Rate constant for a chemical reaction taking place at 500 K is expressed as `k=A*e^(-1000)`. The activation energy of the reaction is `yxx 10^(6)"cal mol"^(-1)`. y is ______. `[R=2"cal mol"^(-1)K^(-1)]`

Explanation

To solve the problem, we need to find the value of \( y \) in the expression for activation energy given as \( y \times 10^6 \) cal/mol. We will use the Arrhenius equation to relate the rate constant \( k \) to the activation energy \( E_a \).

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The rate constant \( k \) is expressed as:
\[ k = A e^{-1000} \]
- The activation energy \( E_a \) is given as:
\[ E_a = y \times 10^6 \text{ cal/mol} \]
- The gas constant \( R \) is:
\[ R = 2 \text{ cal mol}^{-1} \text{ K}^{-1} \]
- The temperature \( T \) is:
\[ T = 500 \text{ K} \]

2. **Use the Arrhenius Equation:**
The Arrhenius equation relates the rate constant \( k \) to the activation energy \( E_a \):
\[ k = A e^{-\frac{E_a}{RT}} \]
By comparing the two expressions for \( k \), we have:
\[ e^{-1000} = e^{-\frac{E_a}{RT}} \]

3. **Equate the Exponents:**
Since the bases are the same, we can equate the exponents:
\[ -1000 = -\frac{E_a}{RT} \]
This simplifies to:
\[ 1000 = \frac{E_a}{RT} \]

4. **Solve for Activation Energy \( E_a \):**
Rearranging gives:
\[ E_a = 1000 \times RT \]
Now substitute the values of \( R \) and \( T \):
\[ E_a = 1000 \times (2 \text{ cal mol}^{-1} \text{ K}^{-1}) \times (500 \text{ K}) \]

5. **Calculate \( E_a \):**
\[ E_a = 1000 \times 2 \times 500 = 1000 \times 1000 = 10^6 \text{ cal/mol} \]

6. **Compare with Given Expression:**
We have found that:
\[ E_a = 10^6 \text{ cal/mol} \]
This can be expressed as:
\[ E_a = 1 \times 10^6 \text{ cal/mol} \]
Therefore, comparing with \( E_a = y \times 10^6 \text{ cal/mol} \), we find:
\[ y = 1 \]

### Final Answer:
The value of \( y \) is \( 1 \).