For a general reaction $A \rightarrow B$, the concentration of A is plotted against time and the plot is a straight line that falls steadily (a constant negative slope) from its initial value toward zero. On the basis of this straight-line concentration-versus-time plot, answer the following: (i) What is the order of the reaction? (ii) What is the slope of the line? (iii) What are the units of the rate constant?

For a general reaction A arrow B, the concentration of A is plotted against time and the plot is a straight li

NCERT Exemplar · Q30

A \rightarrow B

(i) What is the order of the reaction?

(ii) What is the slope of the line?

(iii) What are the units of the rate constant?

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Concept understanding — Zero Order Kinetics

Zero Order Kinetics: The Drug That Doesn't Care How Much You Give It

Imagine you're filling a bathtub. You turn the tap to a fixed flow rate — say, 5 litres per minute. The amount of water in the tub increases by exactly 5 litres every minute, regardless of whether the tub is empty or already half full. That's the core intuition behind zero order kinetics: a constant amount disappears per unit time, no matter how much is left.

Now contrast this with what you probably expect. Most processes in nature follow first order kinetics: the rate depends on how much is present. If you have 100 molecules, 10 might react per second; if you have 10 molecules, only 1 reacts per second. The fraction lost is constant, but the amount lost per second shrinks as the quantity shrinks. Zero order is the opposite — the amount lost per second is fixed, so the fraction lost actually increases as the quantity drops.

The Precise Statement

-\frac{d[A]}{dt} = k_0

Where $[A]$ is the concentration of the substance (or amount, depending on context), $t$ is time, and $k_0$ is the zero order rate constant with units of concentration per time (e.g., mg/L per hour, or simply mg/hour if we're talking about total amount).

The negative sign indicates the substance is being removed. The key point: the rate does not depend on $[A]$ . It's a flat, constant rate.

The Integrated Form and Half-Life

If you integrate the differential equation, you get a straight line:

[A]_t = [A]_0 - k_0 t

This is the equation of a line with slope $-k_0$ and intercept $[A]_0$ . Plot concentration vs. time, and you get a straight line sloping downward until it hits zero.

The half-life — the time for the concentration to fall to half its initial value — is:

t_{1/2} = \frac{[A]_0}{2k_0}

Notice something crucial: the half-life depends on the initial concentration. Double the starting amount, and the half-life doubles. This is completely different from first order kinetics, where half-life is constant regardless of starting concentration.

Watch out

A common mistake: students assume half-life is always constant. For zero order, it is not. The half-life changes with the starting amount. If you start with 100 mg, half-life might be 5 hours; start with 200 mg, half-life becomes 10 hours.

Where Does Zero Order Kinetics Actually Happen?

In pharmacology, zero order kinetics is most famously seen with ethanol (alcohol) and aspirin at high doses. The reason is saturation of enzymes.

Your body metabolises alcohol using an enzyme called alcohol dehydrogenase. At low alcohol levels, the enzyme works efficiently and the rate depends on how much alcohol is present (first order). But at higher concentrations — say, after a few drinks — the enzyme becomes saturated. It's working at maximum speed, like a factory running at full capacity. Adding more raw material (alcohol) doesn't make it work faster. The rate becomes constant: a fixed amount of alcohol is metabolised per hour, regardless of how much is in your blood.

This is why alcohol elimination follows a straight line when you plot blood alcohol concentration vs. time. A typical person eliminates about 0.015 g/dL per hour — a fixed amount, not a fixed fraction.

Note

The same saturation principle applies to some drug transporters in the kidneys. When the transport proteins are working at maximum capacity, drug excretion becomes zero order. This is why high doses of certain drugs (like phenytoin) can lead to unexpectedly long elimination times — the system is overwhelmed.

A Quick Comparison Table

Property	Zero Order	First Order
Rate depends on	Nothing (constant)	Concentration
Rate equation	$-\frac{d[A]}{dt} = k_0$	$-\frac{d[A]}{dt} = k_1[A]$

Why this formula?

Zero Order Kinetics: Why the Formula Holds

Let's build this from the ground up — understanding the why before the what.

The Core Idea

Zero order kinetics describes a process where the rate is constant — it does not depend on the concentration of the reactant.

This is the definition, but why would that ever happen?

Why the Rate is Constant

Imagine a reaction happening on a solid surface (like a catalyst or a tablet dissolving). The reactant molecules must first adsorb onto the surface before reacting.

