Another first-order reaction also has a rate constant of 2.45×10−2 s−1 at 25 ∘C. What is...

Another first-order reaction also has a rate constant of 2.45×10−2 s−1 at 25 ∘C. What is the value of k at 70 ∘C if Ea = 120 kJ/mol ?

Solutions

Expert Solution

CFor a given First order reaction,

K₁ (rate constant) = 2.45 x 10^-2 s^-1 at T₁ = 25^oC = 25+273.15 = 298.15 K

k (rate constant) = ? at T₂ =70 ^oC = 70+273.15 = 343.15 K.

and Activation energy (E_a)= 120 kJ = 120 x 10³ J = 1.2 x 10⁵ J.

If rate constant for a same raction at 2 different temperatures is known then the energy of activation (Ea) can be calculated using the equation,

log(k₂/k₁) = [Ea/2.303R] x [T₂-T₁/T₂T₁]

With the notations we used we get equation as,

log(k/k₁) = [Ea/2.303R] x [T₂-T₁/T₂T₁]

log(k/2.45x10^-2) = [1.2x10⁵/( 2.303x8.314)] x [(343.15-298.15)/(343.15 x298.15)]

= [1.2x10⁵/19.15] x [45 /1.023 x 10⁵]

= [1.2 x 45] / [19.15 x 1.023]...................(10⁵ cancelled mumually)

log(k/2.45x10^-2) = 2.7564

k/2.45x10^-2 = 10^2.7564

k/2.45 x 10^-2 = 5.71 x 10²

k = 5.71x10² x 2.45x10^-2

k = 13.995 s^-1.

= 1.3995x10¹ s^-1

Hence, for the First order reaction rate constant at 70^oC is 13.995 or 1.3995 x10¹s^-1

queen_honey_blossom answered 1 year ago

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