How much does the enzyme staphylococcal nuclease (Rate enhancement = 5.6 X1014) decrease the activation free energy of its reaction at 25°C?
The formula to find out the activation free energy is
Go = -RTln k
Go - The activation free energy
R - 8.3145
J/mol-K (Gas constant)
T -
temperature (298 K )
k - rate
enhancement; 5.6 x 1014
Let's apply the values given and calculate,
Go = - (8.3145 J/mol-K)(298 K)/(1000 J/kJ) ln (5.6 x
1014 )
Go = - 84.2 kJ/mol
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