Part A: Using the heats of formation and standard entropy, calculate H0 and S0 t
ID: 511979 • Letter: P
Question
Part A: Using the heats of formation and standard entropy, calculate H0 and S0 then use these to calculate G0 and determine if the reaction is spontaneous at 1000C. If not, under what temperature conditions will it become spontaneous? and K at the given temperature. Assume H0 and S0 are not affected by temperature.
O2(g) + 2F2(g) <----> 2F2O(g) T= 1000C
Hf0(kJ/mol) 0 0 24.7
S0(J/K-mol) 205.2 202.8 247.43
Part B: Calculate equilibrium constant (K) and state whether products or reactants are favored
Explanation / Answer
dHo reaction = dHo products - dHo reactants
= 2dHO ( F2O) - dHo (O2) - 2dHO ( F2)
= 2( 24.7) - 0 - 2( 0)
= 49.4 KJ/mol
dSo rxn = 2 dSo ( F2O) - dSo ( O2) - 2 dSO ( F2)
= 2( 247.43) - 205.2 - 2 ( 202.8)
= 86.86 J/molK = 0.08686 KJ/molK
T = 1000C = 1000 + 273 K = 1273 K
dGo = dHo - TdSo is formula
dGo = 49.4 KJ/mol - 1273 K ( 0.08686 KJ/molK)
= - 61.173 KJ/mol
Reaction is sponatenous at 1000 C ( since we got dGo = -ve value , hence sponatenous)
we have relation dGo = - RT ln Keq
- 61173 J/mol = - 8.314 J/molK x 1273 K ln ( keq)
ln Keq = 5.7799
Keq = exp ( 5.7799)
= 323.73