Q.For the reversible reaction $A \rightleftharpoons B$, the concentrations of reactant and product change exponentially with time until equilibrium is reached. Four concentration-versus-time plots are proposed:

In chemical kinetics, a first-order reaction is one whose rate depends linearly on the concentration of a single reactant. This means that as the reactant concentration decreases, the reaction rate also decreases proportionally. A key characteristic of first-order reactions is that the time required for a certain fraction of the reactant to be consumed (e.g., half-life, or time for 10% completion) is constant and independent of the initial concentration.

To solve problems involving the time taken for a certain percentage of a first-order reaction to complete, we use the integrated rate law for first-order reactions. This equation relates the concentration of the reactant at any given time to its initial concentration and the rate constant.

1. Recall the Integrated Rate Law for a First-Order Reaction:

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

   $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$

   where:

   • $k$ is the rate constant of the reaction.
   • $t$ is the time elapsed.
   • $[A]_0$ is the initial concentration of the reactant.
   • $[A]_t$ is the concentration of the reactant at time $t$.

2. Calculate the Rate Constant ($k$) using the First Set of Data:

   We are given that 10% completion occurs in 20 minutes.

   • If 10% of the reaction is completed, then 90% of the reactant remains.
   • So, $[A]_t = 0.90 [A]_0$.
   • The time $t = 20$ minutes.

   Substitute these values into the integrated rate law:

We will keep $k$ in this form to avoid rounding errors in intermediate steps.

3. Calculate the Time ($t$) for 19% Completion:

Now we need to find the time taken for 19% completion of the same reaction.

* If 19% of the reaction is completed, then $100\% - 19\% = 81\%$ of the reactant remains.

* So, $[A]_t = 0.81 [A]_0$.

* We use the same rate constant $k$ calculated in the previous step, as $k$ is constant for a given reaction at a specific temperature. …

Concept

For $A(g)\to B(g)+C(g)$, if $x$ is the pressure of A that reacts, total pressure $= P_0 + x$ and $p_A = P_0 - x$. For a first-order reaction $k = \dfrac{2.303}{t}\log\dfrac{P_0}{p_A}$ and $t_{1/2} = \dfrac{0.693}{k}$.

Solution

$\log\tfrac{4}{3} = \log 4 - \log 3 = 0.60 - 0.48 = 0.12$ …

Concept & Intuition

A zero‑order reaction has a constant rate, independent of concentration. That means the concentration decreases linearly with time:

The half‑life for a zero‑order reaction is not constant — it depends on the starting concentration:

Here we are given the half‑life (0.5 h) when the initial concentration is 4 mol L⁻¹. That lets us find the rate constant $k$. Then, because the decrease is linear, the time to go from any concentration to another is simply the concentration difference divided by $k$.

Step‑by‑step solution

1. Find the rate constant $k$ from the given half‑life. For a zero‑order reaction:

Plug in $t_{1/2} = 0.5$ h and $[A]_0 = 4$ mol L⁻¹:

So $k = \frac{2}{0.5} = 4$ mol L⁻¹ h⁻¹.

2. Use the integrated rate law for the interval from 2.0 to 1.0 mol L⁻¹. The zero‑order law:

Here we want the time $\Delta t$ for the concentration to drop from 2.0 to 1.0. Let $t=0$ be the moment when $[A]=2.0$. Then:

Substitute $k = 4$:

$4\,\Delta t = 1.0 \quad\Rightarrow\quad \Delta t = \frac{1}{4} \text{ hour}$ …

Concept & Intuition

In a zero‑order reaction, the rate does not depend on concentration:

This means the concentration decreases linearly with time:

The half‑life for a zero‑order reaction is given by

Here, the half‑life is 0.5 hour when the initial concentration is 4 mol L⁻¹. That lets us find the rate constant $k$. Once we have $k$, the time to go from any concentration $C_1$ to $C_2$ is simply

because the change is linear. No complicated integration needed.

Step‑by‑step solution

1. Find the rate constant $k$ from the given half‑life. For zero‑order:

Plug in $t_{1/2} = 0.5$ hr and $[A]_0 = 4$ mol L⁻¹:

So

2. Use the zero‑order integrated rate law to find the time interval. The law is:

For a change from $[A] = 2.0$ to $[A] = 1.0$ mol L⁻¹, the amount consumed is $2.0 - 1.0 = 1.0$ mol L⁻¹.

Since the rate is constant,

Concept & Intuition

For a first‑order reaction, the concentration decays exponentially:

The time to go from the initial concentration to any fraction $f$ (where $[A] = f[A]_0$) depends only on $f$ and the rate constant $k$. Crucially, the ratio of two such times is independent of $k$ — it depends only on the logarithms of the fractions. That’s why we can compute $t_{1/4}/t_{1/10}$ without knowing $k$ at all.

Step‑by‑step reasoning

1. Write the first‑order integrated law For a first‑order reaction:

If the fraction remaining is $f = [A]/[A]_0$, then

2. Express $t_{1/4}$ and $t_{1/10}$
   • When decomposition is to $\frac{1}{4}$ of initial, $f = \frac{1}{4}$:

• When decomposition is to $\frac{1}{10}$ of initial, $f = \frac{1}{10}$:

3. Take the ratio The $1/k$ cancels:

Using $\ln x = (\log x)(\ln 10)$, we can also write:

since the factor $\ln 10$ cancels top and bottom.

