Q.A first order reaction has a rate constant $1.15\times10^{-3}\ \text{s}^{-1}$. How long will 5 g of this reactant take to reduce to 3 g?

The key idea is First Order Kinetics, where the rate depends only on the concentration of one reactant. For a first order reaction, the integrated rate law relates time, the rate constant $k$, and the ratio of initial and remaining amounts.

Step 1: Write the integrated first order rate law in terms of mass (since mass is proportional to concentration for a given volume):

Step 2: Substitute the given values. Initial mass $[A]_0 = 5\ \text{g}$, remaining mass $[A] = 3\ \text{g}$, and $k = 1.15 \times 10^{-3}\ \text{s}^{-1}$:

Step 3: Calculate $\log(5/3) = \log(1.6667) \approx 0.2218$. Then:

First-order kinetics is one of the simplest and most elegant rate laws in chemistry. The defining property: the rate of reaction is directly proportional to the concentration (or amount) of a single reactant. This means that in equal time intervals, the fraction of reactant remaining is constant — not the absolute amount lost.

Why does that matter here? Because we’re given masses (5 g and 3 g), not concentrations. But for a first-order reaction, the ratio of amounts at two times is all we need. The volume cancels if the reaction is in solution, and if it’s a pure solid decomposing, the mass is directly proportional to the number of moles. So we can treat mass as a proxy for concentration.

The integrated rate law for a first-order reaction is:

where $[A]_0$ is the initial concentration (or amount), $[A]_t$ is the concentration at time $t$, and $k$ is the rate constant.

We want $t$, so rearrange:

Now plug in the numbers.

1. Identify the given values.

   $k = 1.15 \times 10^{-3}\ \text{s}^{-1}$

   Initial mass $m_0 = 5\ \text{g}$

   Final mass $m_t = 3\ \text{g}$

   Since mass is proportional to amount for a pure substance, $\frac{[A]_0}{[A]_t} = \frac{m_0}{m_t} = \frac{5}{3}$.

2. Write the expression for time.

3. Compute the natural logarithm.

   $\frac{5}{3} \approx 1.6667$

   $\ln(1.6667) \approx 0.5108$

   (You can verify: $e^{0.5108} \approx 1.667$.)

4. Divide by the rate constant.

Do the division:

$0.5108 \div 0.00115 = 444.17\ \text{s}$.

5. Round appropriately. The rate constant is given to three significant figures, so the time should be reported to three significant figures as well: $444\ \text{s}$.

Method: Integrated Rate Law for First-Order Kinetics

For a first-order reaction, the rate depends linearly on the concentration of one reactant. The key relationship is:

Where:

• $[A]_0$ = initial concentration (or amount)
• $[A]_t$ = concentration (or amount) at time $t$
• $k$ = rate constant
• $t$ = time

Since mass is proportional to concentration (same volume), we can directly use masses.

Steps

1. Identify given data

   • $k = 1.15 \times 10^{-3} \ \text{s}^{-1}$
   • Initial mass $= 5 \ \text{g}$
   • Final mass $= 3 \ \text{g}$

2. Write the integrated rate law with masses

3. Solve for $t$

4. Calculate
   • $\ln(5/3) = \ln(1.6667) \approx 0.5108$
   • $t = \frac{0.5108}{1.15 \times 10^{-3}}$

Final Answer:

Key insight: In first-order kinetics, the time depends only on the ratio of initial to remaining amount — not on the absolute quantity. That’s why we used grams directly.

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

Mistake 1: Using the wrong formula (Zero Order or Second Order)

The error:

Students often plug numbers into the zero-order equation ($t = \frac{[A]_0 - [A]}{k}$) or the second-order equation ($t = \frac{1}{k} \left( \frac{1}{[A]} - \frac{1}{[A]_0} \right)$) because they memorise formulas without checking the order.

Why it happens:

The problem explicitly says “first order reaction,” but under time pressure, students grab the first formula they recall.

How to avoid:

• Always confirm the order from the question before writing any equation.
• For first order, the integrated rate law is:

• Write this formula down before substituting numbers.

Mistake 2: Confusing mass with concentration

The error:

Students think they need to convert 5 g and 3 g into molar concentrations (mol/L) using molar mass and volume.

Why it happens:

Textbook problems often use concentration (mol/L), so students assume mass cannot be used directly.

How to avoid:

• For a first order reaction, the ratio $\frac{[A]_0}{[A]}$ is dimensionless.
• Since mass is directly proportional to concentration (same volume, same container), you can use mass in grams directly:

• No need for molar mass or volume — just the ratio of initial to remaining mass.

Mistake 3: Using $\log$ instead of $\log_{10}$ (or vice versa)

The error:

Students use natural log ($\ln$) with the constant 2.303, or use $\log_{10}$ without the 2.303 factor.

Why it happens:

The formula $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$ uses base-10 log. Some calculators default to $\ln$.

How to avoid:

• Remember:

• If your calculator has only $\ln$, compute $\ln(5/3)$ and then divide by 2.303 to get $\log_{10}(5/3)$.
• Better: use the $\log$ button (base 10) directly.

Mistake 4: Forgetting to match time units with $k$

The error:

The rate constant $k = 1.15 \times 10^{-3}\ \text{s}^{-1}$ is in $\text{s}^{-1}$, but students report the answer in minutes or hours without converting.

Why it happens:

They compute $t$ in seconds but then write “444 s” as the final answer without checking if the question expects a different unit.

How to avoid:

• Always check the unit of $k$ — here it’s $\text{s}^{-1}$, so $t$ will be in seconds.
• If the question asks for minutes or hours, convert at the end:

Mistake 5: Arithmetic errors in the log calculation

The error:

Students compute $\frac{5}{3} = 1.6667$, then take $\log(1.6667) \approx 0.2218$, but then multiply/divide incorrectly.

Why it happens:

Rushing through calculator steps or misplacing decimal points.

How to avoid:

• Write the calculation step-by-step:

• First compute $\frac{2.303}{1.15 \times 10^{-3}} = 2002.6$ (approx).
• Then $\log(5/3) \approx 0.2218$.
• Multiply: $2002.6 \times 0.2218 \approx 444\ \text{s}$.

Double-check with estimation:

• $\frac{2.303}{0.00115} \approx 2000$
• $\log(1.67) \approx 0.22$
• $2000 \times 0.22 = 440\ \text{s}$ — so 444 s is reasonable.

Quick Summary Checklist

Final answer (for reference):