Mapd Organic Chemistry s & Company Publishers presented by Sapling Learning A re
ID: 559488 • Letter: M
Question
Mapd Organic Chemistry s & Company Publishers presented by Sapling Learning A reaction has a standard free-energy change of-548 kJ/mol at 25°C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively? Number Number Number How would your answers above change if hereaction had a standard free-energy change of +5.48 kJlmor? O There would be more A and B but less C O All concentrations would be lower O There would be no change to the answers. There would be less A and B but more c. O All concentrations would be higher Exil HntExplanation / Answer
for the reaction, A(aq)+B(aq)<---->C(aq)
deltaG0= standard free energy =-RTlnK= -5.48 Kj/mole
where R= 8.314 J/mole, T= 25 deg.c= 25+273= 298K, K= equilibrium constant
lnK= -deltaG/RT= 5.48*1000 J/mole/ (8.314*298)= 2.211, K= 9.13
for the reaction given K= [C]/ [A] [B]
given initial concentration, [A] =0.3, [B]= 0.4 and [C]=0
preparing the ICE table
component initial concentration (M) change equilibrium concentration
A 0.3 -x 0.3-x
B 0.4 - x 0.4- x
C 0 x x
K= 9.13= x/((0.3-x)*0.4-x), when solved for x using excel, x =0.1954
At Equilibrium, [A] =0.3-0.1954= 0.1046, [B]=0.4-0.1954 =0.2046, [C]=0.1954
when free energy change is 5.48 Kj/mole, lnK= exp(-5.48*1000/(8.314*298) = 0.11
Q= Reaction coefficient = [C]/ [A][B]= 0 < K
the reaction proceeds so as to increase [C] and redcue [A] and [B]
K= x/(0.3-x)*(0.4-x)= 0.11, when solved for x using excel, x= 0.0124
[A] =0.3-0.0124= 0.2876, [B] =0.4-0.0124= 0.2876, [C]= 0.0124, there would be more A and B, less C at Equilibrium ( A is correct)