The velocity constant of a reaction at 290 K was found to be `3.2 xx 10^(-3) s^(-1)`. When the temperature is raised to 310 K, it will be about

Options

A. `6.4 xx 10^(-3)`
B. `3.2 xx 10^(-4)`
C. `9.6 xx 10^(-3)`
D. `1.28 xx 10^(-2)`

Explanation

To solve the problem of finding the velocity constant \( K_2 \) at a temperature of 310 K, given that the velocity constant \( K_1 \) at 290 K is \( 3.2 \times 10^{-3} \, s^{-1} \), we can use the concept of the temperature coefficient. Here’s a step-by-step solution:

### Step 1: Understand the Temperature Coefficient
The temperature coefficient is a measure of how the rate constant changes with temperature. A common rule of thumb in chemistry is that for many reactions, the rate constant approximately doubles for every 10°C increase in temperature.

### Step 2: Calculate the Temperature Increase
The temperature increase from 290 K to 310 K is:
\[ \Delta T = 310 \, K - 290 \, K = 20 \, K \]
This is equivalent to a temperature increase of 20°C.

### Step 3: Determine the Doubling Factor
Since the rate constant doubles for every 10°C increase, a 20°C increase would result in the rate constant increasing by a factor of:
\[ 2 \text{ (for 10°C)} \times 2 \text{ (for another 10°C)} = 4 \]

### Step 4: Calculate \( K_2 \)
Now, we can calculate \( K_2 \) using the factor we found:
\[ K_2 = 4 \times K_1 = 4 \times (3.2 \times 10^{-3} \, s^{-1}) \]
Calculating this gives:
\[ K_2 = 12.8 \times 10^{-3} \, s^{-1} \]

### Step 5: Express in Standard Form
To express this in standard scientific notation:
\[ K_2 = 1.28 \times 10^{-2} \, s^{-1} \]

### Final Answer
Thus, the velocity constant \( K_2 \) at 310 K is:
\[ \boxed{1.28 \times 10^{-2} \, s^{-1}} \]