Practice free →
Home › Bihar Board Class 12 (Science) › Chemistry › Chemical Kinetics › Per Arrhenius, raising T:

Per Arrhenius, raising T:

ADecreases k
BIncreases k by raising the fraction of effective collisions
CDoesn't affect k
DChanges the products
Answer & Solution
Correct answer: B. Increases k by raising the fraction of effective collisions
The relationship between the rate constant ($k$) and temperature ($T$) is described by the **Arrhenius Equation**: $$k = A e^{-\frac{E_a}{RT}}$$ Where: - $k$ is the rate constant. - $A$ is the pre-exponential factor (frequency of collisions). - $E_a$ is the activation energy. - $R$ is the universal gas constant. - $T$ is the absolute temperature (in Kelvin). ### Step-by-Step Breakdown: 1. **The Exponential Term:** The term $e^{-\frac{E_a}{RT}}$ represents the fraction of molecules that possess kinetic energy greater than or equal to the activation energy ($E_a$). 2. **Increasing Temperature:** As $T$ increases, the denominator in the exponent increases, making the fraction $\frac{E_a}{RT}$ smaller. Since it is negative, $-\frac{E_a}{RT}$ becomes closer to zero (less negative). 3. **Effect on $k$:** As $-\frac{E_a}{RT}$ becomes less negative, the value of $e^{-\frac{E_a}{RT}}$ increases. 4. **Collision Theory Context:** Physically, this means that at a higher temperature, a larger percentage of collisions between molecules occur with enough energy to overcome the energy barrier ($E_a$). These are known as **effective collisions**. Therefore, raising $T$ increases $k$ because it raises the fraction of effective collisions.
Solve this in the app — Bihar Board Class 12 (Science) practice & 24k+ MCQs →
Related questions