Answer:
e
Explanation:
<em>Provided the reaction that leads to the formation of the products can proceed in both forward and backward directions, the correct answer would be yes because the reaction will proceed backward until equilibrium is reached.</em>
<u>For a reaction that can proceed both forward and backward, the addition of a catalyst increases the rate of reaction in both directions based on the fact that a catalyst cannot alter the equilibrium of a reaction. </u>
Hence, if an enzyme is added to the product of a reaction that has the potential to proceed in both forward and reverse reactions, a substrate would be expected to form because the reaction will proceed backward until an equilibrium is reached.
The correct option is e.
The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:
ln(k2/k1) = Ea/R[1/T1 - 1/T2]
where :
k1 is the rate constant at temperature T1
k2 is the rate constant at temperature T2
R = gas constant = 8.314 J/K-mol
Given data:
k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K
k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K
ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]
2.478 = 2.774 *10^-5 Ea
Ea = 0.8934*10^5 J = 89.3 kJ
What is your question please write properly>-< >_
