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.
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