In a zero order reaction for every `10^@` rise of temperature , the rate is doubled. If the temperature is increased from `10^@C " to " 100^@C` , the rate of the reaction will become

Options

A. 64 times
B. 128 times
C. 256 times
D. 512 times

Answer

To solve the problem, we need to understand the relationship between temperature and the rate of a zero-order reaction. Given that the rate doubles for every 10°C rise in temperature, we can calculate the change in rate when the temperature is increased from 10°C to 100°C.

### Step-by-Step Solution:

1. **Identify the Initial Temperature and Final Temperature**:
- Initial Temperature (T1) = 10°C
- Final Temperature (T2) = 100°C

2. **Calculate the Temperature Increase**:
- Temperature Increase = T2 - T1 = 100°C - 10°C = 90°C

3. **Determine the Number of 10°C Increments**:
- Since the rate doubles for every 10°C increase, we need to find how many 10°C increments are in a 90°C increase.
- Number of Increments = 90°C / 10°C = 9 increments

4. **Calculate the Rate Change**:
- Let the initial rate at 10°C be R.
- After each 10°C increment, the rate doubles. Therefore, after 9 increments, the rate will be:
- After 1 increment: 2R
- After 2 increments: 4R
- After 3 increments: 8R
- After 4 increments: 16R
- After 5 increments: 32R
- After 6 increments: 64R
- After 7 increments: 128R
- After 8 increments: 256R
- After 9 increments: 512R

5. **Final Rate at 100°C**:
- The final rate at 100°C will be 512R.

### Conclusion:
The rate of the reaction when the temperature is increased from 10°C to 100°C will become **512R**.

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