Question

A. The activation energy of a certain reaction is 42.0 kJ/mol . At 29 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

B. Given that the initial rate constant is 0.0190s−1 at an initial temperature of 29  ∘C , what would the rate constant be at a temperature of 120.  ∘C for the same reaction described in Part A?

Please give the correct units too. Thank you!

Answer 1

Explanation-

We can use the Arrhenius equation:

k = A e ^(-Ea/RT)

here,

k = rate constant

A = frequency factor and is constant

R = the gas constant= 8.31 J /K/mol

E_a energy of activation

T is the absolute temperature

A more stable form of the equation is obtained by taking natural log of both sides, so

ln k = ln A - (E_a/RT)

If reaction of occour at two different temperature T₁ and T₂, then we can write,

ln k₁= ln A - ( E_a / RT₁) ( 1)​​​​​

ln k₂= ln A - ( E_a /RT₂) ( 2)

Now we can substract both sides of (1) and (2) ,

ln (k₁ /k₂) = - (E_a/RT₁) + ( E_a/RT₂)

But we know what k₂ = 2k₁ , so by substituting value,

ln (1/2) = E_a/R ( (1/T₂ ) - (1/T₁)

Here T₁ = 29°C , but we have to use in K, so

= 273 + 29

T ₁=302 K , E_a = 42.0 KJ/mol , but we need to use in joules, so

E_a= 42.0 x 10³ J/mol

So by putting values,

ln (1/2) = E_a/R ( (1/T₂) -(1/T1)

0.693 =( 42.0 x 10³/8.31) ( (1/T₂) -(1/302) )

So

Follow hand writing calculation,

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