(i) (5 points) Calculate the equilibrium constant at 60°C and 75°C for glucose i

Question

(i) (5 points) Calculate the equilibrium constant at 60°C and 75°C for glucose isomerase. This enzyme is used extensively in the United States for production of high-fructose syrup. The reaction is: Glucose ?? fructose ?H°rxn for this reaction is 5.73 kJ gmol-1; ?S°rxn is 0.0176 kJ gmol-1K-1 (note: R = 8.3144 J gmol-1 K; T (K) = T (°C) + 273)

(ii) (5 points) Calculate equilibrium constant for the following reactions under standard conditions: Glutamine + water ? glutamate + ammonium ?G°rxn for this reaction is -14.1 kJ gmol-1

Explanation / Answer

delG= -RTlnk

delG= delH- TdelS

at 60 deg.c =60+273.15=333.15K

delG=5 .73*- 333.15*0.0176=-0.13344 Kj/Mol

delG= -RTlnK

0.13344*1000= 8.314*333.15 lnK

lnK= 0.13344*1000/(8.314*333.15)=0.048177

K= exp(0.048)=1.049356

At 70 deg.c =70+273.15= 343.15K

delG= 5.73-343.15*0.0176=-0.30944 Kj/mol

delG=-RTlnK

0.30944*1000= 8.314*343.15lnk

Lnk =0.30944*1000/(8.314*343.15)=0.108463

K= exp(0.108463)=1.11

ii)

delG0= -RTlnK

-14.1*100= -8.314*298.15*lnK

lnK= 14.1*1000/ (8.314*298.15)=5.688

K= exp(5.688)=295.3024

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