Consider the following reaction: € 2NO2(g) N2O4(g) At 298 K for this reaction, H
ID: 884644 • Letter: C
Question
Consider the following reaction: € 2NO2(g) N2O4(g) At 298 K for this reaction, H° = -58.0 kJ and S° = -176.6 J/K. 4.
(a) What is G° for the forward reaction at 298 K? Is the reaction spontaneous at 298 K? Assuming the enthalpy and entropy change do not vary with temperature, at what special temperature, T*, does G° = 0.0 J? 4.
(b) Express the equilibrium constant K for this reaction in terms of equilibrium pressures of the reactant and product gases. What is K at 298 K?
(c) Suppose the initial pressures are: € pNO2 =1.00 atm and pN2O4 =1.00 atm at 298 K. Predict the direction in which the reaction will shift as equilibrium is approached. Show your work and give a brief sentence to explain your reasoning
Explanation / Answer
a)
2NO2(g) N2O4(g)
DG0 = DH0-TDS0
= ((-58*10^3)-(298*(-176.6))
= -5373.2 joule.
As the DG0 is -ve .the reaction is spontaneous.
if T = DH/DS there fore DG0 = 0 Jjoule
T = (58*10^3)/176.6 = 328.42 K
b) DG0 = -2.303RTlog K
-5373.2 = -2.303*8.314*298*logk
K = equilibrium constant = 8.74
c) Reaction quotient(Q) = pN2O4/pNO2^2
Q = 1/1^2 = 1 atm^-1.
if Q<k . forward isfavourable.