If the surface is saturated with reactant molecules, adding more reactant in solution doesn't help — the surface is already full.
The reaction proceeds at a fixed speed determined by how fast the surface can process the adsorbed molecules.

Key insight: The rate is limited by the surface, not by how much reactant is floating around.

Deriving the Zero Order Rate Law

Step 1: Write the rate definition

For a reaction $A \rightarrow \text{products}$ , the rate of disappearance of $A$ is:

-\frac{d[A]}{dt} = k

where $k$ is the zero order rate constant (units: concentration/time, e.g., $\text{mol L}^{-1} \text{s}^{-1}$ ).

Notice: No $[A]$ term on the right side — that's the signature of zero order.

Step 2: Separate variables and integrate

-d[A] = k \, dt

Integrate from initial time $t=0$ (concentration $[A]_0$ ) to time $t$ (concentration $[A]_t$ ):

-\int_{[A]_0}^{[A]_t} d[A] = k \int_0^t dt

-[A]_t + [A]_0 = kt

Step 3: Rearrange to the familiar form

\boxed{[A]_t = [A]_0 - kt}

This is the integrated rate law for zero order kinetics.

What This Formula Tells Us

Linear decrease: Concentration falls linearly with time (not exponentially like first order).
Slope = $-k$ : A plot of $[A]_t$ vs. $t$ gives a straight line with slope $-k$ .
Half-life depends on initial concentration:

Set $[A]_t = \frac{[A]_0}{2}$ :

\frac{[A]_0}{2} = [A]_0 - k t_{1/2}

\boxed{t_{1/2} = \frac{[A]_0}{2k}}

Critical exam point: Unlike first order (where $t_{1/2}$ is constant), zero order half-life increases with higher initial concentration.

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A linear fall of $[A]$ with time means the rate is constant, i.e. a zero order reaction. The slope of the line is $-k$ , and the rate constant has units of $\text{mol L}^{-1}\text{ s}^{-1}$ .

(i) Order of the reaction

For a zero order reaction, $[A]=[A]_0-kt$ , which plots as a straight line of $[A]$ against $t$ . Since the given plot is a straight line, the reaction is zero order.

(ii) Slope of the line

Differentiating $[A]=[A]_0-kt$ gives $\dfrac{d[A]}{dt}=-k$ . So the slope of the concentration-time line is $-k$ (equivalently, $\text{slope}=\dfrac{\Delta[A]}{\Delta t}=-k$ ). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
$\rightarrow$
›Reveal solutionSolution
For a zero-order reaction, the rate is constant and independent of concentration. The integrated rate law gives $k = \frac{[A]_0 - [A]_t}{t}$ , which yields $k = 4.5 \times 10^{-3} \, \mathrm{mol\, L^{-1}\, s^{-1}}$ .

The key idea is that a zero-order reaction proceeds at a constant rate — the concentration of the reactant decreases linearly with time. This is fundamentally different from first-order or second-order reactions, where the rate depends on how much reactant is left.

For a zero-order reaction $\mathrm{A \rightarrow P}$ , the rate law is:
$\text{Rate} = -\frac{d[A]}{dt} = k$
Integrating this from $t = 0$ to $t = t$ gives the integrated rate equation:
$[A]_t = [A]_0 - kt$
This is a straight line: concentration on the y-axis, time on the x-axis, with slope $-k$ .
$k = \frac{[A]_0 - [A]_t}{t}$
Now let's apply it step by step.

Identify the given data

Initial concentration: $[A]_0 = 0.5 \, \mathrm{mol\, L^{-1}}$

Concentration after 100 s: $[A]_t = 0.05 \, \mathrm{mol\, L^{-1}}$

Time: $t = 100 \, \mathrm{s}$

Find the change in concentration

$[A]_0 - [A]_t = 0.5 - 0.05 = 0.45 \, \mathrm{mol\, L^{-1}}$

Apply the zero-order formula

$k = \frac{0.45}{100} = 0.0045 \, \mathrm{mol\, L^{-1}\, s^{-1}}$

Express in scientific notation $k = 4.5 \times 10^{-3} \, \mathrm{mol\, L^{-1}\, s^{-1}}$ …
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TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
$\Delta_f H^\Theta$
›Reveal solutionSolution
The enthalpy change for a reaction is the sum of the enthalpies of formation of products minus reactants, multiplied by their stoichiometric coefficients. For the formation of 2 moles of NH₃ from its elements, ΔᵣH° = 2 × (−46.2 kJ mol⁻¹) = −92.4 kJ.

