K = Ae^(-Ea/RT)

Taking ln on both sides

lnk = lnA + (-Ea/R) x 1/T

Now ATQ: lnk1 = lnA + (-Ea1/R) x 1/T

lnk2 = lnA + (-Ea₂ /R) x 1/T

Subtracting the two equations

ln(k₂ /k₁ ) = (Ea₁ - Ea₂ )/RT

ln(k₂ /k₁ ) = (75 - 50) x 1000 / 8.314 x 293

K₂ /K₁ = e ^{(25000 / 8.314 x 293)}

K₂ /K₁ = e^10.26

This is approximately equal to 30000.

Hence C is correct.

The correct answer is Option B.

For exothermic reaction, ΔH <0
E_b = E_f + ∣ΔH ∣or
E_f b

The ratio of the rate constant of a reaction at two temperatures differing by 10⁰ C is called temperature coefficient of reaction.

The correct answer is Option A.

As the value of activation energy, E_a increases, the value of rate constant, k decreases.
 
So, k₁ >k₂ since E₁ 2

The correct answer is Option C
Activation energy can be determined by evaluation rate constants at different temperatures using equation:

Where, K₂ = rate constant at temperature T₂
       K₁ = rate constant at temperature T₁

The correct answer is option C
Arrhenius equation
⇒K = Ae −Ea/RT
As T →∞,
RT →∞

e^−Ea/RT  →1
Hence K →A as T →∞
∴Value of K as T →∞= 6.0 ×10¹⁴ S⁻¹

The Arrhenius equation is  k=Ae ^{−E_a ​/RT}  

ln k = ln A - E_a /RT

comparing with y = mx+c where y = ln k and x = 1/T, slope m becomes -E_a /R

so when ln k vs 1/T is plotted, the slope comes out to be - E_a /R

B is the right answer.

Rate constant is doubled with 10 degree rise in temperature.

The correct answer is Option C.

The reactions with low activation energies are always fast whereas the reactions with high activation energy are always slow. The process of speeding up a reaction by reducing its activation energy is known as catalysis, and the factor that 's added to lower the activation energy is called a catalyst.