The rate of a reaction becomes 2 times for every `10^@C` rise in temperature . How many times rate of reaction will be increased when temperature is increased from `30^@ C " to " 80^@C` ?

Options

A. 16
B. 32
C. 64
D. 128

Answer

To solve the problem step by step, we will analyze how the rate of reaction changes with temperature according to the given information.

### Step 1: Understand the relationship between temperature and rate of reaction
The problem states that the rate of a reaction doubles for every increase of \(10^\circ C\) in temperature.

### Step 2: Determine the initial and final temperatures
We are given:
- Initial temperature, \(T_1 = 30^\circ C\)
- Final temperature, \(T_2 = 80^\circ C\)

### Step 3: Calculate the change in temperature
The change in temperature, \(\Delta T\), can be calculated as:
\[ \Delta T = T_2 - T_1 = 80^\circ C - 30^\circ C = 50^\circ C \]

### Step 4: Determine the number of \(10^\circ C\) intervals in the change
Next, we need to find out how many \(10^\circ C\) intervals are in the \(50^\circ C\) change:
\[ \text{Number of intervals} = \frac{\Delta T}{10^\circ C} = \frac{50^\circ C}{10^\circ C} = 5 \]

### Step 5: Calculate the increase in the rate of reaction
Since the rate doubles for each \(10^\circ C\) increase, we can express the increase in rate as:
\[ \text{Rate increase} = 2^{\text{number of intervals}} = 2^5 \]
Calculating \(2^5\):
\[ 2^5 = 32 \]

### Step 6: Conclusion
Thus, the rate of reaction at \(80^\circ C\) will be \(32\) times the rate of reaction at \(30^\circ C\).

### Final Answer
The rate of reaction will increase by a factor of \(32\) when the temperature is increased from \(30^\circ C\) to \(80^\circ C\).

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