A reaction has a standard free-energy change of-3.00 kJ/mol at 25°C, What are th

Question

A reaction has a standard free-energy change of-3.00 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 A0.27 B-0.37 [c]- 0.03 How would your answers above change if the reaction had a standard free-energy change of +3.00 kJ/mol? O There would be more A and B but less C O There would be no change to the answers All concentrations would be lower O All concentrations would be higher O There would be less A and B but more C

Explanation / Answer

deltaGo , Gibbs free energy is related to rate constatn K as

deltaGo =-RT lnK, R= gas constant= 8.314 J/mole.K and T= 25 deg.c= 25+273= 298K

deltaGo=-3Kj/mole= -3KJ*1000J/KJ= -3000Kj/mole

lnK= -deltaG/RT =3000/(298*8.314) = 1.21

K= 3.36

for the reaction A(aq)+ B(aq) <--------->C (aq)

Q= reaction coefficient = [C]/[A][B]

given initially, [A]=0.3M, [B]=0.4M and [C]=0

Q= 0/(0.4*0.3)=0, if Q<K, the reaction proceeds towards products side

if Q>K, the reaction proceeds toward reactants side and

if Q= K, the reaction is at equilibrium.

here Q<K, the reaction proceeds towards products side

let x= drop in concentration of A to reach equilibrium

at Equilibrium, [A]=0.3-x, [B]=0.4-x and [C]=x

K= 3.36= x/(0..3-x)*(0.4-x)

when the equation is solved for x, x= 0.14

hence at equilibrium, [C]= 0.14M M , [A]= 0.3-0.14=0.16 and [B]= 0.4-0.14= 0.26

2. when deltaG= 3Kj/mole

K= -deltaG/RT= -3000/(298*8.314)= 0.30

here Q=0 is less than K. So the reaction has to proceed forward toward products side

so if x= concentratino of C at equilibrium

x/(0.3-x)*(0.4-x)=0.3, when sovled for

so at equilibrium, [C]= 0.03, [A]=0.3-0.03= 0.27 , [B]= 0.4-0.03=0.37

so there will be less C and more A and B.

deltaG is-ve indicates spontaneous reaction . Hence when the reaction is spontaneous, there will be more C. and hence less A and B. When deltaG is +ve, the reaction is non-spontaneous and hence there will be less C leading to more A and B.( A is correct)

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