Sure, let's go through each part of the question step-by-step.
### Part A: Determine whether the reaction is first-order or second-order
To decide if a reaction is first-order or second-order, we typically examine the relationship between time and concentration (or its transformations) of the reactant. For a first-order reaction, the natural logarithm of the concentration vs. time should yield a straight line. For a second-order reaction, the reciprocal of the concentration vs. time should yield a straight line.
Given the concentrations of A at different times:
- Time (s): 0, 40, 80, 120, 160
- Molos of A: 0.100, 0.067, 0.045, 0.030, 0.020
It has been determined that the natural log of the concentration of A vs. time yields a straight line, which indicates that this reaction is first-order.
Therefore, the reaction is first-order.
### Part B: What is the rate constant?
For a first-order reaction, the rate constant [tex]\( k \)[/tex] can be determined from the slope of the line obtained by plotting the natural log of the concentration of A against time.
From the results, the rate constant [tex]\( k \)[/tex] is the negative of the slope of this line.
The calculated slope is:
[tex]\[
-0.01005592765649252
\][/tex]
Thus, the rate constant [tex]\( k \)[/tex] is:
[tex]\[
k = 0.01005592765649252 \, \text{s}^{-1}
\][/tex]
### Part C: What is the half-life?
The half-life [tex]\( t_{1/2} \)[/tex] for a first-order reaction is related to the rate constant [tex]\( k \)[/tex] by the formula:
[tex]\[
t_{1/2} = \frac{\ln(2)}{k}
\][/tex]
Using the rate constant calculated:
[tex]\[
t_{1/2} = \frac{\ln(2)}{0.01005592765649252}
\][/tex]
The half-life is:
[tex]\[
t_{1/2} = 68.92921312062354 \, \text{s}
\][/tex]
To summarize:
- The reaction is first-order.
- The rate constant [tex]\( k \)[/tex] is [tex]\( 0.01005592765649252 \, \text{s}^{-1} \)[/tex].
- The half-life [tex]\( t_{1/2} \)[/tex] is [tex]\( 68.92921312062354 \, \text{s} \)[/tex].
