Question

The activation energy for a reaction is 30.8 x 298 x 4.606 cal/mol. If the increase in the rate constant when its temperature is increased from 298 K to 308 K is x % then 'x' would be

Answer 1

900

Answer 2

The problem asks us to calculate the percentage increase in the rate constant of a reaction when the temperature is raised from 298 K to 308 K, given the activation energy.

Relevant Formula: The relationship between the rate constants at two different temperatures is given by the integrated form of the Arrhenius equation:

$$\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$$

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
• $T_1$ and $T_2$ are the absolute temperatures in Kelvin

Given Values:

• Activation energy, $E_a = 30.8 \times 298 \times 4.606 \text{ cal/mol}$
• Initial temperature, $T_1 = 298 \text{ K}$
• Final temperature, $T_2 = 308 \text{ K}$
• Since $E_a$ is in calories, we use the gas constant $R = 2 \text{ cal mol}^{-1} \text{ K}^{-1}$.

Calculation: Substitute the given values into the Arrhenius equation:

$$\log \left( \frac{k_2}{k_1} \right) = \frac{(30.8 \times 298 \times 4.606)}{2.303 \times 2} \left( \frac{308 - 298}{298 \times 308} \right)$$

Notice that $2.303 \times 2 = 4.606$. So, the denominator term $2.303 \times R$ becomes $4.606$.

$$\log \left( \frac{k_2}{k_1} \right) = \frac{(30.8 \times 298 \times 4.606)}{4.606} \left( \frac{10}{298 \times 308} \right)$$

The term $4.606$ cancels out:

$$\log \left( \frac{k_2}{k_1} \right) = (30.8 \times 298) \left( \frac{10}{298 \times 308} \right)$$

The term $298$ cancels out:

$$\log \left( \frac{k_2}{k_1} \right) = 30.8 \times \frac{10}{308}$$ $$\log \left( \frac{k_2}{k_1} \right) = \frac{308}{308}$$ $$\log \left( \frac{k_2}{k_1} \right) = 1$$

To find the ratio $\frac{k_2}{k_1}$, we take the antilog:

$$\frac{k_2}{k_1} = 10^1$$ $$\frac{k_2}{k_1} = 10$$

This means $k_2 = 10k_1$.

Percentage Increase: The increase in the rate constant is $k_2 - k_1$. The percentage increase is calculated as:

$$\text{Percentage increase} = \frac{k_2 - k_1}{k_1} \times 100\%$$

Substitute $k_2 = 10k_1$:

$$\text{Percentage increase} = \frac{10k_1 - k_1}{k_1} \times 100\%$$ $$\text{Percentage increase} = \frac{9k_1}{k_1} \times 100\%$$ $$\text{Percentage increase} = 9 \times 100\%$$ $$\text{Percentage increase} = 900\%$$

The question states that the increase in the rate constant is x %. Therefore, x = 900.

The final answer is $\boxed{900}$.

Solution: The Arrhenius equation for two temperatures is $\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$. Given $E_a = 30.8 \times 298 \times 4.606 \text{ cal/mol}$, $T_1 = 298 \text{ K}$, $T_2 = 308 \text{ K}$, and $R = 2 \text{ cal mol}^{-1} \text{ K}^{-1}$. Substitute these values: $\log \left( \frac{k_2}{k_1} \right) = \frac{(30.8 \times 298 \times 4.606)}{2.303 \times 2} \left( \frac{308 - 298}{298 \times 308} \right)$ Since $2.303 \times 2 = 4.606$: $\log \left( \frac{k_2}{k_1} \right) = \frac{(30.8 \times 298 \times 4.606)}{4.606} \left( \frac{10}{298 \times 308} \right)$ $\log \left( \frac{k_2}{k_1} \right) = (30.8 \times 298) \left( \frac{10}{298 \times 308} \right)$ $\log \left( \frac{k_2}{k_1} \right) = \frac{30.8 \times 10}{308} = \frac{308}{308} = 1$ Therefore, $\frac{k_2}{k_1} = 10^1 = 10$. The percentage increase is $\frac{k_2 - k_1}{k_1} \times 100\% = \frac{10k_1 - k_1}{k_1} \times 100\% = \frac{9k_1}{k_1} \times 100\% = 900\%$. So, 'x' is 900.