Question

How much faster would a reaction proceed at $ 25^\circ C $ than at $ 0^\circ C $ if the activation energy is 65 kJ?

• A. 4 times
• B. 6 times
• C. 12 times
• D. 11 times

Hint:

The Arrhenius equation helps calculate the change in the reaction rate with a change in temperature, depending on the activation energy.

Answer 1

The Correct Option is D

To determine how much faster a reaction proceeds at 25°C compared to 0°C with an activation energy of 65 kJ, we'll use the Arrhenius equation.

1. Arrhenius Equation:
The equation relating rate constant (k) to temperature is:
$$ k = A \cdot e^{-\frac{E_a}{RT}} $$
where:
- A = pre-exponential factor
- E_a = activation energy (65 kJ/mol)
- R = gas constant (8.314 J/mol·K)
- T = temperature in Kelvin

2. Ratio of Rate Constants:
The ratio of rate constants at two temperatures is:
$$ \frac{k_2}{k_1} = e^{\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)} $$

3. Temperature Conversion:
- T₁ (0°C) = 273.15 K
- T₂ (25°C) = 298.15 K

4. Calculation:
Substituting values:
$$ \frac{k_2}{k_1} = e^{\frac{65000}{8.314}\left(\frac{1}{273.15} - \frac{1}{298.15}\right)} $$
$$ = e^{7818.14 \times (0.003661 - 0.003354)} $$
$$ = e^{2.40} $$
$$ \approx 11.02 $$

5. Interpretation:
The reaction proceeds about 11 times faster at 25°C than at 0°C.

Final Answer:
The correct answer is 11 times (Option D).

Answer 2

To determine how much faster the reaction proceeds at the new temperature, we'll use the Arrhenius equation to analyze the rate constant change.

1. Arrhenius Equation Form:
The logarithmic form of the Arrhenius equation is:
$$ 2.303 \log \frac{k_2}{k_1} = \frac{E_a}{R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right] $$

2. Substituting Given Values:
Plugging in the known values:
- Activation energy (E_a) = 65 kJ/mol = 65 × 10³ J/mol
- Gas constant (R) = 8.314 J/mol·K
- Temperature difference = 25K (from 273K to 298K)

$$ 2.303 \log \frac{k_2}{k_1} = \frac{65 \times 10^3}{8.314} \left[ \frac{25}{298 \times 273} \right] $$

3. Calculating the Ratio:
Solving the equation gives:
$$ \frac{k_2}{k_1} = 11.5 $$

4. Interpretation:
This means:
$$ \text{The reaction would proceed 11 times faster.} $$

Final Answer:
The reaction rate increases by a factor of 11 at the higher temperature.