

∴ 0.0776 mol of Na2SO4.10H2O absorbs x kJ

0.791 =

x0776.0

x = 79.0 × 0.0776 = 6.13 kJ

Heat energy absorbed when crystals dissolve = 6.13 kJ

8 1 dm3 of water has a mass of 1000 g

n(H2O) = ==02.18

1000Mm 55.49 mol




From the equation, 1 mol of H2O absorbs 44.0 kJ

∴ 55.49 mol of H2O absorbs x kJ

0.441 =

x49.55

x = 44.0 × 55.49 = 2.44 × 103 kJ

Energy required = 2.44 × 103 kJ

9 a q = m × c × ∆T = 20.0 × 4.18 × (85.0 – 11.0) = 6186 J = 6.19 kJ

b m(ethanol) = density × volume = 0.785 × 45.0 = 35.3 g

q = m × c × ∆T = 35.3 × 2.46 × (37.0 – 4.00) = 2866 J = 2.87 kJ

c q = m × c × ∆T = 1000 × 103 × 0.448 × (1000 – 25.0) = 4.37 × 108 J

10 q = m × c × ∆T ∴ ∆T = 1841200

000200.cm

q

×=

× = 39.9°C

Final temperature of water = 13.3 + 39.9 = 53.2°C

11 n(C) = ==01.120.10

Mm 0.833 mol

From the equation, 1 mol of C releases 393.4 kJ

∴ 0.833 mol of C releases x kJ

4.3931 =

x833.0

x = 393.4 × 0.833 = 328 kJ

Energy released = 328 kJ

12 n(Fe3O4) = ==55.231

250Mm 1.08 mol

From the equation, 1 mol of Fe3O4 absorbs 38.0 kJ

∴ 1.08 mol of Fe3O4 absorbs x kJ

0.381 =

x08.1

x = 38.0 × 1.08 = 41.0 kJ

Energy absorbed = 41.0 kJ




13 n(O2) = 4.22

500.0

m

=VV = 0.0223 mol

From the equation, when 1 mol of O2 is produced, 572 kJ of energy is consumed.

∴ 0.0223 mol of O2 consumes x kJ

5721 =

x0223.0

x = 572 × 0.0223 = 12.8 kJ

Electrical energy required = 12.8 kJ

14 n(CuSO4) = ==61.159

73.5Mm 0.0359 mol

If 0.0359 mol of CuSO4 releases 4.81 kJ

1 mol of CuSO4 releases x kJ

81.4359.0 =

x1

x = 0359.0

81.4 = 134 kJ

CuSO4(s) + 5H2O(l) → CuSO4.5H2O(s) ∆H = –134 kJ mol–1

15 ∆H = 2 × 4158 = –8316 kJ mol–1

16 ∆H = [10DC–C + 28DC–H + 19DO=O] – [24DC=O + 28DH–O]

= [10 × 348 + 28 × 412 + 19 × 496] – [24 × 743 + 28 × 463]

= 24440 – 30796

= –6356 kJ (2 mole of C6H14) Average bond enthalpies are based on a range of compounds containing that bond. The bond enthalpy of a bond is not exactly the same in all compounds containing that bond.

17 Heat of combustion = 21 × 5354 = –2677 kJ mol–1

18 q = m × c × ∆T = 2000 × 4.18 × (100 – 16.6) = 697 kJ

From the equation, 1 mol of propane produces 2217 kJ

∴ x mol of propane produces 697 kJ

22171 =

697x

x = 2217697 = 0.314 mol

m(propane) required = n × M = 0.314 × 44.11 = 13.9 g




19 ∆H = [DC=C + 4DC–H + DH–H] – [DC–C + 6DC–H]

= [612 + 4 × 412 + 436] – [348 + 6 × 412]

= 2696 – 2820

= –124 kJ

20 a n(CH3OH) = 05.320.50 = 1.56 mol

b E = 1.56 × 715 = 1115 kJ

c E = m × c × ∆T

1 115 000 = m × 4.18 × (100 – 15)

m = 3138 g

Volume of water = 3.14 dm3

d The campers will be able to boil less than this amount in reality because some heat from the burning methanol will be lost to the surroundings.

Chapter 6 Test

Part A: Multiple-choice questions

Question Answer Explanation 1 C In exothermic reactions the products have a lower enthalpy than the reactants. 2 C q = m × c × ∆T

therefore, using the symbols given in the question, q = y × c × z = cyz The mass of water is used rather than the mass of sodium hydroxide as the specific heat refers to that of water.