The key concept here is Hess’s Law and the definition of standard enthalpy of formation. The standard enthalpy of formation (Δ_f H°) of a compound is the enthalpy change when one mole of that compound is formed from its elements in their standard states. For ammonia, Δ_f H° = −46.2 kJ mol⁻¹ means:
$\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{NH}_3(g) \quad \Delta_r H^\circ = -46.2\ \text{kJ mol}^{-1}$
But the reaction given is:
$\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$
This is simply twice the formation reaction for NH₃. Since enthalpy is an extensive property (it scales with the amount of substance), multiplying the entire reaction by 2 multiplies Δ_r H° by 2.

Let’s work through it step by step:

Identify the formation reaction for NH₃ The formation reaction for one mole of NH₃ is:

$\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{NH}_3(g) \quad \Delta H = -46.2\ \text{kJ}$

Scale the reaction to match the target The target reaction produces 2 moles of NH₃. Multiply the formation reaction by 2:

$2 \times \left( \frac{1}{2}\text{N}_2 + \frac{3}{2}\text{H}_2 \rightarrow \text{NH}_3 \right) = \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$

Scale the enthalpy change accordingly Since enthalpy is proportional to the amount:
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TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$\dfrac{V}{3}, \dfrac{T}{3}$
›Reveal solutionSolution
Partitioning a vessel into equal parts divides the volume equally but does not change the temperature of the gas in each part — the answer is $\dfrac{V}{3}$ L and $T$ K.

The key idea here is that temperature is an intensive property of a gas, not an extensive one. When you take a sample of gas and simply divide the container into smaller compartments without doing any work or adding heat, the temperature of the gas in each compartment remains the same as the original. Volume, on the other hand, is extensive — it gets divided equally among the parts.

Let’s walk through it carefully.

What happens physically?

The vessel initially contains an ideal gas at temperature $T$ and occupies volume $V$ . A partition is inserted, dividing the vessel into three equal parts. No external work is done on the gas (the partition just slides in or is placed), and no heat is exchanged if the process is quick enough or the walls are insulating. The gas in each part is still the same gas, just in a smaller space.

Volume in each part:

Since the vessel is divided into three equal parts, the volume of each part is simply one-third of the original volume:

$V_{\text{each}} = \frac{V}{3}$

Temperature in each part: Temperature is a measure of the average kinetic energy of the gas molecules. Partitioning the container does not change the speed or energy of the molecules — they just now have less space to move in. No energy has been added or removed, so the temperature remains $T$ in each part.

$T_{\text{each}} = T$

Check the options:

(A) $\dfrac{V}{3}, \dfrac{T}{3}$ — Wrong, temperature doesn’t drop. …
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TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$_3$
›Reveal solutionSolution
In the Ostwald process, NH₃ oxidizes to NO, then NO₂, which disproportionates in water. The stoichiometry shows that 12 moles of NH₃ ultimately produces 8 moles of HNO₃ (504 g).

Understanding the Ostwald Process

The Ostwald process converts ammonia to nitric acid through a series of oxidation steps. Let's trace what happens to our 12 moles of NH₃ through each stage.

The key insight: Not all nitrogen atoms end up as HNO₃ because the final disproportionation reaction regenerates some NO.

Step-by-Step Reaction Sequence

1. First oxidation: Ammonia to nitric oxide
$4\text{NH}_3(g) + 5\text{O}_2(g) \xrightarrow{\text{catalyst, 500K}} 4\text{NO}(g) + 6\text{H}_2\text{O}(g)$
Starting with 12 moles of NH₃:
$12 \text{ moles NH}_3 \rightarrow 12 \text{ moles NO}$
The stoichiometry is 1:1 for NH₃ to NO.

2. Second oxidation: Nitric oxide to nitrogen dioxide
$2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)$
Our 12 moles of NO become:
$12 \text{ moles NO} \rightarrow 12 \text{ moles NO}_2$
Again, 1:1 stoichiometry.

