The rate of reaction becomes 2 times for every `10^@C` rise in temperature. By how much will the rate of reaction increase 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 how much the rate of reaction increases when the temperature is raised from 30°C to 80°C, given that the rate doubles for every 10°C increase in temperature.

### Step-by-Step Solution:

1. **Identify the Temperature Change**:
- The initial temperature (T1) is 30°C, and the final temperature (T2) is 80°C.
- The change in temperature (ΔT) is:
\[ \Delta T = T2 - T1 = 80°C - 30°C = 50°C \]

2. **Determine the Number of 10°C Increments**:
- Since the rate doubles every 10°C, we need to find out how many 10°C increments are in the 50°C increase.
- The number of increments (n) is:
\[ n = \frac{\Delta T}{10°C} = \frac{50°C}{10°C} = 5 \]

3. **Calculate the Rate Increase**:
- The rate of reaction doubles for each increment. Therefore, after 5 increments, the rate of reaction will increase by a factor of \(2^n\):
\[ \text{Rate increase} = 2^n = 2^5 = 32 \]

4. **Final Rate of Reaction**:
- If we denote the initial rate of reaction at 30°C as \(R_{30}\), then the rate at 80°C (\(R_{80}\)) will be:
\[ R_{80} = 32 \times R_{30} \]

### Conclusion:
The rate of reaction increases by a factor of 32 when the temperature is increased from 30°C to 80°C.

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