Unit 12. Thermochemistry

Transcription

1 Unit 12 Thermochemistry

2 A reaction is spontaneous if it will occur without a continuous input of energy However, it may require an initial input of energy to get it started (activation energy)

3 For Thermochemistry standard state is 1 atm 25ºC

4 Enthalpy is the change in heat at a constant pressure Entropy is the disorder of a system

5 Low entropy (more ordered) High entropy (less ordered) 2 pure substances mixture of 2 substances solid liquid liquid gas lower temperature higher temperature crystalline solid amorphous solid smaller molecule larger molecule

6 (Gibb s) free energy is the energy (enthalpy and entropy) released or absorbed by a reaction. ΔG = ΔH TΔS

7 The enthalpy of formation is the energy required to form 1 mole of a substance from its constituent elements at standard conditions. ΔH f º The º denotes standard conditions. The enthalpy of formation for any element in its standard state is zero.

8 The enthalpy for any reaction is the enthalpy (of formation) of the products minus the enthalpy (of formation) of the reactants. ΔHº rxn = ΣΔH f º (products) ΣΔH f º (reactants)

9 Calculate the ΔHº rxn for the reaction: P 4 O 10(s) + 6H 2 O (l) 4H 3 PO 4(aq). ΔHº f : P 4 O 10(s) = 2980 kj/mol; H 2 O (l) = 286 kj/mol; H 3 PO 4(aq) = 1280 kj/mol. ΔHºrxn = ΣΔH f º (products) ΣΔH f º (reactants) ΔHº rxn = (4)ΔH f º (H3PO4(aq) ) [(1)ΔH f º (P4O10(s) ) + (6)ΔH f º (H2O(l) )] ΔHº rxn = (4)( 1280) [(1)( 2980) + (6)( 286)] ΔHº rxn = 424 kj

10 Coefficients in thermochemistry equations are not always whole numbers.

11 Determine the value of ΔHº rxn for the following reaction. 2C (gr) + N 2(g) + H 2(g) 2HCN (g) Bond Bond Energy (kj/mol) H H 432 C H 411 C N 887 N N 167

12 ΔHº rxn = [E N N + E H H ] [(2)E H C + (2)E C N ] ΔHº rxn = ( ) [(2)411 + (2)887)] ΔHº rxn = 1997 kj

13 2C 2 H 6(g) + 7O 2(g) 4CO 2(g) + 6H 2 O (g) ΔH =? (1) C 2 H 4(g) + 3O 2(g) 2CO 2(g) + 2H 2 O (g) ΔH = 1323 kj (2) C 2 H 4(g) + H 2(g) C 2 H 6(g) ΔH = +137 kj (3) H 2(g) + ½O 2(g) H 2 O (g) ΔH = 242 kj (1a) 2C 2 H 4(g) + 6O 2(g) 4CO 2(g) + 4H 2 O (g) ΔH = 2646 kj (2a) 2C 2 H 6(g) 2C 2 H 4(g) + 2H 2(g) ΔH = 274 kj (3a) 2H 2(g) + O 2(g) 2H 2 O (g) ΔH = 484 kj 2C 2 H 6(g) + 7O 2(g) 4CO 2(g) + 6H 2 O (g) ΔH = 3404 kj

14 When ΔG is positive the reaction will NOT be spontaneous When ΔG is negative the reaction WILL be spontaneous When ΔG is equal to zero the reaction is at equilibrium

15 ΔH ΔS ΔG Spontaneity Explanation when ΔS is positive the + term TΔS is negative. + + or spontaneous at high T When TΔS > ΔH then ΔG is negative + + never spontaneous both ΔH and TΔS will always be positive, therefore ΔG will always be positive + always spontaneous both ΔH and TΔS will always be negative, therefore ΔG will always be negative when ΔS is negative the term TΔS is + or spontaneous at low T positive. When TΔS < ΔH then ΔG is negative

16 In most tables ΔH is given in units of kj/mol while ΔS is given in J/mol K. You MUST convert one of them before using them together in the equation. What is typically done is to convert ΔS to kj before use. ΔG will then have units of kj also. Your temperature must also be converted to Kelvin. If you forget to do either of these things your answer WILL BE WRONG!

17 Will the following reaction occur spontaneously at 25ºC? If there is a temperature at which spontaneity changes direction calculate it, if not explain why not. ΔS = 210. J, ΔH = 424 kj. P 4 O 10(s) + 6H 2 O (l) 4H 3 PO 4(aq) ΔG = ΔH TΔS ΔG = ( 424 kj) (298 K)( kj/k) ΔG = 361 kj A negative value of ΔG means it is definitely spontaneous. Spontaneity will change direction when ΔG = 0 ΔG = ΔH TΔS 0 = ( 424 kj) T( kj) T = 2020 K or 1750ºC

18 Free Energy and the Equilibrium Constant Everything we have seen to this point has been done at standard conditions. We can adjust to nonstandard conditions by using the following formula ΔG = ΔGº + RT ln Q Where R is the gas law constant: 8.31 J/K mol Q is the reaction quotient (calculated like the equilibrium constant, only we are not at equilibrium) At equilibrium the equation will reduce to ΔGº = RT ln K

19 Calculate the equilibrium constant at 25ºC. ΔGº = 13.6 kj. 2NH 3(g) + CO 2(g) NH 2 CONH 2(aq) + H 2 O (l) Rearranging the equation we have ln K = G RT ln K = J (8.31 J/K mol)(298 K) Note that we had to change units on ΔGº because they were incompatible with the units of R ln K = 5.49 K = e 5.49 K = 242