In the start of summer, a given sample of milk turns sour at room temperature `(27^(@)C)` in `48` hours. In a refrigerator at `2^(@)C`, milk can be stored there times longer before it sours.
Calculate the rate constant at `310 K`, when rate constant at `300 K` is `1.6 xx 10^(5)`

Options

A. `2.363 xx 10^(5)`
B. `2.4 xx 10^(5)`
C. `2.450 xx 10^(5)`
D. `3.123 xx 10^(5)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between rate constants and temperature
We know that the rate constant \( k \) changes with temperature according to the Arrhenius equation. The relationship can be expressed as:
\[ \log \frac{k_2}{k_1} = \frac{E_A}{2.303R} \cdot \frac{T_2 - T_1}{T_1 T_2} \]
where:
- \( k_1 \) is the rate constant at temperature \( T_1 \)
- \( k_2 \) is the rate constant at temperature \( T_2 \)
- \( E_A \) is the activation energy
- \( R \) is the universal gas constant (8.314 J/mol·K)

### Step 2: Identify the known values
From the problem:
- \( k_1 = 1.6 \times 10^5 \) at \( T_1 = 300 \, K \)
- The milk lasts 48 hours at room temperature (27°C or 300 K).
- In the refrigerator (2°C or 275 K), it lasts 3 times longer, so \( k_2 \) can be expressed as:
\[ k_2 = \frac{k_1}{3} \]
Thus, \( k_2 = \frac{1.6 \times 10^5}{3} = 5.33 \times 10^4 \).

### Step 3: Calculate the activation energy \( E_A \)
Using the values:
- \( T_1 = 275 \, K \)
- \( T_2 = 300 \, K \)
- \( k_1 = 5.33 \times 10^4 \)
- \( k_2 = 1.6 \times 10^5 \)

Substituting these into the equation:
\[ \log \frac{5.33 \times 10^4}{1.6 \times 10^5} = \frac{E_A}{2.303 \times 8.314} \cdot \frac{300 - 275}{275 \times 300} \]

Calculating the left side:
\[ \log \frac{5.33 \times 10^4}{1.6 \times 10^5} = \log \left( \frac{5.33}{1.6} \times 10^{-1} \right) = \log(0.333125) \approx -0.477 \]

Now substituting into the equation:
\[ -0.477 = \frac{E_A}{2.303 \times 8.314} \cdot \frac{25}{275 \times 300} \]

Calculating the denominator:
\[ 275 \times 300 = 82500 \]
Thus:
\[ -0.477 = \frac{E_A}{2.303 \times 8.314} \cdot \frac{25}{82500} \]

Calculating \( 2.303 \times 8.314 \):
\[ 2.303 \times 8.314 \approx 19.1 \]

Now substituting:
\[ -0.477 = \frac{E_A}{19.1} \cdot \frac{25}{82500} \]
\[ -0.477 = \frac{E_A \cdot 25}{1575000} \]

Rearranging gives:
\[ E_A = -0.477 \cdot \frac{1575000}{25} \]
Calculating:
\[ E_A \approx -0.477 \cdot 63000 \approx -30000 \, J/mol \approx 30.147 \, kJ/mol \]

### Step 4: Calculate the rate constant at 310 K
Now we need to find \( k_3 \) at \( T_2 = 310 \, K \):
Using the same formula:
\[ \log \frac{k_3}{k_1} = \frac{E_A}{2.303R} \cdot \frac{T_2 - T_1}{T_1 T_2} \]
Where:
- \( T_1 = 300 \, K \)
- \( T_2 = 310 \, K \)

Substituting the known values:
\[ \log \frac{k_3}{1.6 \times 10^5} = \frac{30.147 \times 10^3}{2.303 \times 8.314} \cdot \frac{10}{300 \times 310} \]

Calculating the right side:
\[ \frac{30.147 \times 10^3}{19.1} \cdot \frac{10}{93000} \]
Calculating:
\[ \frac{30.147 \times 10^3}{19.1} \approx 1575.5 \]
Thus:
\[ \log \frac{k_3}{1.6 \times 10^5} \approx 1575.5 \cdot \frac{10}{93000} \approx 0.1693 \]

Now solving for \( k_3 \):
\[ \frac{k_3}{1.6 \times 10^5} = 10^{0.1693} \approx 1.476 \]
\[ k_3 = 1.6 \times 10^5 \cdot 1.476 \approx 2.36 \times 10^5 \]

### Final Answer:
The rate constant at \( 310 \, K \) is approximately:
\[ \boxed{2.36 \times 10^5} \]