A first-order reaction is 4.0 times faster at 50°C than it is at 25°C. What is the activation energy of this reaction?

Explanation:

Using the expression,

Where,

is the activation energy

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

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

T(K) = T( °C) + 273.15  

So,  

= (25 + 273.15) K = 298.15 K  

= (50 + 273.15) K = 323.15 K  

So,  

the answer is:  

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