3. Disproportionation in water: The critical step
$3\text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2\text{HNO}_3(aq) + \text{NO}(g)$
This is where we must be careful! For every 3 moles of NO₂:

2 moles become HNO₃

1 mole is reduced back to NO (which is mentioned in the problem) …
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TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$_2$
›Reveal solutionSolution
During electrolysis of aqueous CuSO₄ with inert platinum electrodes, Cu²⁺ is preferentially reduced at the cathode (over H⁺) and OH⁻ is preferentially oxidized at the anode (over SO₄²⁻), producing Cu at the cathode and O₂ at the anode.

When an aqueous solution undergoes electrolysis, we need to identify which species present in the solution will actually react at each electrode. The key is comparing the reduction and oxidation potentials of the competing species.

In aqueous CuSO₄, the solution contains:

Cu²⁺ and SO₄²⁻ ions from the salt

H⁺ and OH⁻ ions from water (H₂O ⇌ H⁺ + OH⁻)

At each electrode, the species with the more favorable potential wins the competition.

At the Cathode (Reduction)

The cathode is where reduction occurs. Two species can potentially gain electrons:

Cu²⁺ + 2e⁻ → Cu, with $E° = +0.34 \text{ V}$

2H⁺ + 2e⁻ → H₂, with $E° = 0.00 \text{ V}$

The more positive (less negative) the reduction potential, the easier the reduction. Since copper has a significantly more positive $E°$ , Cu²⁺ ions are preferentially reduced and metallic copper deposits on the platinum cathode.
$\text{Cathode: } \ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$
Tip

Metals with positive reduction potentials (like Cu, Ag, Au) plate out during electrolysis; those with very negative potentials (like Na, K) do not — hydrogen gas forms instead.

At the Anode (Oxidation)

The anode is where oxidation occurs. Two species can potentially lose electrons:

2H₂O → O₂ + 4H⁺ + 4e⁻, with $E° = +1.23 \text{ V}$ (or equivalently, 4OH⁻ → O₂ + 2H₂O + 4e⁻)

2SO₄²⁻ → S₂O₈²⁻ + 2e⁻, with $E° ≈ +2.0 \text{ V}$ …
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TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$Q. Number of moles of nitrogen gas liberated when ammonia is treated with excess of chlorine is (A) 1 (B) 2 (C) 0 (D) 3$
›Reveal solutionSolution
When ammonia reacts with an excess of chlorine, nitrogen trichloride ( $NCl_3$ ) is formed instead of nitrogen gas ( $N_2$ ). Therefore, no nitrogen gas is liberated. The number of moles of nitrogen gas liberated is $\boxed{0}$ .

Ammonia ( $NH_3$ ) is a compound where nitrogen is in a $-3$ oxidation state. It can act as a reducing agent, meaning it can be oxidized to a higher oxidation state. Chlorine ( $Cl_2$ ) is an oxidizing agent. The reaction between ammonia and chlorine is a redox reaction, and its products depend critically on which reactant is in excess.

Here's why the amount of each reactant matters:

If ammonia is in excess, chlorine acts as the limiting reagent. Ammonia is oxidized to elemental nitrogen ( $N_2$ , oxidation state $0$ ), and chlorine is reduced to chloride ions (in $NH_4Cl$ ).

If chlorine is in excess, ammonia acts as the limiting reagent. Chlorine is a strong enough oxidizing agent to oxidize nitrogen in ammonia beyond the elemental state, forming nitrogen trichloride ( $NCl_3$ , where nitrogen is in a $+3$ oxidation state).

The question specifies that ammonia is treated with an excess of chlorine. This dictates the specific reaction pathway.

Identify the correct reaction based on excess reactant:

Since chlorine is in excess, the reaction will lead to the formation of nitrogen trichloride ( $NCl_3$ ) and hydrogen chloride ( $HCl$ ). This is because the abundant chlorine can further oxidize the nitrogen from its $-3$ state in $NH_3$ to a $+3$ state in $NCl_3$ .

Write and balance the chemical equation:

The unbalanced reaction is:

$NH_3 + Cl_2 \rightarrow NCl_3 + HCl$

To balance this equation:

Balance nitrogen: 1 atom on both sides.

