in-a-chemical-reaction-the-rate-constant-k-is-given-by-k-ae-eart-if-ea-is-50-kjm

Question

In a chemical reaction, the rate constant k is given by k = $$Ae^{(-Ea/RT)}. If E$$_{a} is 50 kJ/mol, R = 8.314 J/mol·K, and T = 298 K, what is the value of ln(k) assuming A = 1?

Accepted Answer

ln(k) = -20.13

Answer

ln(k) = -5.13

Answer

ln(k) = -50.00

Answer

ln(k) = -10.00

In a chemical reaction, the rate constant k is given by k = Ae(-Ea/RT). If Ea is 50 kJ/mol, R = 8.314 J/mol·K, and T = 298 K, what is the value of ln(k) assuming A = 1?

In a chemical reaction, the rate constant k is given by k = Ae^(-Ea/RT). If E_a is 50 kJ/mol, R = 8.314 J/mol·K, and T = 298 K, what is the value of ln(k) assuming A = 1?