The kinetics of double-strand formation for a DNA oligonucleotide was measured b

Question

The kinetics of double-strand formation for a DNA oligonucleotide was measured by relaxation techniques. The reaction can bc represented by 2 singlet duplex The following data were obtained: Enter the values of E_a, delta H^0 and delta S^0 for the forward reaction in that order to two significant figures. For E_a, and delta H^0 , enter the values in kJ/mol and for delta S^0 t in J K^-1 mol^-1 but do not enter the units. To obtain the correct values, use linear regression for the appropriate linear plot (for example, the intercept and slope functions in Excel). Also, in formulas requiring a value of the temperature, use the midpoint of the temperature range for the above data. Enter the value of the equilibrium constant at 37 degree C to two significant figures (see c) below). As in part a), enter the values of E_a, delta H^0 and delta S^0 but for the reverse reaction in that order and to two significant figures. For Ea, and delta H^0, enter the values in kJ/mol and for delta S^0 in J K^-1 mol^-1 but do not enter the units. Recall that in class we used the fact that delta G degree = degree G^0 _forward - delta G^0 _reverse. Use similar relationships and the values in part a) and c) to determine delta H degree in kJ/mol and delta S degree in J K^-1 mol^-1 for the reaction. Enter the values in that order, to two significant figures and do not enter the units.

Explanation / Answer

(a): For forward reaction:

T1 = 31.8 DegC = 31.8 + 273 = 304.8 K

k1 = 0.8x105 M-1s-1

Applying Arrhenius equation

k1 = A x (exp)(-Ea / RT)  

=> 0.8x105 M-1s-1 =  A x (exp)(-Ea / Rx304.8K) -------- (1)

T2 = 36.8 DegC = 36.8 + 273 = 309.8 K

k2 = 2.3x105 M-1s-1

Applying Arrhenius equation

k2 = A x (exp)(-Ea / RT2)  

=> 2.3x105 M-1s-1 =  A x (exp)(-Ea / Rx309.8K) -------- (2)

Now we can find the value of Ea by dividing eqn(1) and eqn(2)

=> (0.8 / 2.3) = (exp)(Ea/309.8R) - (Ea/304.8R)

=> ln(0.8 / 2.3) = (Ea/R)x[1/309.8 - 1/304.8] = - (Ea / 8.314 JK-1mol-1) x [ 5 / 309.8x304.8]

=> Ea = 165814 J/mol = 170 kJ/mol (answer)

(c): Reverse reaction:

T1 = 31.8 DegC = 31.8 + 273 = 304.8 K

k1 = 1.00 s-1

Applying Arrhenius equation

k1 = A x (exp)(-Ea / RT1)  

=> 1.00 s-1 =  A x (exp)(-Ea / Rx304.8K) -------- (1)

T2 = 36.8 DegC = 36.8 + 273 = 309.8 K

k2 = 3.20 s-1

Applying Arrhenius equation

k2 = A x (exp)(-Ea / RT2)  

=> 3.20 s-1 =  A x (exp)(-Ea / Rx309.8K) -------- (2)

Now we can find the value of Ea for reverse reation by dividing eqn(1) and eqn(2)

=> (1.00 / 3.20) = (exp)(Ea/309.8R) - (Ea/304.8R)

=> ln(1.00/3.20) = (Ea/R)x[1/309.8 - 1/304.8] = - (Ea / 8.314 JK-1mol-1) x [ 5 / 309.8x304.8]

=> Ea = 182630 J/mol = 180 kJ/mol (answer)

(b): Given T = 37 DegC = 37 + 273 = 310 K

At T = 310 K, forward rate constant, k1 = A x (exp)(-Ea / Rx310K)

=>  k1 = A x (exp)(-165814 Jmol-1 / Rx310K)

reverse rate constant, k-1 =  A x (exp)(-Ea / Rx310K)

=> k-1 =  A x (exp)(-182630 Jmol-1 / Rx310K)

Equilibrium constant K = k1 / k-1 = (exp)^(182630 Jmol-1 / Rx310K - 165814 Jmol-1 / Rx310K)

=> K = (exp)^(16816 / 310x8.314) = 680 (answer)

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