Q.The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k=2.5×10−4 mol−1Ls−1?
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
−dtd[A]=k0
Where [A] is the concentration of the substance (or amount, depending on context), t is time, and k0 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−k0t
This is the equation of a line with slope −k0 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:
t1/2=2k0[A]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.
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.
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 | −dtd[A]=k0 | −dtd[A]=k1[A] |
| Units of rate constant | concentration/time | 1/time |
| Plot of [A] vs. t | Straight line | Curved (exponential decay) |
| Half-life | [A]0/2k0 (depends on starting amount) | ln2/k1 (constant) |
| Real example | Alcohol elimination at high doses | Most drug metabolism at therapeutic doses |
The Intuition Check
If someone asks you "Is this process zero order?", ask yourself: does the rate stay the same even when the amount drops? If you have a machine that destroys exactly 5 units per hour, and you start with 100 units, after 10 hours you'll have 50 units left. After another 10 hours, you'll have 0. The machine doesn't slow down as the pile shrinks — that's zero order.
If instead the machine destroys 5% of what's left per hour, then in the first hour it destroys 5 units (5% of 100), but in the tenth hour it destroys only about 3 units (5% of 60). The amount destroyed per hour keeps dropping — that's first order.
Zero order is the exception, not the rule. It happens when something is saturated — enzymes, transporters, or any system that has a maximum processing capacity. When you meet it in an exam problem, the dead giveaway is a straight line on a concentration-time graph, or a half-life that changes with the starting dose.
Zero order kinetics is a distinctive case within the NCERT/CBSE Class 12 Chemistry Chemical Kinetics chapter, and ‘zero order reaction graph’ or ‘zero order kinetics examples’ are recurring important-question searches for board exams as well as JEE Main and NEET. Recognising a zero-order reaction from its straight-line concentration-time graph is a skill directly tested in competitive-exam MCQs.
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→products, the rate of disappearance of A is:
−dtd[A]=k
where k is the zero order rate constant (units: concentration/time, e.g., mol L−1s−1).
Notice: No [A] term on the right side — that's the signature of zero order.
Step 2: Separate variables and integrate
−d[A]=kdt
Integrate from initial time t=0 (concentration [A]0) to time t (concentration [A]t):
−∫[A]0[A]td[A]=k∫0tdt
−[A]t+[A]0=kt
Step 3: Rearrange to the familiar form
[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=2[A]0:
2[A]0=[A]0−kt1/2
t1/2=2k[A]0
Critical exam point: Unlike first order (where t1/2 is constant), zero order half-life increases with higher initial concentration.
Real-World Examples (for context)
| Example | Why it's zero order |
|---|---|
| Drug dissolution (e.g., a sustained-release tablet) | Surface area is constant; drug saturates the boundary layer |
| Enzyme-catalyzed reactions (at high substrate) | Enzyme active sites are fully occupied (saturation) |
| Photochemical reactions (with constant light) | Light intensity (not reactant concentration) limits the rate |
Quick Summary for Exams
| Property | Zero Order |
|---|---|
| Rate law | −dtd[A]=k |
| Integrated form | [A]t=[A]0−kt |
| Plot for straight line | [A]t vs. t |
| Slope | −k |
| Half-life | t1/2=2k[A]0 |
| Units of k | concentration⋅time−1 |
Remember: The reason for zero order is always a saturation or surface limitation — the rate can't go faster because something else (not the reactant) is the bottleneck.
The key idea is that for a zero order reaction, the rate is constant and independent of concentration. The stoichiometric coefficients relate the rate of disappearance of NH3 to the rates of appearance of products.
Step 1: Write the balanced equation:
2NH3(g)PtN2(g)+3H2(g)
Step 2: For a zero order reaction, the rate law is:
Rate=−21dtd[NH3]=k=2.5×10−4 mol L−1s−1
Step 3: Relate to product formation rates:
dtd[N2]=k=2.5×10−4 mol L−1s−1
dtd[H2]=3k=7.5×10−4 mol L−1s−1
The rate of production of N2 is 2.5×10−4 mol L−1s−1 and of H2 is 7.5×10−4 mol L−1s−1.
For a zero-order reaction, the rate is constant and equals the rate constant k. Using the stoichiometry of 2NH3→N2+3H2, the rate of production of N2 is 2.5×10−4 mol L−1s−1 and that of H2 is 7.5×10−4 mol L−1s−1.
The key to this problem lies in understanding what "zero order" means — and then connecting that to the stoichiometric coefficients of the reaction.
In a zero-order reaction, the rate of the reaction does not depend on the concentration of the reactant. The rate is constant throughout the reaction, and the given rate constant k is the rate of the reaction itself (not merely the raw rate of disappearance of one particular species).
