Q.Consider a certain reaction $A \rightarrow$ Products with $k = 2.0\times10^{-2}\ \text{s}^{-1}$. Calculate the concentration of A remaining after 100 s if the initial concentration of A is $1.0\ \text{mol L}^{-1}$.

The key idea is First Order Kinetics, where the rate depends linearly on the concentration of one reactant.

For a first-order reaction, the integrated rate law is:

Step 1: Identify the given values.

$k = 2.0 \times 10^{-2}\ \text{s}^{-1}$, $t = 100\ \text{s}$, $[A]_0 = 1.0\ \text{mol L}^{-1}$.

Step 2: Substitute into the equation.

Step 3: Solve for $[A]$. …

The problem gives us a reaction $A \rightarrow$ Products with a rate constant $k = 2.0 \times 10^{-2}\ \text{s}^{-1}$. The units of $k$ — per second — are the first big clue. For a reaction, the units of the rate constant tell you the order. If $k$ has units of $\text{s}^{-1}$, the reaction is first order. That’s a non-negotiable fact in chemical kinetics.

Why does that matter? Because each order has its own integrated rate law — the equation that tells you how concentration changes with time. For a first-order reaction, the rate depends only on the concentration of one reactant: $\text{rate} = k[A]$. Integrating that differential equation gives a clean, exponential decay.

We know $[A]_0 = 1.0\ \text{mol L}^{-1}$, $k = 2.0 \times 10^{-2}\ \text{s}^{-1}$, and $t = 100\ \text{s}$. Let’s work through it.

1. Plug into the logarithmic form.

So $\ln\frac{1.0}{[A]_t} = 2.0$.

2. Exponentiate both sides.

Therefore,

3. Evaluate $e^{-2.0}$. $e^{2.0} \approx 7.389$, so $e^{-2.0} \approx 0.1353$. Hence, $[A]_t \approx 0.135\ \text{mol L}^{-1}$ …

Method: Integrated Rate Law for First-Order Kinetics

This method uses the first-order integrated rate equation to directly calculate remaining concentration at a given time.

Steps

1. Identify the reaction order and relevant equation

For a first-order reaction $A \rightarrow \text{Products}$:

where:

• $k$ = rate constant ($\text{s}^{-1}$)
• $t$ = time elapsed (s)
• $[A]_0$ = initial concentration
• $[A]_t$ = concentration at time $t$

2. Substitute the given values

• $k = 2.0 \times 10^{-2}\ \text{s}^{-1}$
• $t = 100\ \text{s}$
• $[A]_0 = 1.0\ \text{mol L}^{-1}$

3. Solve for $\log \frac{[A]_0}{[A]_t}$

4. Take antilog to find the ratio

$\frac{1.0}{[A]_t} = 10^{0.868} \approx 7.38$ …

Here are the common mistakes students make on this First Order Kinetics problem, along with how to avoid each.

Mistake 1: Using the wrong integrated rate law

The error:

Students often plug numbers into the zero-order or second-order equation by mistake.

For example, using $[A] = [A]_0 - kt$ (zero-order) or $\frac{1}{[A]} = \frac{1}{[A]_0} + kt$ (second-order).

Why it happens:

They don’t check the units of $k$.

Here, $k = 2.0 \times 10^{-2}\ \text{s}^{-1}$ — the unit $\text{s}^{-1}$ is a dead giveaway for first-order kinetics.

How to avoid:

Always check the unit of $k$ first:

• $\text{s}^{-1}$ → first order
• $\text{mol L}^{-1} \text{s}^{-1}$ → zero order
• $\text{L mol}^{-1} \text{s}^{-1}$ → second order

Then write the correct integrated law:

Mistake 2: Forgetting to take the natural log

The error:

Students write $[A] = [A]_0 e^{-kt}$ but then plug numbers directly without using $\ln$, or they use $\log_{10}$ without converting.

Example of wrong calculation:

They compute $kt = 2.0 \times 10^{-2} \times 100 = 2.0$, then say $[A] = 1.0 \times e^{-2.0}$ — which is actually correct if they evaluate $e^{-2.0}$ properly. But many forget the exponential step entirely and just do $[A] = 1.0 - 2.0$ (wrong).

How to avoid:

Use the logarithmic form step-by-step:

Then:

1. Compute $kt = 2.0 \times 10^{-2} \times 100 = 2.0$
2. So $\ln \frac{1.0}{[A]} = 2.0$
3. Exponentiate: $\frac{1.0}{[A]} = e^{2.0}$
4. Therefore $[A] = \frac{1.0}{e^{2.0}}$

Final answer:

Mistake 3: Mis-handling the exponential calculation

The error:

Students approximate $e^{2.0} \approx 7.4$ (correct) but then do $1.0 / 7.4 \approx 0.135$ — that’s fine.

But some use $e^{2.0} \approx 7.389$ and round poorly, or they forget the negative sign and compute $e^{2.0}$ instead of $e^{-2.0}$.

How to avoid:

• Use the exponential function on your calculator correctly.
• Remember: $e^{-2.0} = \frac{1}{e^{2.0}}$
• Keep at least 3 significant figures: $e^{-2.0} \approx 0.1353$

Mistake 4: Confusing concentration with amount or pressure

The error: …