At 407 K the rate constant of a chemical reaction is `9.5xx10^(-5)s^(-1)` and at 420 K, the rate constant is `1.9xx10^(-4)s^(-1)`. The frequency factor of the reaction is `x xx 10^(5)s^(-1)`. The value of 'x' is. Report your answer by rounding it up to nearest whole number.

Answer

To solve the problem, we will use the Arrhenius equation, which relates the rate constants of a reaction at two different temperatures to the activation energy and the frequency factor. The equation can be expressed as:

\[ \frac{k_2}{k_1} = e^{\frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)} \]

Where:
- \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \) respectively.
- \( E_a \) is the activation energy.
- \( R \) is the universal gas constant (8.314 J/(mol·K)).
- \( T_1 \) and \( T_2 \) are the absolute temperatures in Kelvin.

### Step 1: Identify the given values
- \( k_1 = 9.5 \times 10^{-5} \, s^{-1} \) at \( T_1 = 407 \, K \)
- \( k_2 = 1.9 \times 10^{-4} \, s^{-1} \) at \( T_2 = 420 \, K \)

### Step 2: Calculate the activation energy \( E_a \)
Using the logarithmic form of the Arrhenius equation:

\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Substituting the known values:

\[ \log \left( \frac{1.9 \times 10^{-4}}{9.5 \times 10^{-5}} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{407} - \frac{1}{420} \right) \]

### Step 3: Calculate the left-hand side
Calculating \( \frac{k_2}{k_1} \):

\[ \frac{1.9 \times 10^{-4}}{9.5 \times 10^{-5}} = 2 \]

Now, taking the logarithm:

\[ \log(2) \approx 0.301 \]

### Step 4: Calculate the right-hand side
Calculating \( \frac{1}{407} - \frac{1}{420} \):

\[ \frac{1}{407} \approx 0.002454 \quad \text{and} \quad \frac{1}{420} \approx 0.002381 \]

So,

\[ \frac{1}{407} - \frac{1}{420} \approx 0.002454 - 0.002381 = 0.000073 \]

### Step 5: Substitute back to find \( E_a \)
Now substituting back into the equation:

\[ 0.301 = \frac{E_a}{2.303 \times 8.314} \times 0.000073 \]

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

\[ 0.301 = \frac{E_a}{19.1} \times 0.000073 \]

Rearranging gives:

\[ E_a = \frac{0.301 \times 19.1}{0.000073} \]

Calculating \( E_a \):

\[ E_a \approx 75782.3 \, J/mol \]

### Step 6: Calculate the frequency factor \( A \)
Using the Arrhenius equation in its logarithmic form:

\[ \log k = \log A - \frac{E_a}{2.303RT} \]

For \( k_1 \):

\[ \log(9.5 \times 10^{-5}) = \log A - \frac{75782.3}{2.303 \times 8.314 \times 407} \]

Calculating \( 2.303 \times 8.314 \times 407 \approx 75782.3 \):

\[ \log(9.5 \times 10^{-5}) \approx -4.021 \]

Now substituting:

\[ -4.021 = \log A - \frac{75782.3}{75782.3} \]

This simplifies to:

\[ \log A \approx -4.021 + 1 \]

So:

\[ \log A \approx -3.021 \]

Calculating \( A \):

\[ A \approx 0.000975 \quad \text{or} \quad 0.0975 \times 10^{-5} \, s^{-1} \]

### Step 7: Express \( A \) in the required form
The frequency factor \( A \) can be expressed as:

\[ A \approx 0.04 \times 10^5 \, s^{-1} \]

Thus, \( x = 4 \).

### Final Answer
The value of \( x \) is **4**.