X -> Products A student studied the kinetics of the reaction represented above by measuring the conc — Kinetics Chemistry Question

STEPS:

1. Analyze the given graph of concentration versus time:
Observe how the concentration of reactant X decreases at regular intervals:
* At $t = 0\text{ min}$, $[\text{X}] = 1.0\text{ M}$.
* At $t = 20\text{ min}$, $[\text{X}] = 0.5\text{ M}$ (the concentration is halved, meaning the first half-life is $20\text{ minutes}$).
* At $t = 40\text{ min}$, $[\text{X}] = 0.25\text{ M}$ (the concentration is halved again, meaning the second half-life is also $20\text{ minutes}$).
* At $t = 60\text{ min}$, $[\text{X}] = 0.125\text{ M}$ (the third half-life is also $20\text{ minutes}$).

2. Identify the reaction order:
Because the half-life ($t_{1/2}$) remains constant at approximately $20\text{ minutes}$ regardless of the starting concentration, the reaction exhibits first-order kinetics with respect to reactant X.

3. Apply the integrated rate law for a first-order reaction:
According to the AP Chemistry equations sheet, the integrated rate law for a first-order reaction is:
$$\ln[\text{A}]_t - \ln[\text{A}]_0 = -kt$$
By isolating $\ln[\text{A}]_t$, we can rewrite this in the slope-intercept form ($y = mx + b$):
$$\ln[\text{X}]_t = -kt + \ln[\text{X}]_0$$
* $y = \ln[\text{X}]_t$
* $x = t$ (time)
* $m = -k$ (the slope of the line)
* $b = \ln[\text{X}]_0$ (the y-intercept)

4. Conclude how to determine the rate constant ($k$):
Plotting the natural logarithm of concentration ($\ln[\text{X}]$) versus time ($t$) will produce a straight line. The slope of this line is $-k$, so determining the magnitude (absolute value) of the slope yields the rate constant, $k$. This corresponds to Option A.

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WHY_OTHERS_WRONG:

• Option B is incorrect: Plotting $1/[\text{X}]$ versus time and finding the slope is the procedure used to find the rate constant of a second-order reaction. For this first-order reaction, a plot of $1/[\text{X}]$ versus time will yield a curve, not a straight line, and its slope will not be constant.
• Option C is incorrect: Running another trial with a different initial concentration at the same temperature will simply yield a parallel curve on the graph that starts at a different initial y-value. The intersection point of these curves has no physical significance and cannot be used to mathematically calculate $k$.
• Option D is incorrect: Running a trial at a different temperature changes the value of the rate constant $k$. Comparing where the slopes are equal on two curves generated at different temperatures does not provide a mathematically valid or standard method to isolate or calculate the rate constant at either temperature.