The rate constant for a certain chemical reaction is 77.0 s-1 at a temperature of 20.0°c. if the activation energy for this reaction is 52. kj/mole, what is the rate constant at a temperature of 30.0°c?

 is the rate constant at a temperature of 30.0 °C.

Explanation:

Using the expression,

Wherem  

is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, = 52 kJ/mol = 52000 J/mol (As 1 kJ = 1000 J)

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (20 + 273.15) K = 293.15 K  

T = (30 + 273.15) K = 303.15 K  

So,

 is the rate constant at a temperature of 30.0 °C.

the answer is c. nacl