Question

Calculate the standard entropy, Delta S_rxn^degree, of the following reaction at 25.0 degree C using the data in this table. The standard enthalpy of the reaction, Delta H_rxn^degree, is -633.1 kJ middot mol^-1. 3C_2 H_2(t) rightarrow C_6 H_6 (l) Delta S_rxn^degree = Number J middot K^-1 middot mol^-1 Then, calculate the standard Gibbs free energy of the reaction, Delta G_rxn^degree. Delta G_rxn^degree = Number kJ middot mol^-1 Finally, determine which direction the reaction is spontaneous as written at 25.0 degree C and standard pressure. forward reverse both neither

Answer 1

For reaction we need:

SRxn = Sproduct - Sreactants

Srxn = Benzen e- 3*Ethylene = 173.4 -3*219.4 = -484.8J/molK

b)

Grxn:

dG = dH - T*dS

dG = -633.1*1000 - 298*(-484.8) = -488629.6 J/mol = -488.63 kJ/mol

c)

this is negative in dG change, so the process must be SPONTENEOUS