The rate constant of a reaction is given by `k = 2.1 xx 10^(10) exp(-2700//RT)`. It means that

Options

A. `log k` versus `1//T` will be a straight line with slope `= (-2700)/(2.303 R)`
B. `log k` versus `1//T` will be straight line with intercept on `log k` axis `= log(2.1 xx 10^(10)`).
C. The number of effective colliisons are `2.1 xx 10^(10) cm^(-3) s^(-1)`.
D. Half-life of the reaction increase with increase of temperature.

Explanation

To analyze the given rate constant expression \( k = 2.1 \times 10^{10} \exp\left(-\frac{2700}{RT}\right) \), we can derive important information regarding the reaction kinetics using the Arrhenius equation.

### Step-by-Step Solution:

1. **Identify the Form of the Rate Constant**:
The rate constant \( k \) is given in the form of the Arrhenius equation:
\[ k = A \exp\left(-\frac{E_a}{RT}\right) \]
where \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

2. **Extract Parameters**:
From the given equation, we can identify:
- Pre-exponential factor \( A = 2.1 \times 10^{10} \)
- Activation energy \( E_a = 2700 \, \text{J/mol} \)

3. **Taking the Natural Logarithm**:
To analyze the relationship between \( \log k \) and \( \frac{1}{T} \), we take the logarithm of both sides:
\[ \log k = \log A - \frac{E_a}{2.303R} \cdot \frac{1}{T} \]

4. **Rearranging into Linear Form**:
This can be rearranged into the linear form \( y = mx + c \):
- Let \( y = \log k \)
- Let \( x = \frac{1}{T} \)
- Slope \( m = -\frac{E_a}{2.303R} \)
- Intercept \( c = \log A \)

5. **Calculate the Slope**:
Substituting the values:
\[ m = -\frac{2700}{2.303R} \]

6. **Calculate the Intercept**:
The intercept can be calculated as:
\[ c = \log(2.1 \times 10^{10}) = \log 2.1 + \log(10^{10}) = \log 2.1 + 10 \]

### Conclusion:
From the analysis, we can conclude:
- The plot of \( \log k \) versus \( \frac{1}{T} \) will yield a straight line.
- The slope of this line is \( -\frac{2700}{2.303R} \).
- The intercept on the \( \log k \) axis is \( \log(2.1 \times 10^{10}) \).

### Options Analysis:
1. **Option A**: Correct - The slope is \( -\frac{2700}{2.303R} \).
2. **Option B**: Correct - The intercept is \( \log(2.1 \times 10^{10}) \).
3. **Option C**: Incorrect - The number of effective collisions is not directly represented by \( A \).
4. **Option D**: Incorrect - The half-life of the reaction does not necessarily increase with temperature.