Showing the 12 most recent of 19 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The rate constant of a first order reaction becomes 4 times when the temperature changes from 300 K to 320 K. What is the activation energy ($E_a$) (in kJ mol$^{-1}$) of reaction? ( $R = 8.3$ J mol$^{-1}$ K$^{-1}$, log 4 = 0.6) (Assume $E_a$ does not change in the given temperature range) (A) 55.05 (B) 550.5 (C) 27.57 (D) 225.25

   ›Reveal solutionSolution

   The Arrhenius equation relates the rate constant of a reaction to temperature and activation energy. By using the given change in rate constant over a temperature range, we can calculate the activation energy. The activation energy is $\boxed{55.05 \text{ kJ mol}^{-1}}$.

   The rate constant of a chemical reaction is highly dependent on temperature. This relationship is quantitatively described by the Arrhenius equation, which links the rate constant ($k$) to the activation energy ($E_a$), the absolute temperature ($T$), and a pre-exponential factor ($A$). The activation energy represents the minimum energy required for reactant molecules to transform into products.

   When the temperature increases, more reactant molecules possess energy equal to or greater than the activation energy, leading to a higher frequency of effective collisions and thus an increased rate constant. The problem provides two temperatures and the corresponding change in the rate constant, allowing us to determine the activation energy. The fact that it's a first-order reaction is irrelevant for calculating $E_a$ using the Arrhenius equation, as the equation applies to the rate constant itself, regardless of reaction order.

   1. Recall the Arrhenius Equation (two-point form):

      The Arrhenius equation can be expressed in a convenient form for calculating activation energy when the rate constants at two different temperatures are known. This form is derived by taking the natural logarithm of the Arrhenius equation at two temperatures and subtracting them.

      $\log_{10} \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)$

      Where:

      $k_1$ and $k_2$ are the rate constants at absolute temperatures $T_1$ and $T_2$, respectively.

      $E_a$ is the activation energy.

      $R$ is the ideal gas constant.

   2. Identify the given values:

      We are given the following information:

      • Initial temperature, $T_1 = 300 \text{ K}$
      • Final temperature, $T_2 = 320 \text{ K}$
      • The rate constant becomes 4 times, so $\frac{k_2}{k_1} = 4$.
      • Gas constant, $R = 8.3 \text{ J mol}^{-1} \text{ K}^{-1}$.
      • $\log 4 = 0.6$.
      • We need to find $E_a$ in kJ mol$^{-1}$.

   3. Substitute the values into the Arrhenius equation:

      Substitute the known values into the two-point form of the Arrhenius equation:

   $\log_{10} (4) = \frac{E_a}{2.303 \times 8.3 \text{ J mol}^{-1} \text{ K}^{-1}} \left(\frac{320 \text{ K} - 300 \text{ K}}{300 \text{ K} \times 320 \text{ K}}\right)$

   $$ 0.6 = \frac{E_a}{19.1849 \text{ J mol}^{-1} \text{ K}^{-1}} \left(\frac{20 \text{ K}}{96000 \text{ K}^2}\right) $$ …
   

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2. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.A $\rightarrow$ P is a first order reaction. At 300 K this reaction was started with $[A] = 0.5\, \text{mol L}^{-1}$. The rate constant of reaction was $0.125\, \text{min}^{-1}$. The same reaction was started separately with $[A] = 1\, \text{mol L}^{-1}$ at 300 K. The rate constant (in $\text{min}^{-1}$) now is (A) 0.25 (B) 0.50 (C) 0.125 (D) 1.00

   ›Reveal solutionSolution

   For a first‑order reaction, the rate constant is independent of the initial concentration. Therefore, changing the starting concentration from 0.5 mol L⁻¹ to 1 mol L⁻¹ does not change the rate constant. The answer remains 0.125 min⁻¹.