4. Plug in the given value $\log 2 = 0.3$, so $\log 4 = \log(2^2) = 2 \log 2 = 0.6$. Also $\log 10 = 1$. …

The decomposition of HCOOH is a first-order reaction, meaning its rate depends linearly on the concentration of HCOOH. This characteristic allows us to use specific integrated rate laws to relate the concentration of the reactant to time. The key idea is that the time required for a certain fraction of the reactant to decompose is independent of the initial concentration.

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

Let's apply this to the given problem.

1. Identify the given values:

   We are given the rate constant, $k = 3.465 \times 10^{-3} \ \text{s}^{-1}$.

   We need to find the time ($t$) for 87.5% of the initial quantity of HCOOH to decompose.

2. Determine the initial and final concentrations:

   Let the initial quantity (or concentration) of HCOOH be $[A]_0$.

   If 87.5% of HCOOH decomposes, then the amount remaining, $[A]_t$, is $100\% - 87.5\% = 12.5\%$ of the initial quantity.

   So, $[A]_t = 0.125 \times [A]_0$.

3. Substitute values into the integrated rate law:

   Now, we plug these values into the first-order integrated rate law:

   $t = \frac{2.303}{3.465 \times 10^{-3} \ \text{s}^{-1}} \log \frac{[A]_0}{0.125 [A]_0}$

   The $[A]_0$ terms cancel out, simplifying the expression:

   $t = \frac{2.303}{3.465 \times 10^{-3}} \log \left(\frac{1}{0.125}\right)$

   Since $\frac{1}{0.125} = 8$, the equation becomes:

   $t = \frac{2.303}{3.465 \times 10^{-3}} \log 8$

4. Calculate the time:

   We know that $\log 8$ can be written as $\log (2^3) = 3 \log 2$.

   The problem provides $\log 2 = 0.30$.

   Therefore, $\log 8 = 3 \times 0.30 = 0.90$.

   Substitute this value back into the equation for $t$:

   $t = \frac{2.303}{3.465 \times 10^{-3}} \times 0.90$

   $t = \frac{2.0727}{3.465 \times 10^{-3}}$

   $t = \frac{2072.7}{3.465}$

   $t \approx 598.2 \ \text{s}$

   Rounding this to the nearest hundred, we get 600 seconds. …

The key idea is that for a first-order reaction, the time to reach a certain percentage completion is independent of the initial concentration. The half-life formula $t_{1/2} = \frac{\ln 2}{k}$ lets us find the rate constant $k$ from the given half-life. Then we use the integrated first-order equation $\ln \frac{[A]_0}{[A]} = kt$ to find the time when 90% of the reactant has reacted — meaning only 10% remains.

A common mistake is to think that 90% completion takes three half-lives (since 50% → 75% → 87.5% → 93.75% in three half-lives), but that’s only approximate. We need the exact calculation.

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

Given $t_{1/2} = 30$ minutes and $\log 2 = 0.30$ (which means $\ln 2 = 2.303 \times 0.30 = 0.6909$, but we can work directly with logs to base 10 if we prefer).

(We’ll keep it symbolic for now.)

2. Write the integrated rate law for first-order kinetics.

If 90% of the reaction is complete, then 10% of the reactant remains: $[A] = 0.10 [A]_0$.

So $\ln \frac{[A]_0}{0.10 [A]_0} = \ln 10 = kt$

3. Substitute $k$ from step 1.

4. Convert to base-10 logs using the given $\log 2 = 0.30$. Recall $\ln x = 2.303 \log x$, so the ratio $\frac{\ln 10}{\ln 2} = \frac{\log 10}{\log 2} = \frac{1}{0.30} = \frac{10}{3}$. Therefore: …

In chemical kinetics, the rate constant ($k$) quantifies the speed of a reaction, while the half-life ($t_{1/2}$) is the time required for the concentration of a reactant to decrease to half of its initial value. For a first-order reaction, these two quantities are directly related by a simple formula. Understanding this relationship is crucial because it allows us to determine one if the other is known, without needing to know the initial concentration of the reactant. This is a unique characteristic of first-order reactions, as for other reaction orders, the half-life depends on the initial concentration.

Here's how to find the specific rate constant:

1. Recall the integrated rate law for a first-order reaction: The integrated rate law describes how the concentration of a reactant changes over time. For a first-order reaction, it is given by:

where $[A]_t$ is the concentration of reactant A at time $t$, $[A]_0$ is the initial concentration of reactant A, and $k$ is the specific rate constant. This can also be written as:

2. Define half-life ($t_{1/2}$): By definition, at the half-life ($t = t_{1/2}$), the concentration of the reactant becomes half of its initial concentration.

3. Derive the relationship between $t_{1/2}$ and $k$ for a first-order reaction: Substitute the conditions for half-life into the integrated rate law:

Using the logarithm property $\ln(1/x) = -\ln(x)$:

Multiplying both sides by $-1$:

Rearranging to solve for $k$: …