The rate of reaction is doubled for every `10^(@)C` rise in temperature. The increase in rate as result of an increase in temperature from `10^(@)C` to `100^(@)C` is how many times of the original rate?

Options

A. 112
B. 512
C. 400
D. 256

Answer

To solve the problem, we need to determine how many times the rate of reaction increases when the temperature rises from \(10^\circ C\) to \(100^\circ C\), given that the rate doubles for every \(10^\circ C\) increase in temperature.

### Step-by-Step Solution:

1. **Identify the Temperature Change**:
The temperature changes from \(10^\circ C\) to \(100^\circ C\). The total change in temperature is:
\[ 100^\circ C - 10^\circ C = 90^\circ C \]

2. **Determine the Number of 10°C Increments**:
Since the rate doubles for every \(10^\circ C\) increase, we need to find how many \(10^\circ C\) increments fit into the \(90^\circ C\) change:
\[ \frac{90^\circ C}{10^\circ C} = 9 \text{ increments} \]

3. **Calculate the Rate Increase**:
The rate doubles with each increment. Therefore, after \(n\) increments, the rate increases by a factor of \(2^n\). Here, \(n = 9\):
\[ \text{Rate increase} = 2^9 \]

4. **Compute \(2^9\)**:
\[ 2^9 = 512 \]

5. **Conclusion**:
The rate of reaction increases by a factor of \(512\) when the temperature rises from \(10^\circ C\) to \(100^\circ C\).

### Final Answer:
The increase in rate as a result of the temperature increase from \(10^\circ C\) to \(100^\circ C\) is **512 times** the original rate.

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