The decomposition reaction is:
2NH3(g)PtN2(g)+3H2(g)
For every 2 molecules of NH3 that disappear, 1 molecule of N2 and 3 molecules of H2 appear.
For a general reaction aA→bB+cC, the rate of reaction is:
−a1dtd[A]=b1dtd[B]=c1dtd[C]
For a zero-order process, this common value equals the rate constant k.
Let’s apply this step by step.
- Write the rate of reaction in terms of NH3. Since the reaction is zero order, the rate of reaction is:
−21dtd[NH3]=k=2.5×10−4 mol L−1s−1
This means the actual rate of disappearance of NH3 is −dtd[NH3]=2k=5.0×10−4 mol L−1s−1.
- Relate this to the rate of production of N2. From the balanced equation:
−21dtd[NH3]=dtd[N2]
Both sides equal k, so:
dtd[N2]=k=2.5×10−4 mol L−1s−1
- Relate to the rate of production of H2. From the balanced equation:
−21dtd[NH3]=31dtd[H2]
So:
dtd[H2]=3k=3×(2.5×10−4)=7.5×10−4 mol L−1s−1
A common mistake is to treat the given k as the raw rate of disappearance of NH3 (−d[NH3]/dt=k) and then divide by the coefficient again when finding the rate of production of N2 and H2 — that double-counts the stoichiometric factor. By the standard definition, k (the zero-order rate constant) already equals the overall rate of reaction −21dtd[NH3], so the rate of production of N2 is simply k, and of H2 is 3k.
The rate of production of N2 is 2.5×10−4 mol L−1s−1 and that of H2 is 7.5×10−4 mol L−1s−1.
Method: Stoichiometric Rate Analysis for Zero-Order Reactions
Why this method?
In a zero-order reaction, the rate is independent of concentration and equals the rate constant k. For a balanced chemical equation, the rates of appearance/disappearance of species are linked by stoichiometric coefficients.
Step 1: Write the balanced equation
The decomposition of ammonia on platinum is:
2NH3(g)→N2(g)+3H2(g)
Step 2: Write the rate expression in terms of NH3
For a zero-order reaction:
Rate=−21dtd[NH3]=k
Given:
k=2.5×10−4 mol L−1s−1
So:
−dtd[NH3]=2k=5.0×10−4 mol L−1s−1
Step 3: Relate rates of production of N2 and H2
From stoichiometry:
−21dtd[NH3]=dtd[N2]=31dtd[H2]
Since each equals k:
dtd[N2]=k=2.5×10−4 mol L−1s−1
dtd[H2]=3k=7.5×10−4 mol L−1s−1
Final Answer
- Rate of production of N2 = 2.5×10−4 mol L−1s−1
- Rate of production of H2 = 7.5×10−4 mol L−1s−1
Key Exam Tip
In zero-order kinetics, the rate constant k directly gives the rate of reaction. Multiply by the stoichiometric coefficient of the product (as written in the balanced equation) to get its rate of appearance.
Here are the common mistakes students make on this exact type of zero-order kinetics problem, and how to avoid each.
1. Mistake: Forgetting the Stoichiometric Ratios
The error:
Students often assume the rate of disappearance of NH3 equals the rate of appearance of N2 or H2. They write:
−dtd[NH3]=dtd[N2]=dtd[H2]
This is wrong because the balanced equation shows different coefficients.
How to avoid:
Always write the balanced chemical equation first:
2NH3(g)PtN2(g)+3H2(g)
Then use the stoichiometric relationship:
−21dtd[NH3]=dtd[N2]=31dtd[H2]
2. Mistake: Treating the Given k as the Rate of Disappearance of NH3 Instead of the Rate of Reaction
The error:
For a zero-order reaction, the rate constant k is the overall rate of reaction:
Rate=−21dtd[NH3]=k
Some students instead assume k=−dtd[NH3] directly (i.e., that k is the raw rate of disappearance of NH3, ignoring its own coefficient of 2). This introduces an extra, incorrect factor of 21 into every downstream answer.
How to avoid:
By the standard definition used throughout NCERT and Indian board exams, for a reaction aA→bB+cC, the rate constant of a zero-order reaction is the common value:
k=−a1dtd[A]=b1dtd[B]=c1dtd[C]
So here, k=−21dtd[NH3]=dtd[N2]=31dtd[H2]=2.5×10−4 mol L−1s−1.
3. Mistake: Confusing Rate of Reaction with Rate of Appearance/Disappearance
The error:
Students write:
Rate=−dtd[NH3]=dtd[N2]=dtd[H2]
This ignores coefficients.
