1.For the reaction 2 A(aq) <---> B(aq) + C(aq) a) the standard Gibbs free enthal
ID: 1025258 • Letter: 1
Question
1.For the reaction 2 A(aq) <---> B(aq) + C(aq)
a) the standard Gibbs free enthalpy change is 1.38 kJ at 25oC. The initial concentration of A(aq) is 0.503 M, the initial concentration of B(aq) is 0.341 M, and the initial concentration of C(aq) is 0.201 M. What would be the concentration of A(aq) (in mol/L) once we attain equilibrium (we are still at 25oC)?
b) assume the reaction is 2 A(g) <---> B(g) + C(g). The total pressure is 9.26 atm and the temperature is 25.0oC. The partial pressure of A(g) is 5.70 atm. Calculate the value of the standard Gibbs free enthalphy change (in kJ) for this reaction at 25.0oC.
Explanation / Answer
deltaGo=-RT lnK, K =equilibrium constant
lnK= -deltaG/RT= -1.38*1000/(8.314 j/mole.K*298)
K=1.745
for the reaction 2A<------>B+C, K= [B][C]/[A]2=
Q= [B][C]/[A]2, for the initial conditions given, Q= 0.341*0.201/(0.503)2=0.271<K
so the reaction proceeds towards products side so as to increase Q to become equal to K
let x= increase in concentration of B due to reaction to reach equilibrium
at equilibrium, [B]=0.341+x, [C]=0.201+x, [A]= 0.503-2x
K= (0.341+x)*(0.201+x)/(0.503-2x)2= 1.745, when solved for x using excel, x=0.1098M
So at equilibrium, [B]= 0.341+0.1098 =0.4508M , [C]=0.201+0.1098=0.3108M and [A]= 0.503-2*0.1098=0.2834M
2. for the reaction 2A<----->B+C, there is no change in moles during the course of reaction. So pressure does not change. So initial pressure = 9.26 atm , partial pressure of A= 5.7 atm
partial pressure of B and C = 9.26-5.7= 3.56 atm
since moles of c =moles of B, partial pressure of B =partial pressure of C= 3.56/2= 1.78 atm
since KP= Kc*(RT)deltan, deltan = change in moles during the reaction =2-2=0
Kp= KC= PB*PC/(PA)2=1.78*1.78/(5.7)2 = 0.098
deltaGo= -RT lnK= -8.314*298* ln(0.098)=5767 joules= 5767/1000Kj= 5.767 Kj/mole
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