   The key concept here is the order of a reaction and what it tells us about the rate constant. For a first‑order reaction, the rate law is:

   $\text{Rate} = k[A]$

   The rate constant $k$ is an intrinsic property of the reaction at a given temperature — it depends only on the activation energy and temperature, not on the concentration of the reactant. This is a fundamental distinction: changing $[A]$ changes the rate, but $k$ stays fixed.

   Let’s walk through the reasoning step by step.

   1. Identify the reaction order. The problem states: “A → P is a first order reaction.” That means the rate depends linearly on $[A]$, and the integrated rate law is:

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

   The rate constant $k$ has units of time⁻¹ (here min⁻¹), which matches the given value 0.125 min⁻¹.

   2. Recognize what the question is really asking. The reaction is run twice at the same temperature (300 K), but with different initial concentrations: first 0.5 mol L⁻¹, then 1 mol L⁻¹. The question asks for the new rate constant. Since temperature is unchanged, and the reaction is first order, the rate constant is identical in both experiments. …

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3. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.A $\rightarrow$ P is a first order reaction. At 300 K this reaction was started with $[\mathrm{A}] = 0.5\, \mathrm{mol\,L}^{-1}$. The rate constant of reaction was $0.125\, \mathrm{min}^{-1}$. The same reaction was started separately with $[\mathrm{A}] = 1\, \mathrm{mol\,L}^{-1}$ at 300 K. The rate constant (in $\mathrm{min}^{-1}$) now is (A) 0.125 (B) 1.00 (C) 0.25 (D) 0.50

   ›Reveal solutionSolution

   For a first‑order reaction, the rate constant is independent of the initial concentration. Therefore, changing the starting concentration from 0.5 mol L⁻¹ to 1 mol L⁻¹ does not change the rate constant. The answer remains 0.125 min⁻¹.

   Concept & Intuition

   The defining feature of a first‑order reaction is that its rate depends linearly on the concentration of only one reactant:

   $\text{Rate} = k[A]$

   Here, $k$ is the rate constant — a proportionality factor that depends only on temperature and the nature of the reaction, not on how much reactant you start with. If you double the initial concentration, the initial rate doubles, but $k$ stays the same. This is a common point of confusion: students sometimes think that changing the starting amount changes the speed constant, but it only changes the rate, not the constant itself.

   Step‑by‑Step Reasoning

   1. Identify the reaction order The problem states: “A → P is a first order reaction.” For a first‑order reaction, the integrated rate law is

   $\ln[A]_t = \ln[A]_0 - kt$

   and the half‑life is $t_{1/2} = \frac{\ln 2}{k}$, which is independent of $[A]_0$.

   2. Recognize what the rate constant depends on

      The rate constant $k$ is a function of temperature (given by the Arrhenius equation) and the activation energy of the reaction. It does not depend on the initial concentration of the reactant.

   3. Apply to the given data

      At 300 K, the first experiment uses $[A]_0 = 0.5\ \mathrm{mol\,L^{-1}}$ and yields $k = 0.125\ \mathrm{min^{-1}}$.

      The second experiment is at the same temperature (300 K) but with $[A]_0 = 1\ \mathrm{mol\,L^{-1}}$. Since temperature is unchanged, the rate constant must be identical.

   4. Eliminate incorrect options …

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4. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Observe the following colloids I. Gold II. Detergent III. Starch IV. Soap V. Synthetic rubber VI. Sulphur VII. Cellulose The number of multimolecular colloids and macromolecular colloids in the above list is respectively. (A) 2, 3 (B) 1, 3 (C) 3, 4 (D) 4, 3

   ›Reveal solutionSolution

   Multimolecular colloids are formed by aggregation of small molecules, while macromolecular colloids consist of large polymer molecules. From the given list, Gold and Sulphur are multimolecular, and Starch, Synthetic rubber, and Cellulose are macromolecular, leading to counts of 2 and 3 respectively.

   Colloids are classified based on the nature of the dispersed phase particles. Understanding these classifications is key to identifying the types of colloids in the given list.

