Consider the following reaction: 2 NO2 (g) <-> N2O4 (g) For this reaction at 298
ID: 783107 • Letter: C
Question
Consider the following reaction: 2 NO2 (g) <-> N2O4 (g) For this reaction at 298 K, deltaH = %u201356.8 kJ and deltaS = %u2013175 J /K.
In a container at 298 K, N2O4 (g) is mixed with NO2 (g). The initial partial pressures are: PN2O4 = 2.4 atm and PNO2 = 0.42 atm. Which of the following statements is correct?
a) Some N2O4 (g) will decompose into NO2 (g)
b) Some NO2 (g) will react to produce N2O4 (g).
c) The system is at equilibrium at these initial pressures.
d) The final total pressure must be known to answer this question.
e) None of these
Please explain why!!!
Explanation / Answer
dH = 56.8 KJ
dS = 0.175 KJ/K
dG = dH - T*dS
dG = 56.8 - 298*0.175 = 4.65 KJ = 4650 J
dG = -RTlnK
R = 8.314
T = 298 K
K = 0.153...............this is equilibrium constant
2 NO2 (g) <-> N2O4 (g)
Ki = p(N2O4)/p(NO2)^2
Ki = 2.4/0.42^2 = 13.60
as Ki>k the reaction will move in backward direction so ans is
a) Some N2O4 (g) will decompose into NO2 (g)