Question

Consider the following reaction at 25 ^oC:

N_2(g) + 3H_2(g) --> 2NH_3(g) ∆H^o_rxn= -92.6 kJ/mol and ∆S^o_rxn= -199 J/molK  

Calculate the change in entropy of the universe, ∆S_univ. Does the result indicate that the reaction is spontaneous

Answer 1

Given delta H rxn = -92.6 kJ /mol

and delta S rxn = -199J/mol.K

temperature = 25 c = 298 K

Thus delta G rxn = delta H - T.delta S

= -92600J /mol- 298K (-199J/mol.K)

= -33298 j

= -33.298 kJ

Thus the reaction is spontaneous at the given temperature

delta S surr = actual heat transferred to the surroundings /temperature

heat transferred to the surroundings = - DeltaH rxn

= + 92.6kJ/mol

= 92600J

Thus delta S surr = 92600J /298

= + 310.74 J

Hence delta S univ = deltaSrxn + delta S surr

= -199 J + 310.74 J

= + 111.74 J

Thus the result (delta S univ is positive) indicates the reaction is SPONTANEOUS.