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

Explanation

To solve the problem, we need to determine how the rate of a zero-order reaction changes when the temperature is increased from 10°C to 100°C, given that the rate doubles for every 10°C rise in temperature.

### Step-by-Step Solution:

1. **Identify the Temperature Change**:
- The initial temperature is 10°C and the final temperature is 100°C.
- The change in temperature (ΔT) is calculated as:
\[ \Delta T = 100°C - 10°C = 90°C \]

2. **Determine the Number of 10°C Increments**:
- Since the rate doubles for every 10°C increase, we need to find out how many 10°C increments are in a 90°C change.
- This is calculated as:
\[ n = \frac{90°C}{10°C} = 9 \]

3. **Calculate the Rate Increase**:
- The rate of the reaction doubles for each increment, which can be expressed as:
\[ \text{Rate increase} = 2^n \]
- Substituting the value of \( n \):
\[ \text{Rate increase} = 2^9 \]

4. **Calculate \( 2^9 \)**:
- Now we compute \( 2^9 \):
\[ 2^9 = 512 \]

5. **Conclusion**:
- Therefore, the rate of the reaction will become 512 times greater when the temperature is increased from 10°C to 100°C.

### Final Answer:
The rate of the reaction will become **512 times**.

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