To find the rate constant \( K \) for the reaction \( A_2(g) + B_2(g) \rightarrow 2AB(g) \), we can follow these steps:
### Step 1: Write the rate law expression
The rate law for the reaction can be expressed as:
\[
\text{Rate} = K [A_2]^m [B_2]^n
\]
where \( m \) and \( n \) are the orders of the reaction with respect to \( A_2 \) and \( B_2 \), respectively. Since we need to determine \( K \), we will assume the reaction is first order with respect to both reactants (i.e., \( m = 1 \) and \( n = 1 \)).
### Step 2: Use the provided data
We have the following initial concentrations and rates:
1. For \( [A_2]_0 = 0.10 \, M \), \( [B_2]_0 = 0.10 \, M \), Rate = \( 2.5 \times 10^{-4} \, M/s \)
2. For \( [A_2]_0 = 0.20 \, M \), \( [B_2]_0 = 0.10 \, M \), Rate = \( 5.0 \times 10^{-4} \, M/s \)
3. For \( [A_2]_0 = 0.20 \, M \), \( [B_2]_0 = 0.20 \, M \), Rate = \( 10.0 \times 10^{-4} \, M/s \)
### Step 3: Calculate the rate from the first data point
From the first data point, the rate of appearance of \( AB \) is given as \( 2.5 \times 10^{-4} \, M/s \). The rate of the reaction can be expressed as:
\[
\text{Rate} = \frac{1}{2} \frac{d[AB]}{dt} = K [A_2][B_2]
\]
Thus, we can express the rate as:
\[
\text{Rate} = \frac{1}{2} \cdot 2.5 \times 10^{-4} = 1.25 \times 10^{-4} \, M/s
\]
### Step 4: Substitute values into the rate law
Using the first data point:
\[
1.25 \times 10^{-4} = K \cdot (0.10) \cdot (0.10)
\]
\[
1.25 \times 10^{-4} = K \cdot 0.01
\]
### Step 5: Solve for \( K \)
Rearranging gives:
\[
K = \frac{1.25 \times 10^{-4}}{0.01} = 1.25 \times 10^{-2} \, M^{-1}s^{-1}
\]
### Step 6: Determine the units of \( K \)
Since the reaction is second order (first order in \( A_2 \) and first order in \( B_2 \)), the units of \( K \) are:
\[
K \text{ units} = M^{-1} s^{-1}
\]
### Final Answer
The value of the rate constant \( K \) for the reaction is:
\[
K = 1.25 \times 10^{-2} \, M^{-1}s^{-1}
\]
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