Test: Temperature Dependence of the Rate of a Reaction(8 Nov) - JEE MCQ

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 < E_b

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

• Definition: The Arrhenius equation describes the relationship between the rate constant of a reaction and the temperature at which the reaction occurs.


• Formula: The Arrhenius equation is given by: k = A * e^(-Ea/RT), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the ideal gas constant, and T is the temperature in Kelvin.


• Effect of Temperature: As per the Arrhenius equation, an increase in temperature leads to an increase in the rate constant of the reaction. This is because higher temperatures provide the reactant molecules with more energy, allowing them to overcome the activation energy barrier more easily.


• Application: The Arrhenius equation is widely used in chemical kinetics to predict how changes in temperature will affect the rate of a 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 < E₂

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 → ∞ = 6.0×10¹⁴S⁻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.