Q.The decomposition of $NH_3$ on platinum surface is zero order reaction. What are the rates of production of $N_2$ and $H_2$ if $k = 2.5\times10^{-4}\ \text{mol}^{-1}\,\text{L}\,\text{s}^{-1}$?

Concept and Intuition

In chemical kinetics, the half‑life $t_{1/2}$ is the time required for the concentration of a reactant to fall to half its initial value. For a zero‑order reaction, the rate is constant — it does not depend on the concentration of the reactant. That means the reactant disappears at a steady, unchanging speed.

Think of it like draining a tank of water at a constant rate: if you start with twice as much water, it will take twice as long to drain half of it. Similarly, for a zero‑order reaction, doubling the starting amount doubles the time needed to consume half of it. This is fundamentally different from first‑order reactions (where half‑life is constant) or second‑order reactions (where half‑life is inversely proportional to initial concentration).

Step‑by‑Step Derivation

1. Write the integrated rate law for a zero‑order reaction. For a reaction $A \rightarrow \text{products}$ with rate law $\text{rate} = k$ (where $k$ is the rate constant), the concentration of $A$ at time $t$ is:

This is a straight line with slope $-k$.

2. Define the half‑life condition. At $t = t_{1/2}$, the concentration is half the initial:

3. Substitute into the integrated law.

4. Solve for $t_{1/2}$. Rearranging:

5. Interpret the result. The half‑life is directly proportional to the initial concentration $[A]_0$. If $[A]_0$ is doubled, then:

So the half‑life doubles.

Concept & Intuition

In a zero‑order reaction, the rate does not depend on the concentration of the reactant. That means the reactant disappears at a constant speed — like water draining from a tank at a fixed rate. The concentration vs. time graph is a straight line with slope $-k$. So if you know how much concentration has dropped and the constant rate, you can directly find the time by dividing the change in concentration by the rate constant.

Step‑by‑Step Solution

1. Recall the zero‑order integrated rate law For a reaction $A \rightarrow \text{products}$ that is zero order in $A$:

where $[A]_0$ is the initial concentration, $[A]_t$ is the concentration at time $t$, and $k$ is the rate constant (with units $\text{mol L}^{-1} \text{s}^{-1}$).

2. Identify the given quantities

   • Initial concentration: $[A]_0 = 0.2\ \text{M}$
   • Final concentration: $[A]_t = 0.024\ \text{M}$
   • Rate constant: $k = 6.0 \times 10^{-3}\ \text{mol L}^{-1} \text{s}^{-1}$

3. Rearrange the equation to solve for time

   From $[A]_t = [A]_0 - k t$, we get:

4. Plug in the numbers

5. Calculate

Rounding to two decimal places gives $29.33\ \text{s}$.

Concept & Intuition

Half‑life ($t_{1/2}$) is the time required for the concentration of a reactant to fall to half its initial value. For a zero‑order reaction, the rate is constant — it does not depend on concentration. That means the reactant is consumed at a steady pace, so the more you start with, the longer it takes to reach half that amount. Also, a larger rate constant means faster consumption, so the half‑life gets shorter. Hence $t_{1/2}$ should be proportional to $[A]_0$ and inversely proportional to $k$.

Step‑by‑step derivation

1. Write the integrated rate law for a zero‑order reaction For a reaction $A \rightarrow \text{products}$ with rate $-\frac{d[A]}{dt} = k$, integration gives:

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

2. Define the half‑life condition

   At $t = t_{1/2}$, the concentration is half the initial: $[A]_{t_{1/2}} = \frac{[A]_0}{2}$.

3. Substitute into the integrated law

4. Solve for $t_{1/2}$ Rearranging:

5. Interpret the relationship
   • $t_{1/2}$ is directly proportional to $[A]_0$ (double the initial concentration → double the half‑life).
   • $t_{1/2}$ is inversely proportional to $k$ (double the rate constant → half the half‑life).

The key idea is that a zero‑order reaction has a constant rate that does not depend on the concentration of the reactant. This leads to two characteristic plots:

• Concentration vs. time: $[R] = [R]_0 - kt$ → a straight line with slope $-k$.
• Rate vs. concentration: $\text{Rate} = k$ → a horizontal line.

Let’s examine each graph in turn.