How to avoid:
Remember the definition:
Rate of reaction=−a1dtd[A]=b1dtd[B]
For 2NH3→N2+3H2:
Rate=−21dtd[NH3]=dtd[N2]=31dtd[H2]=k
So:
- dtd[N2]=k
- dtd[H2]=3k
4. Mistake: Using Integrated Rate Laws Instead of Differential Rate
The error:
Students try to use [A]=[A]0−kt to find rates, but the question asks for instantaneous rates at any time (since zero order, rate is constant).
How to avoid:
For zero order, the rate is constant and equal to k (the rate constant for the reaction). You do not need initial concentration or time. Just use the differential rate law:
Rate=k
Then apply stoichiometry.
5. Mistake: Not Checking the Units of the Final Answer
The error:
Students give answers without units, or with wrong units (e.g., s−1 instead of mol L−1s−1).
How to avoid:
Always include units. For zero-order, rates of appearance/disappearance have units of concentration per time:
mol L−1s−1
✓ Final Correct Answer (Summary)
Given k=2.5×10−4 mol L−1s−1, which is the rate of the reaction (−21dtd[NH3]):
- Rate of production of N2:
dtd[N2]=k=2.5×10−4 mol L−1s−1
- Rate of production of H2:
dtd[H2]=3k=7.5×10−4 mol L−1s−1
Key takeaway: Always start with the balanced equation, define the rate of reaction, and use stoichiometric coefficients to convert between species — and remember that the given rate constant k for a zero-order reaction already IS the rate of reaction, not the raw disappearance rate of one specific reactant.
- TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQQ.A → P, is a zero-order reaction. At 300 K, this reaction was started with [A]=0.5molL−1. After 100 s, the concentration of A was 0.05molL−1. What is the rate constant (in molL−1s−1) of this reaction? (A) 2.303×10−2 (B) 2.303×10−3 (C) 4.5×10−2 (D) 4.5×10−3
›Reveal solutionSolution
For a zero-order reaction, the rate is constant and independent of concentration. The integrated rate law gives k=t[A]0−[A]t, which yields k=4.5×10−3molL−1s−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 A→P, the rate law is:
Rate=−dtd[A]=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=t[A]0−[A]t
Now let's apply it step by step.
-
Identify the given data
Initial concentration: [A]0=0.5molL−1
Concentration after 100 s: [A]t=0.05molL−1
Time: t=100s
-
Find the change in concentration
[A]0−[A]t=0.5−0.05=0.45molL−1
- Apply the zero-order formula
k=1000.45=0.0045molL−1s−1
- Express in scientific notation
k=4.5×10−3molL−1s−1
Watch outA common mistake is to use the first-order formula k=t2.303log[A]t[A]0 here. That would give 2.303×10−2, which is option (A) — a deliberate distractor. Always check the order of the reaction before choosing the formula.
TipFor zero-order reactions, the units of k are always molL−1s−1 (concentration/time). This is a quick sanity check: if the options have different units, you can eliminate wrong ones immediately.
✓Final answerThe rate constant is 4.5×10−3molL−1s−1, which corresponds to option (D).
-
- TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQQ.The standard enthalpy of formation (ΔfHΘ) of ammonia is −46.2 kJ mol−1. What is the ΔrHΘ of the following reaction? N2(g) + 3H2(g) → 2NH3(g) (A) −46.2 kJ (B) +46.2 kJ (C) −92.4 kJ (D) −184.8 kJ
›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:
21N2(g)+23H2(g)→NH3(g)ΔrH∘=−46.2 kJ mol−1
But the reaction given is:
N2(g)+3H2(g)→2NH3(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:
21N2(g)+23H2(g)→NH3(g)ΔH=−46.2 kJ
- Scale the reaction to match the target The target reaction produces 2 moles of NH₃. Multiply the formation reaction by 2:
2×(21N2+23H2→NH3)=N2+3H2→2NH3
- Scale the enthalpy change accordingly Since enthalpy is proportional to the amount:
ΔrH∘=2×(−46.2 kJ)=−92.4 kJ
- Check the units The answer is in kJ (not kJ mol⁻¹) because the reaction as written consumes specific amounts — the total enthalpy change for the reaction as balanced is −92.4 kJ.
Watch outA common mistake is to forget that Δ_f H° is per mole of compound formed. Here, the reaction forms 2 moles of NH₃, so you must multiply by 2, not leave it as −46.2 kJ.
TipYou can also think of it as: Δ_r H° = Σ(Δ_f H° of products) − Σ(Δ_f H° of reactants). For elements in their standard states (N₂, H₂), Δ_f H° = 0. So Δ_r H° = [2 × (−46.2)] − [0 + 0] = −92.4 kJ.
