The first order rate constant for the decomposition of ethyl iodide by the reaction `C_2 H_5 I_((g)) to C_2 H_4 (g) + HI_((g))` at 600 K is `1.60 xx 10^(-5) s^(-1)` . Its energy of activation is 209 kJ/mol. Calculate the rate constant of the reaction at 700 K.

Explanation

To calculate the rate constant of the reaction at 700 K, we can use the Arrhenius equation in the form of the following equation:

\[ \ln\left(\frac{K_2}{K_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Where:
- \( K_1 \) is the rate constant at temperature \( T_1 \) (600 K),
- \( K_2 \) is the rate constant at temperature \( T_2 \) (700 K),
- \( E_a \) is the activation energy (209 kJ/mol),
- \( R \) is the universal gas constant (8.314 J/mol·K),
- \( T_1 \) is the initial temperature (600 K),
- \( T_2 \) is the final temperature (700 K).

### Step 1: Convert Activation Energy
Convert the activation energy from kJ/mol to J/mol:
\[ E_a = 209 \, \text{kJ/mol} = 209 \times 10^3 \, \text{J/mol} = 209000 \, \text{J/mol} \]

### Step 2: Substitute Values into the Equation
Substituting the known values into the equation:
\[ \ln\left(\frac{K_2}{1.60 \times 10^{-5}}\right) = \frac{209000}{8.314} \left(\frac{1}{600} - \frac{1}{700}\right) \]

### Step 3: Calculate the Right Side
First, calculate \(\frac{1}{600} - \frac{1}{700}\):
\[ \frac{1}{600} - \frac{1}{700} = \frac{700 - 600}{600 \times 700} = \frac{100}{420000} = \frac{1}{4200} \]

Now, substitute this back into the equation:
\[ \ln\left(\frac{K_2}{1.60 \times 10^{-5}}\right) = \frac{209000}{8.314} \times \frac{1}{4200} \]

Calculating the right side:
\[ \frac{209000}{8.314} \approx 25105.7 \]
\[ \frac{25105.7}{4200} \approx 5.973 \]

### Step 4: Solve for \( K_2 \)
Now we have:
\[ \ln\left(\frac{K_2}{1.60 \times 10^{-5}}\right) \approx 5.973 \]
To find \( K_2 \), exponentiate both sides:
\[ \frac{K_2}{1.60 \times 10^{-5}} = e^{5.973} \]
Calculating \( e^{5.973} \):
\[ e^{5.973} \approx 396.5 \]
Thus,
\[ K_2 \approx 396.5 \times 1.60 \times 10^{-5} \]
Calculating \( K_2 \):
\[ K_2 \approx 6.344 \times 10^{-3} \, \text{s}^{-1} \]

### Final Answer
The rate constant \( K_2 \) at 700 K is approximately:
\[ K_2 \approx 6.34 \times 10^{-3} \, \text{s}^{-1} \]