Entropy change question help

Question

N2+3H2=2NH3 is an equilibrium reaction and is 7.07x10-4 mol L-1 at 623k.given that the enthalpy of the reaction is -92kjmole-1 what is the entropy change at this temperature. Note R= 8.134J k-1mol-1

Answer 1

K = 7.07 x 10^-4

T = 623 K

ΔH = -92 kJ mol^-1

R = 8.134 J K^-1 mol^-1

ΔG = - RT lnK = - 8.314 J K^-1 mol^-1 x 623 K x ln (7.07 x 10^-4) = 37575 J mol^-1 ≅ 37.6 kJ mol^-1

ΔG = ΔH - TΔS

ΔS = (ΔH - ΔG) / T

= (- 92 kJ mol^-1 - 37.6 kJ mol^-1) / 623 K

= - 0.208 kJ mol^-1 K^-1 (or - 208 J mol^-1 K^-1)