HESS’S LAW what is it ? how is it used ? AS Chemistry

H1H1 HESS’S LAW THE TOTAL ENTHALPY CHANGE OF A REACTION H2H2 REACTANTS PRODUCTS INTERMEDIATES H3H3 By Hess’s Law:  H 1 = Intermediates = IS INDEPENDENT OF THE REACTION ROUTE. oxides if ΔH C known, or gaseous atoms if E[X-Y] known elements if ΔH f known, Alternative route  H 2 +  H 3 Direct measurement of ΔH 1 may not be possible because : 1. Reaction incomplete 2. Other reactions occur 3. Reaction too slow

LEARN!!! STANDARD ENTHALPY CHANGE OF FORMATION,  H  f The heat change when ONE MOLE of a substance is FORMED from its ELEMENTS in their standard states at 298K and 100kPa NB  H o f [ELEMENT in its standard state] = ZERO Q P C STANDARD ENTHALPY CHANGE OF COMBUSTION,  H  C The heat produced when ONE MOLE of a substance is burned in excess oxygen measured at 298K and 100kPa Q P C BOND ENTHALPY or BOND ENERGY E[X-Y] The average heat needed when ONE MOLE of covalent bonds are broken, measured in the gaseous state at 298K and 100kPa Q P C ELEMENTS OXIDES GAS ATOMS

1 One mole of carbon burns to give one mole of carbon dioxide, releasing kJ. One mole of carbon burns to give one mole of carbon monoxide, releasing kJ. Calculate the energy from burning one mole of carbon monoxide. By Hess’s Law:(+110.5)  H = + (-393.5) CO(g) + ½O 2 (g) CO 2 (g) = HH = kJ mole -1 = Alternative route C(s) + O 2 (g) Elements here because  H  f given -  H  f [CO(g)]  H  f [CO 2 (g)]   H  f [CO 2 (g)] = kJ mole -1 and  H  f [CO(g)] = kJ mole -1

2.Use the bond energy data table to calculate heat of reaction for: CH 4 (g) + 2Br 2 (g)  CH 2 Br 2 (g) + 2HBr(g) By Hess’s Law:  H = CH 4 (g) + 2Br 2 (g) CH 2 Br 2 (g) + 2HBr(g) = kJ mole -1 HH +4(413) + 2(193)-2(413) - 2(290) -2(366) E data Alternative route BONDS BROKEN - ENDOTHERMIC BONDS FORMED - EXOTHERMIC C(g) + 4H(g) + 4Br(g) +4E[C-H] +2E[Br-Br] -2E[C-H] -2E[C-Br] -2E[H-Br] Atoms here because bond energies given

3 Use the  H  f values given to calculate  H  of : CH 3 COCH 3 (l) + 4O 2 (g)  3CO 2 (g) + 3H 2 O(l)  H  f [CO 2 (g)] = kJmol -1  H  f [CH 3 COCH 3 (l)] = kJmol -1  H  f [H 2 O(l)] = kJmol -1 = 3(-394) + 3(-286) HH CH 3 COCH 3 (l) + 4O 2 (g) 3CO 2 (g) + 3H 2 O(l) = +248   H = + 3  H  f [H 2 O(l)] = kJ mole -1 Elements here because  H  f given 3C(s) + 3H 2 (g) + 4.5O 2 (g) -  H  f [CH 3 COCH 3 (l)] 3  H  f [CO 2 (g)] (-394) + 3(-286)

4 Calculate  H  f [CH 4 (g)], given  H  f [CO 2 (g)] = kJ mole -1  H  f [H 2 O(l)] = kJ mole -1  H  c [CH 4 (g)] = kJ mole -1 HH C(s) + 2H 2 (g) CH 4 (g) = (-393.5) + 2(-285.8) =   H = (-285.8) = kJ mole -1 Oxides here because  H  C given CO 2 (g) + 2H 2 O(l) =  H  C [C(s)] =  H  C [H 2 (g)]  H  C [C(s)] + 2  H  C [H 2 (g)] -  H  c [CH 4 (g)]

5 Using the table of bond energies, calculate the energy (enthalpy) change for : CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) HH CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g)   H = 4E[C-H] -2E[C=O] 4(+413) + 2(498) -2(+695) - 4(+464) = kJ mole -1 E data C(g) + 4H(g) + 4O(g) Atoms here because bond energies given + 2E[O=O] -4E[O-H]

6 Calculate  H  R for : C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g), given  H  c [C 2 H 4 (g)] = kJ mole -1  H  c [H 2 (g)] = kJ mole -1  H  c [C 2 H 6 (g)] = kJ mole -1 HH C 2 H 4 (g) + H 2 (g) C 2 H 6 (g)   H = = ( ) + (-285.8) = = kJ mole -1 ( ) + (-285.8) + ( ) 2CO 2 (g) + 3H 2 O(l) Oxides here because  H  C given  H  c [C 2 H 4 (g)] +  H  c [H 2 (g)] -  H  c [C 2 H 6 (g)] + 3.5O 2 (g)

NB  H values calculated from bond energies are 1.Average values used 2.Gaseous state may not apply. Also, you may care to remember :  H  r =  H  f [PRODUCTS] -  H  f [REACTANTS] APPROXIMATE because:  H  r =  H  C [REACTANTS] -  H  C [PRODUCTS]  H  r = E [REACTANTS] - E [PRODUCTS]

The End

H C N O S F Cl Br I H C N O 144 S F Cl Br I 151 C=C 612 C  C 838 O=O 498 N  N 945 N=N 410 C  O 1077 C=O 695 C  N 887 SOME BOND ENERGY VALUES (all in kJ mole -1 ) For SINGLE covalent bonds :