The rate of a reaction increases by 2.5 times when the temperature is raised from 300K to 310K. If K is the rate constant at 310 K will be equal to

Options

A. K
B. 2K
C. 2.5K
D. 3K

Answer

To solve the problem, we need to use the Arrhenius equation and the concept of the temperature coefficient. The Arrhenius equation relates the rate constant (k) to temperature (T) as follows:

\[ k = A e^{-\frac{E_a}{RT}} \]

Where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor
- \( E_a \) = activation energy
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin

Given that the rate of the reaction increases by 2.5 times when the temperature is raised from 300 K to 310 K, we can express this relationship as:

\[ \frac{k_2}{k_1} = 2.5 \]

Where:
- \( k_1 \) = rate constant at 300 K
- \( k_2 \) = rate constant at 310 K

Using the Arrhenius equation, we can write:

\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{R \cdot 310}}}{A e^{-\frac{E_a}{R \cdot 300}}} \]

This simplifies to:

\[ \frac{k_2}{k_1} = e^{-\frac{E_a}{R \cdot 310} + \frac{E_a}{R \cdot 300}} \]

This can be further simplified to:

\[ \frac{k_2}{k_1} = e^{\frac{E_a}{R} \left(\frac{1}{300} - \frac{1}{310}\right)} \]

Now, we can calculate \( \frac{1}{300} - \frac{1}{310} \):

\[ \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300} \]

Substituting this back into the equation gives:

\[ \frac{k_2}{k_1} = e^{\frac{E_a}{R \cdot 9300}} \]

Now, we know that \( \frac{k_2}{k_1} = 2.5 \), so we can write:

\[ 2.5 = e^{\frac{E_a}{R \cdot 9300}} \]

Taking the natural logarithm on both sides:

\[ \ln(2.5) = \frac{E_a}{R \cdot 9300} \]

Now, we can express \( k_2 \) in terms of \( k_1 \):

\[ k_2 = k_1 \cdot 2.5 \]

If we assume \( k_1 \) is the rate constant at 300 K, we can use this relationship to find \( k_2 \) once we know \( k_1 \).

However, since we are not given \( k_1 \), we can conclude that \( k_2 \) is 2.5 times \( k_1 \).

### Final Answer:
If \( k_1 \) is the rate constant at 300 K, then \( k_2 = 2.5 \cdot k_1 \).