For the reaction `A_((g))rarr(B)_((g))`, the value of the equilibrium constant at 300 K and I atm is equal to 100.0. The value of `Delta_rG` for the reaction at 300 K and I atm in J `mol^(-1)` is - xR, where x is `"_______"`
(Rounded of to the nearest integer) (R=8.31 J `mol^(-1)K^(-1) and "In" 10 = 2.3)`

Answer

To solve the problem, we need to find the value of \( x \) in the equation \( \Delta_r G = -xR \) for the reaction \( A_{(g)} \rightleftharpoons B_{(g)} \) at equilibrium, given that the equilibrium constant \( K \) is 100.0 at 300 K and 1 atm.

### Step-by-Step Solution:

1. **Understand the relationship between Gibbs free energy and equilibrium constant:**
The relationship between the standard Gibbs free energy change (\( \Delta_r G^\circ \)) and the equilibrium constant (\( K \)) is given by the equation:
\[ \Delta_r G^\circ = -RT \ln K \]
where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

2. **Convert \( K \) to natural logarithm:**
Since we are given \( K = 100.0 \), we can convert it to natural logarithm using the relationship:
\[ \ln K = \log_{10} K \times 2.303 \]
We know that \( \log_{10} 100 = 2 \) because \( 100 = 10^2 \). Therefore:
\[ \ln K = 2 \times 2.303 = 4.606 \]

3. **Substitute the values into the Gibbs free energy equation:**
Now we can substitute \( R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1} \), \( T = 300 \, \text{K} \), and \( \ln K = 4.606 \) into the Gibbs free energy equation:
\[ \Delta_r G^\circ = - (8.31 \, \text{J mol}^{-1} \text{K}^{-1})(300 \, \text{K})(4.606) \]

4. **Calculate \( \Delta_r G^\circ \):**
Performing the multiplication:
\[ \Delta_r G^\circ = - (8.31 \times 300 \times 4.606) \]
First, calculate \( 8.31 \times 300 = 2493 \).
Then, calculate \( 2493 \times 4.606 \approx 11449.58 \).
Therefore:
\[ \Delta_r G^\circ \approx -11449.58 \, \text{J mol}^{-1} \]

5. **Express \( \Delta_r G^\circ \) in terms of \( -xR \):**
We have \( \Delta_r G^\circ = -xR \). Thus:
\[ -xR = -11449.58 \]
Since \( R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1} \), we can express \( x \):
\[ x = \frac{11449.58}{8.31} \approx 1377.4 \]

6. **Round to the nearest integer:**
Rounding \( 1377.4 \) gives us \( x \approx 1377 \).

### Final Answer:
The value of \( x \) is \( \boxed{1377} \).