Q.Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with $t_{1/2} = 3.00$ hours. What fraction of sample of sucrose remains after 8 hours?

The key idea is First Order Kinetics: for a first-order reaction, the fraction remaining after time $t$ is given by $N/N_0 = e^{-kt}$, where $k = \ln 2 / t_{1/2}$.

Step 1: Find the rate constant $k$ from the half-life.

Step 2: Use the integrated first-order rate law for the fraction remaining after $t = 8$ hours. …

First-order kinetics is one of the cleanest models in chemical kinetics. The defining property: the rate of decomposition is directly proportional to the concentration of the reactant itself. That means the fraction of reactant remaining after a given time depends only on the rate constant and time — not on the starting amount. This is why we can solve the problem without ever needing an initial concentration.

The half-life for a first-order reaction is constant: $t_{1/2} = \frac{\ln 2}{k}$. Given $t_{1/2} = 3.00$ hours, we can find $k$, then use the integrated rate law to find the fraction remaining after 8 hours.

1. Find the rate constant $k$ from the half-life. For a first-order reaction:

So

(Using $\ln 2 \approx 0.693$.)

2. Write the integrated first-order rate law. The standard form is:

where $[A]_0$ is the initial concentration and $[A]_t$ is the concentration after time $t$.

The fraction remaining is $\frac{[A]_t}{[A]_0}$.

3. Plug in the values.

So

Therefore

4. Compute the numerical value. $e^{-1.848}$ — you can do this with a calculator or recall that $e^{-1.848} = \frac{1}{e^{1.848}}$. $e^{1.848} \approx 6.35$ (since $e^{1.8} \approx 6.05$ and $e^{1.85} \approx 6.36$). So …

Method: First-Order Integrated Rate Law (Exponential Decay)

This is the standard method for any first-order kinetics problem where you know the half-life and need the fraction remaining after a given time.

Steps

Step 1: Recall the first-order integrated rate law

For a first-order reaction, the fraction of reactant remaining is:

where:

• $[A]_0$ = initial concentration
• $[A]$ = concentration at time $t$
• $k$ = rate constant
• $t$ = time elapsed

Step 2: Find the rate constant $k$ from half-life

For first-order kinetics, the half-life formula is:

Given $t_{1/2} = 3.00$ hours:

Step 3: Substitute into the integrated rate law

Time given: $t = 8$ hours

Step 4: Calculate the fraction

Final Answer …

Common Mistakes & How to Avoid Them

Mistake 1: Using the wrong formula for fraction remaining

The error: Students often plug time directly into the half-life formula without converting to the correct exponential form. They might write:

Why it happens: Confusing "number of half-lives" with a simple ratio.

How to avoid: Remember — for first order kinetics, the fraction remaining is always:

where $n = \frac{\text{time elapsed}}{\text{half-life}} = \frac{8}{3} \approx 2.667$

So the correct expression is:

Mistake 2: Forgetting to use the rate constant $k$ properly

The error: Students jump straight to the half-life formula without first finding $k$, then get lost.

Why it happens: They memorise $t_{1/2} = \frac{0.693}{k}$ but don't connect it to the integrated rate law.

How to avoid: Always follow this two-step process:

1. Find $k$ from half-life:

2. Use the first order integrated law:

Mistake 3: Incorrect calculator use for exponentials

The error: Typing $e^{-1.848}$ incorrectly — either pressing wrong keys or misreading the result.

Why it happens: Students confuse $e^x$ with $10^x$ or misplace decimal points.

How to avoid:

• Use the $e^x$ button (not $10^x$)
• Double-check: $e^{-1.848} \approx 0.157$
• Final answer: $\boxed{0.157}$ or $\boxed{15.7\%}$

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