The rate of a chemical reaction doubles for every `10^(@)C` rise of temperature. If the temperature is raised by `50^(@)C`, the rate of the reaction increases by about

Options

A. 24 times
B. 32 times
C. 64 times
D. 10 times

Explanation

To solve the problem, we need to determine how much the rate of a chemical reaction increases when the temperature is raised by \(50^\circ C\), given that the rate doubles for every \(10^\circ C\) increase in temperature.

### Step-by-Step Solution:

1. **Understand the relationship between temperature change and rate of reaction**:
- It is given that the rate of the reaction doubles for every \(10^\circ C\) increase in temperature. This means that if we denote the rate at a certain temperature as \(k_1\), then after a \(10^\circ C\) increase, the rate becomes \(k_2 = 2k_1\).

2. **Calculate the total temperature increase**:
- The total increase in temperature is \(50^\circ C\).

3. **Determine how many \(10^\circ C\) intervals are in \(50^\circ C\)**:
- We can find the number of \(10^\circ C\) intervals in \(50^\circ C\) by dividing \(50\) by \(10\):
\[ \Delta t = \frac{50}{10} = 5 \]
- This indicates that there are 5 intervals of \(10^\circ C\) in a \(50^\circ C\) increase.

4. **Calculate the increase in rate**:
- Since the rate doubles for each \(10^\circ C\) increase, we can express the increase in rate mathematically:
\[ k_2 = k_1 \cdot 2^{\Delta t} = k_1 \cdot 2^5 \]
- Now, calculate \(2^5\):
\[ 2^5 = 32 \]
- Therefore, the new rate \(k_2\) is:
\[ k_2 = 32k_1 \]

5. **Conclusion**:
- The rate of the reaction increases by a factor of \(32\) when the temperature is raised by \(50^\circ C\).

### Final Answer:
The rate of the reaction increases by about **32 times**.

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