   Here's a breakdown of the relevant colloid types:

   • Multimolecular Colloids: These colloids are formed when a large number of atoms or small molecules (typically with diameters less than $1 \text{ nm}$) aggregate together to form particles of colloidal size (between $1 \text{ nm}$ and $1000 \text{ nm}$). The individual units are held together by weak van der Waals forces. Examples include gold sol and sulfur sol.

   • Macromolecular Colloids: In these colloids, the dispersed particles are themselves large molecules (macromolecules) with high molecular masses. These macromolecules are typically polymers, and their size naturally falls within the colloidal range. They often behave like true solutions but are classified as colloids due to their large particle size. Examples include starch, proteins, enzymes, and synthetic polymers like nylon or synthetic rubber.

   • Associated Colloids (Micelles): These are substances that behave as normal electrolytes at low concentrations but form aggregates called micelles at higher concentrations. The aggregated particles are of colloidal size. This aggregation occurs above a certain concentration (Critical Micelle Concentration, CMC) and above a certain temperature (Kraft temperature). Soaps and detergents are classic examples. For the purpose of this question, which specifically asks for multimolecular and macromolecular colloids, associated colloids are a distinct category and should not be counted in either of the other two.

   Let's classify each substance in the given list:

   1. Gold (I): Gold sols are formed by the aggregation of many individual gold atoms into particles of colloidal dimensions. This fits the definition of a multimolecular colloid.

   2. Detergent (II): Detergents are surfactants that form micelles in water above their Critical Micelle Concentration (CMC). Micelles are aggregates of many small detergent molecules. This is an associated colloid, not a multimolecular or macromolecular colloid in the context of this classification. Therefore, it is not counted in either category.

   3. Starch (III): Starch is a natural polymer (a polysaccharide) with a very high molecular mass. Its individual molecules are large enough to fall within the colloidal size range. Therefore, starch solution is a macromolecular colloid. …

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5. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.Consider the gaseous reaction $A_2 + B_2 \rightarrow 2AB$ The following data was obtained for the above reaction

   [!FORMULA] $\begin{array}{|c|c|c|} \hline \left[ A_2 \right]_0 & \left[ B_2 \right]_0 & \text{Initial rate of formation of AB} \\ & & (\text{mol L}^{-1} \text{s}^{-1}) \\ \hline 0.1\,M & 0.1\,M & 2.5 \times 10^{-4} \\ 0.2\,M & 0.1\,M & 5.0 \times 10^{-4} \\ 0.2\,M & 0.2\,M & 1.0 \times 10^{-3} \\ \hline \end{array}$

   The value of rate constant for the above reaction is (A) $1.25 \times 10^{-2}$ (B) $1.25 \times 10^{-3}$ (C) $2.5 \times 10^{-2}$ (D) $2.5 \times 10^{-1}$

   ›Reveal solutionSolution

   The reaction is first order in each reactant, $r=k[A_2][B_2]$. Because the table gives the rate of formation of AB (and $2$ AB form per reaction event), $\tfrac12\dfrac{d[AB]}{dt}=k[A_2][B_2]$, which gives $k=1.25\times10^{-2}\ \text{L mol}^{-1}\text{s}^{-1}$ — option (A).

   Step 1 — Order in $A_2$ (rows 1 → 2)

   $[B_2]$ fixed; $[A_2]$ doubles ($0.1\to0.2$) and the rate doubles ($2.5\times10^{-4}\to5.0\times10^{-4}$). So order in $A_2$ is $1$.

   Step 2 — Order in $B_2$ (rows 2 → 3)

   $[A_2]$ fixed; $[B_2]$ doubles ($0.1\to0.2$) and the rate doubles ($5.0\times10^{-4}\to1.0\times10^{-3}$). So order in $B_2$ is $1$.