Balance hydrogen: 3 atoms on the left ( $NH_3$ ), 1 atom on the right ( $HCl$ ). Place a coefficient of 3 in front of $HCl$ : $NH_3 + Cl_2 \rightarrow NCl_3 + 3HCl$ …
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TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
$_3$
›Reveal solutionSolution
The key is to use the balanced chemical equation and apply Avogadro’s law under isobaric (constant pressure) conditions: 2 volumes of PH₃ produce 3 volumes of H₂, so 100 ml of PH₃ yields 150 ml of H₂ gas, and the final volume is 150 ml.

Concept and intuition:

When a gas decomposes at constant temperature and pressure (isobaric conditions), the volume of gas is directly proportional to the number of moles (Avogadro’s law). So we don’t need to worry about temperature or pressure values — we just need the stoichiometric ratio of gaseous reactants to gaseous products. Here, PH₃ is a gas, and the products are solid phosphorus (P) and hydrogen gas (H₂). Only the hydrogen gas contributes to the final volume. The solid phosphorus takes negligible volume.

Step-by-step reasoning:

Write the balanced chemical equation Phosphine (PH₃) decomposes into phosphorus and hydrogen:

$2\,\text{PH}_3(g) \rightarrow 2\,\text{P}(s) + 3\,\text{H}_2(g)$
This is the standard decomposition reaction.

Identify which species are gases

PH₃ is a gas.

P(s) is a solid — its volume is negligible compared to gases.

H₂ is a gas. So only the hydrogen gas will occupy volume in the container after decomposition.

Apply Avogadro’s law (constant T and P)

Under isobaric and isothermal conditions, volume is proportional to moles of gas. From the equation:

2 volumes of PH₃ produce 3 volumes of H₂.

Therefore, the volume ratio is:

$\frac{V_{\text{H}_2}}{V_{\text{PH}_3}} = \frac{3}{2}$

Calculate the volume of H₂ produced Given initial volume of PH₃ = 100 ml:
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TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
$2\mathrm{HI} \rightarrow \mathrm{H}_2 + \mathrm{I}_2$
›Reveal solutionSolution
A zero-order reaction's rate is independent of reactant concentration. Among the given options, the photochemical reaction between hydrogen and chlorine is a zero-order reaction.

The order of a reaction describes how the rate of the reaction depends on the concentration of its reactants. For a zero-order reaction, the rate of reaction does not depend on the concentration of the reactants. This means that even if you change the amount of reactant present, the speed at which the reaction proceeds remains constant.

Let's examine each option to determine which one fits the description of a zero-order reaction.

Understanding Reaction Order:

The order of a reaction is an experimentally determined quantity. It cannot generally be predicted from the stoichiometry of the balanced chemical equation alone, especially for complex reactions. However, for elementary reactions, the order with respect to each reactant is equal to its stoichiometric coefficient. Many reactions, particularly those involving surfaces or light, can exhibit zero-order kinetics under specific conditions.

Analyzing Option (A): $2\mathrm{HI} \rightarrow \mathrm{H}_2 + \mathrm{I}_2$

This reaction, the decomposition of hydrogen iodide on a gold surface, is a classic example of a first-order reaction at high concentrations of HI, and can approach zero-order at very high concentrations where the surface is saturated. However, in the gas phase, it is typically second order. Without specific conditions, it's generally not considered zero-order.

Analyzing Option (B): $\mathrm{H}_2 + \mathrm{Br}_2 \xrightarrow{\Delta} 2\mathrm{HBr}$

This is a complex chain reaction. The rate law for this reaction is experimentally found to be:

$\text{Rate} = k[\mathrm{H}_2][\mathrm{Br}_2]^{1/2} / (1 + k'[\mathrm{HBr}]/[\mathrm{Br}_2])$
This rate law is clearly not zero-order with respect to either reactant, nor overall. It's a fractional order reaction.

4. Analyzing Option (C): $2\mathrm{N}_2\mathrm{O}_5 \rightarrow 4\mathrm{NO}_2 + \mathrm{O}_2$

The decomposition of dinitrogen pentoxide is a well-known example of a first-order reaction. Its rate law is:
$\text{Rate} = k[\mathrm{N}_2\mathrm{O}_5]$
This reaction is first-order with respect to $\mathrm{N}_2\mathrm{O}_5$ and overall first-order.

5. Analyzing Option (D): $\mathrm{H}_2 + \mathrm{Cl}_2 \xrightarrow{h\nu} 2\mathrm{HCl}$ …
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(i) Order of the reaction

(ii) Slope of the line

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