The temperature coefficient of a reaction is `2`, by what factor the rate of reaction increases when temperature is increased from `30^(@)C` to `80^(@)C` ?

Options

A. `16`
B. `32`
C. `64`
D. `128`

Answer

To solve the problem, we need to determine the factor by which the rate of a reaction increases when the temperature is raised from \(30^\circ C\) to \(80^\circ C\), given that the temperature coefficient of the reaction is \(2\).

### Step-by-Step Solution:

1. **Identify the initial and final temperatures**:
- Initial temperature (\(T_1\)) = \(30^\circ C\)
- Final temperature (\(T_2\)) = \(80^\circ C\)

2. **Calculate the change in temperature**:
- Change in temperature (\(\Delta T\)) = \(T_2 - T_1 = 80^\circ C - 30^\circ C = 50^\circ C\)

3. **Determine the number of 10-degree intervals in the temperature change**:
- Since the temperature coefficient (Q10) is defined for a 10-degree change, we need to find how many 10-degree intervals are in the total temperature change.
- Number of intervals (\(n\)) = \(\frac{\Delta T}{10} = \frac{50^\circ C}{10} = 5\)

4. **Calculate the factor by which the rate increases**:
- The rate of reaction increases by a factor of \(2^n\), where \(n\) is the number of 10-degree intervals.
- Therefore, the increase in rate = \(2^n = 2^5 = 32\)

5. **Final Answer**:
- The rate of reaction increases by a factor of **32** when the temperature is increased from \(30^\circ C\) to \(80^\circ C\).

### Summary:
The factor by which the rate of reaction increases is **32**.