✓Final answerThe correct option is (C).
ANSWER: C
- TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQQ.At T(K), a vessel contains V litres of an ideal gas. The vessel was partitioned into three equal parts. The volume (in L) and temperature (in K) in each part are respectively (A) 3V,3T (B) 3V,T (C) 3V,T (D) 3V,3T
›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 3V 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:
Veach=3V
- 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.
Teach=T
- Check the options:
- (A) 3V,3T — Wrong, temperature doesn’t drop.
- (B) 3V,T — Correct.
- (C) 3V,T — Wrong, volume doesn’t increase.
- (D) 3V,3T — Wrong, temperature doesn’t triple.
Watch outA common mistake is to think that because pressure or density changes, temperature must also change. But temperature only changes if work is done (compression/expansion) or heat is exchanged. Simply partitioning without moving the gas does neither.
TipRemember: intensive properties (temperature, pressure if equilibrium is maintained, density in a uniform gas) stay the same when you take a sample. Extensive properties (volume, number of moles, mass) scale with the size of the sample.
✓Final answerThe correct option is (B): each part has volume 3V L and temperature T K.
-
- TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQQ.In Ostwald process of manufacture of nitric acid, 12 moles of NH3 was completely oxidized in air by a catalyst at 500 K and 9 bar. The resultant NO(g) was completely oxidized to NO2(g) and dissolved in water to form nitric acid and NO(g). What is the weight (in g) of nitric acid formed? (N = 14 u; O = 16 u; H = 1 u) (A) 756 (B) 378 (C) 504 (D) 252
›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
4NH3(g)+5O2(g)catalyst, 500K4NO(g)+6H2O(g)
Starting with 12 moles of NH₃:
12 moles NH3→12 moles NO
The stoichiometry is 1:1 for NH₃ to NO.
2. Second oxidation: Nitric oxide to nitrogen dioxide
2NO(g)+O2(g)→2NO2(g)
Our 12 moles of NO become:
12 moles NO→12 moles NO2
Again, 1:1 stoichiometry.
3. Disproportionation in water: The critical step
3NO2(g)+H2O(l)→2HNO3(aq)+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)
Watch outThe disproportionation reaction means only ⅔ of the nitrogen atoms end up as nitric acid. Students often miss this and assume all 12 moles convert to HNO₃.
4. Calculating HNO₃ produced
From 12 moles of NO₂:
Moles of HNO3=12×32=8 moles
5. Converting to mass
Molar mass of HNO₃:
MHNO3=1+14+3(16)=1+14+48=63 g/mol
Mass of HNO₃:
Mass=8 moles×63 g/mol=504 g
Overall conversion efficiency:
moles NH3moles HNO3=32
This comes directly from the disproportionation stoichiometry.
✓Final answerThe correct option is (C).
ANSWER: C
- TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQQ.The products formed at cathode and anode when aqueous copper sulphate solution is electrolyzed using platinum electrodes are (A) Cu, O2 (B) H2, O2 (C) Cu, H2 (D) H2, SO2
›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 V
- 2H⁺ + 2e⁻ → H₂, with E°=0.00 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.
Cathode: CuX2+(aq)+2eX−Cu(s)
TipMetals 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 V (or equivalently, 4OH⁻ → O₂ + 2H₂O + 4e⁻)
- 2SO₄²⁻ → S₂O₈²⁻ + 2e⁻, with E°≈+2.0 V
For oxidation, we need to consider which is easier to oxidize (lower oxidation potential). Water/hydroxide ions are much easier to oxidize than sulfate ions. The platinum electrode is inert, so it doesn't participate. Hydroxide ions (or water) are oxidized, releasing oxygen gas.
Anode: 4OHX−(aq)OX2(g)+2HX2O(l)+4eX−
or equivalently:
2HX2O(l)OX2(g)+4HX+(aq)+4eX−
Watch outSulfate ions are extremely stable and resist oxidation under normal electrolysis conditions. They remain as spectator ions in the solution.
The overall process deposits reddish-brown copper metal on the cathode while bubbles of colorless, odorless oxygen gas evolve at the anode.
✓Final answerThe correct option is (A): Cu forms at the cathode and O₂ at the anode.
- TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQQ.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 (NCl3) is formed instead of nitrogen gas (N2). Therefore, no nitrogen gas is liberated. The number of moles of nitrogen gas liberated is 0.
Ammonia (NH3) 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 (Cl2) 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 (N2, oxidation state 0), and chlorine is reduced to chloride ions (in NH4Cl).