3 B Enthalpy of reaction = ∑D(bonds broken) – ∑D(bonds formed) = 2DH–Cl – [DH–H + DCl–Cl] = 2 × 431 – [436 + 242] = 862 – 678 = +184 kJ

4 D Temperature decrease indicates an endothermic reaction for which ∆H is positive. 5 A 2H2O2(l) → 2H2(g) + 2O2(g) ∆H = +375.2 kJ

2H2(g) + O2(g) → 2H2O(l) ∆H = –571.6 kJ 2H2O2(l) → 2H2O(l) + O2(g) ∆H = (+375.2 – 571.6) = –196.4 kJ

6 D When chemical bonds are formed, energy is released and when chemical bonds are broken, energy is absorbed.

7 A When 3 mol of hydrogen is formed, 210 kJ of energy is absorbed; therefore, when 1 mol of hydrogen is formed, 210 ÷ 3 = 70 kJ of energy is absorbed.

8 D 2S(s) + 2O2(g) → 2SO2(g) ∆H = –600 kJ 2SO2(g) + O2(g) → 2SO3(g) ∆H = –200 kJ 2S(s) + 3O2(g) → 2SO3(g) ∆H = –600 – 200 = –800 kJ

9 C Substances with a lower enthalpy are more stable. In endothermic reactions, ∆H is positive because the reactants are more stable (have lower enthalpy) than the products.

10 D Ionization (A) & (C) and bond breaking (B) are endothermic reactions. Sublimation (D) from gas to solid releases energy.




Note that key information is indicated in bold type.

Part B: Short-answer questions

1 a Average bond enthalpy is the energy needed to break 1 mol of a bond in gaseous molecules. The value is averaged over similar compounds.

(2 marks) b CH3OH(g) + NH3(g) → CH3NH2(g) + H2O(g)

∆H = [DC–O + 3DC–H + DO–H + 3DN–H] – [DC–N + 3DC–H + 2DN–H + 2DO–H]

= [360 + 3 × 412 + 463 + 3 × 338] – [305 + 3 × 412 + 2 × 338 + 2 × 463]

= 3223 – 3243

= –20 kJ mol–1 (4 marks)

2 The equation is C + 2H2 + 21 O2 → CH3OH ∆H4

Reverse this equation: CH3OH + 1 21 O2 → CO2 + 2H2O ∆H1 = –676 kJ mol–1

Use this equation as is: C + O2 → CO2 ∆H2 = –394 kJ mol–1

Double this equation: H2 + 21 O2 → H2O ∆H3 = –242 kJ mol–1

CO2 + 2H2O → CH3OH + 1 21 O2 –∆H1 = +676 kJ mol–1

C + O2 → CO2 ∆H2 = –394 kJ mol–1

2H2 + O2 → 2H2O 2∆H3 = –484 kJ mol–1

Add all the enthalpies and equations together:

C + 2H2 + 21 O2 → CH3OH ∆H4 = +676 – 394 – 484 = –202 kJ mol–1

(4 marks)

3 CH4(g) + Br2(g) → CH3Br(g) + HBr(g) ∆H = [4DC–H + DBr–Br] – [3DC–H + DC–Br + DH–Br]

= (4 × 412 + 193) – (3 × 412 + 276 + 366)

= 1841 – 1878

= –37 kJ mol–1 (3 marks)




b

(2 marks)

c The enthalpy change for the reaction given should be about the same as the reaction in part a because there are the same number and type of bonds being broken and formed.

(2 marks)

Part C: Data-based question

a CuSO4 + Zn → ZnSO4 + Cu

b n(Zn) = 37.65

20.1 = 0.018 mol

n(Cu2+) = 1000

50 × 0.200 = 0.010 mol

Ratio: Zn:Cu2+ is 1:1, therefore Zn is in excess. (2 marks)

c At point A the heat being given out by the reaction is equal to the heat being lost to the surroundings.

(1 mark)

d The temperature rise can be determined by extrapolating the line to when the zinc was added: Temperature rise = 26.7 – 17.0 = 9.7°C

(2 marks)

e Heat evolved = m × c × ∆T = 50 × 4.18 × 9.7 = 2027.3 J = 2.03 kJ (2 marks)

f n(CuSO4) = 0.200 × 0.050 = 0.010 mol

∆H = 010.0

03.2− = –203 kJ mol–1

(1 mark)

g Percentage error = 218

203218− × 1

100 = 6.9%

(1 mark)




h The disagreement between the experimental value and the accepted value may have occurred because the solution was assumed to have the same specific heat capacity as water. The heating of the metals (excess zinc and evolved copper) and of the thermometer were ignored.

(1 mark)

Part D: Extended-response question

a Standard enthalpy change of reaction is the enthalpy difference between products and reactants at 1 atm (101 kPa) pressure.

(1 mark)

b • Take a known volume of sodium hydroxide solution of known concentration.

• Place solution in an insulated vessel and record its temperature.

• Add a solution of HCl of known temperature so that an equal number of moles or a known quantity is added.

• Stir.

• Record the final temperature.

• Write the equation for the reaction taking place: HCl + NaOH → NaCl + H2O

• Measure temperature rise.

• ∆H = total mass × specific heat capacity × temperature rise

• Divide by number of moles of limiting reactant if excess or other reactant used. (9 marks)

c

(4 marks)

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