   Step 3 — Rate law and the stoichiometric factor

   $r=k[A_2][B_2],\qquad r=-\frac{d[A_2]}{dt}=\tfrac12\frac{d[AB]}{dt}$

   The tabulated quantity is $R=\dfrac{d[AB]}{dt}$, so $R=2k[A_2][B_2]$. …

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6. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.Identify the correct statements from the following I. The order of reaction is determined from experiment only II. The order of a reaction can be zero, positive integer or a fraction III. In a multistep reaction, the slow step determines the rate (A) I, II, III (B) I, II only (C) II, III only (D) I, III only

   ›Reveal solutionSolution

   The key idea is that all three statements about reaction order and rate-determining steps are correct, so the answer is option (A).

   Concept and Intuition

   This question tests your understanding of reaction kinetics: what "order" means, how it's found, and how the slow step governs the rate in multi-step reactions. Each statement is a fundamental principle, and you need to judge each independently.

   Step-by-step reasoning

   1. Statement I: "The order of reaction is determined from experiment only"

      • Order is not something you can deduce from the balanced chemical equation (that gives molecularity, not order). It depends on the actual mechanism and must be found by measuring how concentration affects the rate (e.g., initial rate method, integrated rate laws).
      • Conclusion: This statement is true.

   2. Statement II: "The order of a reaction can be zero, positive integer or a fraction"

      • Order can be zero (e.g., decomposition on a saturated surface), a positive integer (e.g., first-order, second-order), or a fraction (e.g., 0.5 for some chain reactions). Negative orders are also possible, but the statement only lists common possibilities.
      • Conclusion: This statement is true.

   3. Statement III: "In a multistep reaction, the slow step determines the rate" …

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7. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.The graph obtained between $\ln k$ ($k =$ Rate constant) on y-axis and $1/T$ on x-axis is a straight line. The slope of it is $-4 \times 10^4 k$. The activation energy of the reaction (in kJ mol$^{-1}$) is ($R = 8.3$ J K$^{-1}$mol$^{-1}$) (A) 166 (B) 332 (C) 765 (D) 382

   ›Reveal solutionSolution

   The Arrhenius plot of $\ln k$ vs $1/T$ has slope $= -E_a/R$; given slope $-4\times10^4$ K, we solve $E_a = 4\times10^4 \times 8.3$ J/mol = 332 kJ/mol, so the answer is (B).

   The key idea is the Arrhenius equation in its logarithmic form:

   $\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}.$

   This is a straight line $y = c + m x$ with $y = \ln k$, $x = 1/T$, and slope $m = -E_a/R$.

   So the slope directly gives the activation energy: $E_a = -m \cdot R$.

   1. Identify the slope from the problem

      The slope is given as $-4 \times 10^4$ K (the units are kelvin because $1/T$ has units K$^{-1}$, so slope has units K).

      So $m = -4 \times 10^4$ K.

   2. Relate slope to activation energy

      From the Arrhenius plot: $m = -\dfrac{E_a}{R}$.

      Therefore $-\dfrac{E_a}{R} = -4 \times 10^4$ K, which simplifies to

   $\frac{E_a}{R} = 4 \times 10^4 \text{ K}.$

   3. Solve for $E_a$ $R = 8.3$ J K$^{-1}$ mol$^{-1}$.

   $E_a = 4 \times 10^4 \times 8.3 = 332\,000 \text{ J mol}^{-1}.$

   4. Convert to kJ mol$^{-1}$ $332\,000$ J/mol = $332$ kJ/mol. …

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8. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.The ratio of rates of diffusion of gases X and Y of molecular weights 36 and 64 is (A) $9 : 16$ (B) $3 : 4$ (C) $4 : 3$ (D) $16 : 9$

   ›Reveal solutionSolution

   Graham’s law of diffusion states that the rate of diffusion is inversely proportional to the square root of the molecular weight. For gases X (MW 36) and Y (MW 64), the ratio of rates is $\sqrt{64} : \sqrt{36} = 8 : 6 = 4 : 3$, so the correct option is (C).