- 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 (NCl3, 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.
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Identify the correct reaction based on excess reactant:
Since chlorine is in excess, the reaction will lead to the formation of nitrogen trichloride (NCl3) and hydrogen chloride (HCl). This is because the abundant chlorine can further oxidize the nitrogen from its −3 state in NH3 to a +3 state in NCl3.
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Write and balance the chemical equation:
The unbalanced reaction is:
NH3+Cl2→NCl3+HCl
To balance this equation:
- Balance nitrogen: 1 atom on both sides.
- Balance hydrogen: 3 atoms on the left (NH3), 1 atom on the right (HCl). Place a coefficient of 3 in front of HCl: NH3+Cl2→NCl3+3HCl
- Balance chlorine: There are 2 atoms on the left (Cl2) and 3 (from NCl3) +3 (from 3HCl) =6 atoms on the right. Place a coefficient of 3 in front of Cl2: NH3+3Cl2→NCl3+3HCl
This equation is now balanced.
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Identify the nitrogen-containing product and determine moles of nitrogen gas:
From the balanced equation, the nitrogen-containing product formed is nitrogen trichloride (NCl3). Nitrogen gas (N2) is not formed in this reaction.
Therefore, the number of moles of nitrogen gas liberated is 0.
Watch outIt is a common mistake to confuse this reaction with the one where ammonia is in excess. If ammonia were in excess, the reaction would be 8NH3+3Cl2→6NH4Cl+N2, and 1 mole of nitrogen gas (N2) would be liberated for every 8 moles of ammonia. Always pay close attention to which reactant is in excess.
The correct option is (C).
✓Final answerThe number of moles of nitrogen gas liberated when ammonia is treated with excess of chlorine is 0.
- TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQQ.100 ml of PH3 on decomposition produces P(s) and H2(g) under isobaric condition. The final volume of the container is (A) 50 ml (B) 100 ml (C) 150 ml (D) 250 ml
›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:
2PH3(g)→2P(s)+3H2(g)
This is the standard decomposition reaction.
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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.
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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:
VPH3VH2=23
- Calculate the volume of H₂ produced Given initial volume of PH₃ = 100 ml:
VH2=100×23=150 ml
- Determine the final volume of the container Since the container is isobaric (the piston or walls move to keep pressure constant), the final volume is just the volume of gas present after reaction. The solid P takes up negligible space. So final volume = 150 ml.
Watch outA common mistake is to forget that solid phosphorus does not contribute to gas volume, or to incorrectly balance the equation as 1 PH₃ → 1 P + 1.5 H₂, then forget the factor of 2. Always use whole-number coefficients.
TipIn isobaric gas reactions, you can directly use the mole ratio of gaseous products to gaseous reactants as a volume ratio — no need for ideal gas law calculations.
✓Final answerThe correct option is (C).
ANSWER: C
- TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQQ.Which of the following is a zero order reaction? (A) 2HI→H2+I2 (B) H2+Br2Δ2HBr (C) 2N2O5→4NO2+O2 (D) H2+Cl2hν2HCl
›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.
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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.
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Analyzing Option (A): 2HI→H2+I2
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.
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Analyzing Option (B): H2+Br2Δ2HBr
This is a complex chain reaction. The rate law for this reaction is experimentally found to be:
Rate=k[H2][Br2]1/2/(1+k′[HBr]/[Br2])
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): 2N2O5→4NO2+O2
The decomposition of dinitrogen pentoxide is a well-known example of a first-order reaction. Its rate law is:
Rate=k[N2O5]
This reaction is first-order with respect to $\mathrm{N}_2\mathrm{O}_5$ and overall first-order.5. Analyzing Option (D): H2+Cl2hν2HCl
This is a photochemical reaction, initiated by light (hν). In such reactions, the rate often depends on the intensity of the light absorbed rather than the concentration of the reactants, especially if the light intensity is the limiting factor. The mechanism involves chain propagation steps:
Cl2hν2Cl⋅
Cl⋅+H2→HCl+H⋅
H⋅+Cl2→HCl+Cl⋅
Under typical conditions, the rate of this reaction is found to be independent of the concentrations of $\mathrm{H}_2$ and $\mathrm{Cl}_2$, and instead depends on the intensity of the incident light. Therefore, it is a zero-order reaction with respect to the reactants. > [!IMPORTANT] > Photochemical reactions, enzyme-catalyzed reactions (when the enzyme is saturated), and reactions occurring on a solid surface (when the surface is saturated with reactants) are common examples of zero-order reactions.✓Final answerThe photochemical reaction H2+Cl2hν2HCl is a zero-order reaction. The correct option is (D).
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