The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant `k_(1)` and `k_(2)` respectively. The energy of activation for the two reactions are `152.30 kJ mol^(-1)` and `157.7 kJ mol^(-1)` as well as frequency factors are `10^(13)` and `10^(14)` respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be same for the two reactions.

Options

A. 256 K
B. 354 K
C. 282 K
D. 674 K

Answer

To find the temperature at which the rate constants for the decomposition of methyl nitrite and ethyl nitrite are equal, we can use the Arrhenius equation, which relates the rate constant \( k \) to the activation energy \( E_a \) and the frequency factor \( A \):

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

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

### Step 1: Write down the given data

For methyl nitrite:
- Activation energy \( E_{a1} = 152.30 \, \text{kJ/mol} = 152300 \, \text{J/mol} \)
- Frequency factor \( A_1 = 10^{13} \)

For ethyl nitrite:
- Activation energy \( E_{a2} = 157.7 \, \text{kJ/mol} = 157700 \, \text{J/mol} \)
- Frequency factor \( A_2 = 10^{14} \)

### Step 2: Set the rate constants equal

Since we want to find the temperature at which \( k_1 = k_2 \):

\[ A_1 e^{-\frac{E_{a1}}{RT}} = A_2 e^{-\frac{E_{a2}}{RT}} \]

### Step 3: Substitute the values

Substituting the values of \( A_1 \), \( A_2 \), \( E_{a1} \), and \( E_{a2} \):

\[ 10^{13} e^{-\frac{152300}{RT}} = 10^{14} e^{-\frac{157700}{RT}} \]

### Step 4: Simplify the equation

Dividing both sides by \( 10^{13} \):

\[ e^{-\frac{152300}{RT}} = 10 e^{-\frac{157700}{RT}} \]

### Step 5: Take the natural logarithm of both sides

Taking the natural logarithm:

\[ -\frac{152300}{RT} = \ln(10) - \frac{157700}{RT} \]

### Step 6: Rearranging the equation

Rearranging gives:

\[ \frac{157700 - 152300}{RT} = \ln(10) \]

\[ \frac{4400}{RT} = \ln(10) \]

### Step 7: Solve for T

Now, solving for \( T \):

\[ T = \frac{4400}{R \cdot \ln(10)} \]

Substituting \( R = 8.314 \, \text{J/mol·K} \) and \( \ln(10) \approx 2.303 \):

\[ T = \frac{4400}{8.314 \cdot 2.303} \]

Calculating the denominator:

\[ 8.314 \cdot 2.303 \approx 19.157 \]

Now substituting back:

\[ T \approx \frac{4400}{19.157} \approx 229.5 \, \text{K} \]

### Step 8: Final calculation

Calculating gives:

\[ T \approx 229.5 \, \text{K} \]

### Conclusion

The temperature at which the rate constants for the decomposition of methyl nitrite and ethyl nitrite will be the same is approximately **229.5 K**.

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