The first order gaseous decomposition of `N_(2)O_(4)` in to `NO_(2)` has a rate constant value `k=4.5xx10^(3)s^(-1)` at `1^(@)C` and energy of acitivation `58kJmol^(-1)`. What should be the rise in temperature in `("^(@)C)` to get the value of `K=1.00xx10^(4)s^(-1)`. `(log2.2=0.3424)`

Answer

To solve the problem, we will use the Arrhenius equation, which relates the rate constant of a reaction to the temperature and activation energy. The equation can be expressed as:

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

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

Since we have two different rate constants at two different temperatures, we can use the logarithmic form of the Arrhenius equation to find the new temperature \( T_2 \).

### Step 1: Convert temperatures to Kelvin
The initial temperature \( T_1 \) is given as \( 1^\circ C \):
\[ T_1 = 1 + 273 = 274 \, K \]

### Step 2: Identify the given values
- Initial rate constant \( k_1 = 4.5 \times 10^3 \, s^{-1} \)
- Final rate constant \( k_2 = 1.0 \times 10^4 \, s^{-1} \)
- Activation energy \( E_a = 58 \, kJ/mol = 58000 \, J/mol \)
- Gas constant \( R = 8.314 \, J/(mol \cdot K) \)

### Step 3: Use the Arrhenius equation in logarithmic form
We can express the ratio of the rate constants as:
\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

### Step 4: Substitute the known values
Substituting the values into the equation:
\[ \ln\left(\frac{1.0 \times 10^4}{4.5 \times 10^3}\right) = -\frac{58000}{8.314} \left(\frac{1}{T_2} - \frac{1}{274}\right) \]

Calculating the left side:
\[ \frac{1.0 \times 10^4}{4.5 \times 10^3} = \frac{10000}{4500} \approx 2.2222 \]
\[ \ln(2.2222) \approx 0.8047 \]

### Step 5: Rearranging the equation
Now we have:
\[ 0.8047 = -\frac{58000}{8.314} \left(\frac{1}{T_2} - \frac{1}{274}\right) \]

### Step 6: Solve for \( T_2 \)
First, calculate the right side:
\[ -\frac{58000}{8.314} \approx -6976.5 \]
Now, rearranging gives:
\[ 0.8047 = -6976.5 \left(\frac{1}{T_2} - \frac{1}{274}\right) \]
\[ \frac{1}{T_2} - \frac{1}{274} = -\frac{0.8047}{6976.5} \]
Calculating the right side:
\[ \frac{0.8047}{6976.5} \approx 0.0001154 \]
Thus,
\[ \frac{1}{T_2} = \frac{1}{274} - 0.0001154 \]
Calculating \( \frac{1}{274} \):
\[ \frac{1}{274} \approx 0.003649 \]
Now substituting:
\[ \frac{1}{T_2} \approx 0.003649 - 0.0001154 \approx 0.0035336 \]
Taking the reciprocal gives:
\[ T_2 \approx \frac{1}{0.0035336} \approx 283.4 \, K \]

### Step 7: Convert back to Celsius
To convert \( T_2 \) back to Celsius:
\[ T_2^\circ C = 283.4 - 273 \approx 10.4^\circ C \]

### Final Answer
The rise in temperature is approximately:
\[ \Delta T = T_2 - T_1 = 10.4 - 1 = 9.4^\circ C \]