Consider the following oxidation reaction at 1.00 atm and 25.0 degree C: What is
ID: 844622 • Letter: C
Question
Consider the following oxidation reaction at 1.00 atm and 25.0 degree C: What is the heat of reaction q, associated with the oxidation of 4.00 mol of SO2(g) at constant pressure? Is the reaction endothermic or exothermic. Is this reaction spontaneous? Justify your answer by calculating the total entropy change of the reaction for 1.00 mol of SO2(g). What is the standard Gibbs free energy change Delta G degree associated with this reaction per mole of SO2(g)? What is the equilibrium constant K for this reaction at 25.0 degree C? Based on the value of K, is the forward reaction favorable? Briefly justify with 1-2 sentences.Explanation / Answer
2SO2 + O2 ---> SO3
Heat of reaction = 2 x delta H (SO3) - delta H (O2) - 2 x delta H (SO2)
=> Heat of reaction = 2 x -396 - 0 - 2 x -297 = -198 KJ/mole of reaction (Associated with 2 moles of SO2)
=> Heat of reaction associated with 4 moles of SO2 = 2 x -198 = -396 KJ
The reaction is exothermic
2) delta S = 2 x 257 - 205 - 2 x 248 = -187 J/K mol
delta G = delta H - T x delta S
=> delta G = -198000 - 298 x 187 = -142274 J
delta G = -ve => Process is spontaneous
3) delta G for 2 moles of SO2 = -142274 J
=> for 1 mole of SO2 = -142274/2 = -71137 J = -71.137 KJ
4) delta G = -RT ln K
=> -71137 = - 8.314 x 298 x ln K
=> K = 2.95 x 10^12
K > 1 => Reaction is forward favourable