Calculate the value of the equilibrium constant, Keq , for the following reaction at 298 Kelvin.?

Question

Calculate the value of the equilibrium constant, Keq , for the following reaction at 298 Kelvin.
(Use the reaction free energy given below.)


2NO + O2 = 2NO2 in the gaseous state


ΔGo = −55.6 kJ/mol

IT IS NOT 1.02.

Answer 1

delta G = -RT ln K
K = e ^ (-deltaG/RT) = e ^ (55600/8.314 * 298)
K = e ^ 22.14 = 5.57 * 10^9

Answer 2

Lancenigo di Villorba (television), Italy i will practice the "van't Hoff's Isothermal Equation", this is DeltaG = DeltaG° + R * T * SUM[nu,i * LN(a,i)] the place nu,i means the Stoichiometric Coefficient of the i-th Chemical Stuff on a similar time as a,i means the Chemical pastime of the i-th Chemical Stuff. As you recognize, the Chemical Equilibrium is a undertaking the place it verify right here info : -) SUM[nu,i * LN(a,i)] = LN(Keq) -) DeltaG = 0.0 i will derive that 0.0 = DeltaG = DeltaG° + R * T * LN(Keq) this is Keq = EXP(0 - DeltaG° / (R * T)) CALCULATIONs via making use of the won relation Keq = EXP(0 - DeltaG° / (R * T)) = = EXP(0 - (+3.16E+4) / (8.3 * 298)) = 2.8E-6 i'm hoping this helps you.