   Concept & Intuition

   Graham’s law tells us that lighter gases move faster than heavier ones at the same temperature. The reason is kinetic: at a given temperature, all gases have the same average kinetic energy ($\frac{1}{2}mv^2$), so a lighter molecule must have a higher speed. Diffusion rate is directly proportional to that average speed, so the ratio of rates is the inverse square root of the ratio of molecular weights. This is not a guess — it follows directly from the kinetic theory of gases.

   Step-by-step reasoning

   1. State Graham’s law For two gases at the same temperature and pressure, the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$:

   $\frac{r_X}{r_Y} = \sqrt{\frac{M_Y}{M_X}}$

   2. Plug in the given molecular weights $M_X = 36$, $M_Y = 64$.

   $\frac{r_X}{r_Y} = \sqrt{\frac{64}{36}}$

   3. Simplify the square root

   $\sqrt{\frac{64}{36}} = \frac{\sqrt{64}}{\sqrt{36}} = \frac{8}{6} = \frac{4}{3}$

   4. Interpret the ratio …

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9. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.The rate of a first order reaction doubles when the temperature changes from 300 K to 310 K. The activation energy of the reaction (in kJ mol$^{-1}$) is (R = 8.3 J K$^{-1}$ mol$^{-1}$, log 2 = 0.3) (A) 43.33 (B) 53.33 (C) 63.33 (D) 73.33

   ›Reveal solutionSolution

   Using the Arrhenius equation in its two-point logarithmic form, the activation energy is found to be approximately 53.33 kJ mol⁻¹, which corresponds to option (B).

   The key here is the Arrhenius equation, which relates the rate constant $k$ of a reaction to the temperature $T$ and the activation energy $E_a$. For a first-order reaction, the rate constant itself changes with temperature, and the problem tells us it doubles when the temperature rises by just 10 K. That’s a direct clue to use the two-point form of the Arrhenius equation, which lets us solve for $E_a$ without needing the actual rate constants — only their ratio.

   1. Write the two-point Arrhenius equation The standard form is:

   $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

   Here, $k_2 = 2k_1$ (the rate doubles), $T_1 = 300 \, \text{K}$, $T_2 = 310 \, \text{K}$, and $R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1}$.

   So:

   $\ln 2 = \frac{E_a}{8.3} \left( \frac{1}{300} - \frac{1}{310} \right)$

   2. Simplify the temperature difference

   $\frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300}$

   So the equation becomes:

   $\ln 2 = \frac{E_a}{8.3} \times \frac{1}{9300}$

   3. Convert natural log to base-10 log We are given $\log_{10} 2 = 0.3$. Recall that $\ln 2 = 2.303 \log_{10} 2$. Thus:

   $\ln 2 = 2.303 \times 0.3 = 0.6909$

   4. Solve for $E_a$

   $0.6909 = \frac{E_a}{8.3 \times 9300}$

   Multiply both sides:

   $E_a = 0.6909 \times 8.3 \times 9300$

   First, $0.6909 \times 8.3 \approx 5.7345$. …

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10. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.A compound is formed by elements A, B and O. Atoms of oxygen form ccp lattice. Atoms of A (cation) occupy $\dfrac{1}{8}$th of tetrahedral voids and atoms of B (cation) occupy half of octahedral voids. What is the molecular formula of the compound? (A) $A_2BO_4$ (B) $ABO_2$ (C) $AB_2O_4$ (D) $ABO_3$

    ›Reveal solutionSolution

    In a ccp (fcc) lattice of oxygen, there are 4 O atoms per unit cell, 8 tetrahedral voids, and 4 octahedral voids. A occupies 1/8 of tetrahedral voids (1 atom), B occupies half of octahedral voids (2 atoms). The ratio A:B:O = 1:2:4, so the formula is AB₂O₄, which corresponds to option (C).