1. Graph [A] – vertical axis is $\ln[R]$, horizontal is time.

   For a first‑order reaction, $\ln[R] = \ln[R]_0 - kt$, which gives a straight line with slope $-k$.

   This is not zero‑order; it’s first‑order decay.

2. Graph [B] – vertical axis is $[R]$, horizontal is time.

   The curve starts high, falls steeply, then flattens asymptotically.

   This is the exponential decay of a first‑order reaction ($[R] = [R]_0 e^{-kt}$).

   Not zero‑order.

3. Graph [C] – vertical axis is $[R]$, horizontal is time.

   The plot is a straight line with constant negative slope, annotated $K = -\text{Slope}$.

   This matches $[R] = [R]_0 - kt$ exactly.

   ✓ Zero‑order.

4. Graph [D] – vertical axis is Rate, horizontal is $[R]$.

   The plot is a horizontal line – rate does not change as $[R]$ increases.

   For zero‑order, $\text{Rate} = k$, independent of $[R]$.

   ✓ Zero‑order.

Thus the two graphs that represent a zero‑order reaction are [C] and [D].

For an $n^{\text{th}}$-order reaction the half-life depends on the initial concentration (here pressure) as

Given data:

So

The half-life is directly proportional to the initial pressure, which holds only for a zero-order reaction ($t_{1/2}=[A]_0/2k$).

Step 1 — Determine the overall order.

The rate law is given experimentally as

Order $=$ sum of the exponents $= 1 + 1 = \mathbf{2}$. The reaction (saponification of ethyl acetate) is second order overall, first order in each reactant.

Step 2 — Do the algebra on the units.

Rearrange the rate law:

Now substitute the units. Rate is always a concentration change per unit time, $\mathrm{mol\,L^{-1}\,s^{-1}}$, and each concentration is $\mathrm{mol\,L^{-1}}$:

Step 3 — Simplify.

Step 4 — The general rule (learn this once, use it always).

For a reaction of overall order $n$ with concentrations in $\mathrm{mol\,L^{-1}}$:

Each distractor is simply the answer for a different order, so the whole question turns on correctly reading the order as 2.

ANSWER: A

The Relevant Concept

The Arrhenius equation is the backbone of chemical kinetics for temperature dependence:

Here:

• $k$ is the rate constant.
• $A$ is the Arrhenius constant (or pre-exponential factor), related to collision frequency and orientation.
• $E_a$ is the activation energy — the minimum energy reactant molecules must have for a reaction to occur.
• $R$ is the gas constant.
• $T$ is the absolute temperature.

The fraction of molecules with energy at least $E_a$ is given by the Boltzmann factor $e^{-E_a / RT}$. Taking natural logs gives a linear form:

A catalyst works by providing a different reaction pathway with a lower activation energy, which directly changes $E_a$. Let's check each statement.

Step-by-Step Analysis

1. Statement (A): "Rate constant $k =$ Arrhenius constant $A$ : if $E_a = 0$"

If $E_a = 0$, then $e^{-E_a / RT} = e^0 = 1$. The Arrhenius equation becomes $k = A \cdot 1 = A$. This is mathematically correct. A reaction with zero activation energy would proceed at every collision, so the rate constant equals the collision-frequency factor. This statement is true.

2. Statement (B): "$\ln k$ vs $\frac{1}{T}$ plot is a straight line."

From $\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$, this is of the form $y = c + mx$ (with $y = \ln k$, $x = 1/T$, slope $m = -E_a/R$, intercept $c = \ln A$). Over the temperature ranges where $E_a$ and $A$ are constant, this is indeed a straight line. This statement is true.

3. Statement (C): "$e^{-E_a/RT}$ gives the fraction of reactant molecules that are activated at the given temp"

The Boltzmann distribution tells us that the fraction of molecules with energy $\ge E_a$ is proportional to $e^{-E_a/RT}$. This is the fraction that has enough energy to overcome the activation barrier when they collide. This statement is true.

4. Statement (D): "presence of catalyst will not alter the value of $E_a$"

A catalyst works by providing an alternative reaction pathway with a lower activation energy. This is the fundamental definition of catalysis. The catalyst does not change the reactants or products, but it changes the energy barrier (the $E_a$) that must be crossed. Therefore, the presence of a catalyst does alter the value of $E_a$. This statement is false.