At 300 K rate constant for `Ato` products at t=50 min is `0.02 s^(-1)`, then rate constant at t=75 min and 310 K will be (in `s^(-1)`)

Options

A. `0.04/25`
B. `0.04xx25`
C. `0.04`
D. `(0.02/25)`

Answer

To solve the problem, we need to find the rate constant at a different time and temperature using the information provided. Here's the step-by-step solution:

### Step 1: Understand the given data
- Rate constant at 300 K (k₁) = 0.02 s⁻¹
- Time (t₁) = 50 minutes = 3000 seconds (since 1 minute = 60 seconds)
- We need to find the rate constant (k₂) at:
- Time (t₂) = 75 minutes = 4500 seconds
- Temperature = 310 K

### Step 2: Use the first-order reaction formula
For a first-order reaction, the relationship between the rate constant (k), time (t), and the concentration of reactants can be expressed as:
\[ \ln\left(\frac{[A]_0}{[A]}\right) = k \cdot t \]
Where:
- \([A]_0\) = initial concentration of reactant A
- \([A]\) = concentration of reactant A at time t

### Step 3: Calculate the change in concentration from t₁
Using the formula for the first-order reaction:
\[ \ln\left(\frac{[A]_0}{[A]}\right) = k_1 \cdot t_1 \]
Substituting the known values:
\[ \ln\left(\frac{[A]_0}{[A]}\right) = 0.02 \cdot 3000 \]
\[ \ln\left(\frac{[A]_0}{[A]}\right) = 60 \]

### Step 4: Calculate the concentration change at t₂
Now, we need to find the concentration change at t₂ (75 minutes or 4500 seconds):
\[ \ln\left(\frac{[A]_0}{[A]}\right) = k_2 \cdot t_2 \]
Where \( t_2 = 4500 \) seconds.

### Step 5: Relate the two rate constants
We can relate the two rate constants using the time difference:
\[ k_2 = \frac{\ln\left(\frac{[A]_0}{[A]}\right)}{t_2} \]
Substituting the value we found from t₁:
\[ k_2 = \frac{60}{4500} \]
\[ k_2 = 0.01333 \, \text{s}^{-1} \]

### Step 6: Adjust for temperature change using Arrhenius equation
To find the rate constant at 310 K, we can use the Arrhenius equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
However, for small temperature changes, we can use the approximate relationship:
\[ \frac{k_2}{k_1} = \frac{T_2}{T_1} \]
Where:
- \( T_1 = 300 \, K \)
- \( T_2 = 310 \, K \)

Calculating \( k_2 \):
\[ k_2 = k_1 \cdot \frac{T_2}{T_1} = 0.02 \cdot \frac{310}{300} \]
\[ k_2 = 0.02 \cdot 1.0333 \]
\[ k_2 \approx 0.02067 \, \text{s}^{-1} \]

### Final Answer
The rate constant at 75 minutes and 310 K is approximately \( 0.02067 \, \text{s}^{-1} \).

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