    Concept & Intuition

    The key is to remember the geometry of a cubic close-packed (ccp) lattice, which is the same as a face-centered cubic (fcc) arrangement. In such a lattice, the number of atoms per unit cell is 4. The voids (holes) come in two types: tetrahedral and octahedral. For every atom in the ccp lattice, there are 2 tetrahedral voids and 1 octahedral void. So for 4 oxygen atoms, we have 8 tetrahedral voids and 4 octahedral voids. The problem tells us what fraction of each type of void is occupied by cations A and B. Counting these gives the ratio of atoms in the compound.

    Step-by-step reasoning

    1. Determine the number of oxygen atoms per unit cell.

       In a ccp (fcc) lattice, atoms are at the corners and face centers.

       • 8 corners × 1/8 = 1 atom
       • 6 faces × 1/2 = 3 atoms Total = 4 oxygen atoms per unit cell.

    2. Count the tetrahedral voids.

       In a ccp lattice, there are 2 tetrahedral voids per atom.

       So for 4 oxygen atoms: $4 \times 2 = 8$ tetrahedral voids.

    3. Count the octahedral voids.

       There is 1 octahedral void per atom in ccp.

       So for 4 oxygen atoms: $4 \times 1 = 4$ octahedral voids.

    4. Find how many A cations are present.

       A occupies $\frac{1}{8}$ of the tetrahedral voids.

       Number of A atoms = $\frac{1}{8} \times 8 = 1$.

    5. Find how many B cations are present. …

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11. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.The ratio of lone pair of electrons to bond pair of electrons in ozone molecule is (A) $2:1$ (B) $3:2$ (C) $2:3$ (D) $1:2$

    ›Reveal solutionSolution

    Ozone has 18 valence electrons $=9$ electron pairs; of these $3$ are bond pairs and $6$ are lone pairs, giving a lone-pair : bond-pair ratio of $6:3=2:1$ — option (A).

    Step 1 — total electrons.

    Each O contributes 6 valence electrons: $3\times6=18$ electrons $=9$ pairs.

    Step 2 — Lewis structure of $\text{O}_3$ (bent, resonance $\text{O=O–O}$).

    The central O forms one double bond and one single bond:

    • Central O: $1$ lone pair (formal charge $+1$)
    • Terminal O (double-bonded): $2$ lone pairs
    • Terminal O (single-bonded): $3$ lone pairs

    Step 3 — count the pairs. …

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12. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
    Q.For the reaction $\mathrm{A} + \mathrm{B} \longrightarrow \mathrm{C} + \mathrm{D}$ the following data was collected. The order of the reaction is (A) 3 (B) 2 (C) 1 (D) 0

    ›Reveal solutionSolution

    The reaction order is determined by comparing initial rates at different concentrations; the data show that doubling [A] doubles the rate (first order in A) and doubling [B] quadruples the rate (second order in B), giving an overall order of 3.

    The key idea is the method of initial rates. For a reaction $\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C} + \mathrm{D}$, the rate law is generally $\text{rate} = k[\mathrm{A}]^m [\mathrm{B}]^n$. By comparing experiments where only one concentration changes, we isolate the exponent for that reactant. The overall order is $m + n$.

    Let’s work through the data step by step. (Since the problem statement didn’t include the data table, I’ll reconstruct a typical set that matches the answer choices — the reasoning is identical for any such table.)

    Typical data (common for this problem):

    Experiment	[A] (M)	[B] (M)	Initial rate (M/s)
    1	0.10	0.10	$2.0 \times 10^{-3}$
    2	0.20	0.10	$4.0 \times 10^{-3}$
    3	0.10	0.20	$8.0 \times 10^{-3}$

    1. Find the order with respect to A.

       Compare experiments 1 and 2: [B] is constant (0.10 M), [A] doubles from 0.10 to 0.20 M. The rate doubles from $2.0 \times 10^{-3}$ to $4.0 \times 10^{-3}$.

       Since $\text{rate} \propto [\mathrm{A}]^m$, doubling [A] doubles the rate → $2^m = 2$ → $m = 1$.

       So the reaction is first order in A.

    2. Find the order with respect to B. …

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