Showing the 12 most recent of 35 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$A(g) \rightarrow$ products, follows first order kinetics. Initial concentration of A is $8 \times 10^{-3} \; \mathrm{mol\,L^{-1}}$. For 80% decomposition of A, the time taken was 80 minutes. What is the rate constant for that reaction? (log 2 = 0.3, log 3 = 0.48, log 4 = 0.60, log 5 = 0.70) (A) 0.05 $min^{-1}$ (B) 0.04 $min^{-1}$ (C) 0.02 $min^{-1}$ (D) 0.06 $min^{-1}$

   ›Reveal solutionSolution

   This tests the integrated first-order rate law; the answer is $k\approx0.02\ min^{-1}$.

   Concept and Intuition

   For a first-order reaction, the rate constant only depends on the RATIO of initial to remaining concentration, not on the absolute concentration — this is why the initial concentration $8\times10^{-3}$ mol/L given in the question is a distractor and isn't actually needed.

   Step-by-Step Solution

   1. Integrated first-order equation: $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]_t}$.
   2. 80% decomposition means 80% of A is consumed, so 20% remains: $[A]_t = 0.2\,[A]_0$, giving $\dfrac{[A]_0}{[A]_t} = \dfrac{1}{0.2}=5$.
   3. Substitute $t=80$ min and $\log 5 = 0.70$ (given): $k = \dfrac{2.303}{80}\times0.70 = \dfrac{1.6121}{80}$.
   4. $k \approx 0.02015\ \text{min}^{-1} \approx 0.02\ \text{min}^{-1}$.

   Common Mistakes …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.At $T(K)$, the following first order reaction takes place $A(g) \rightarrow B(g) + C(g)$ Initial pressure ($t = 0$ min) of $A(g)$ is $p_A^0$. After 10 min of the reaction, the pressure is $p_t$. The correct equation for the rate constant of this reaction is (A) $k = \dfrac{1}{t}\ln\dfrac{p_A^0}{(2p_A^0 - p_t)}$ (B) $k = \dfrac{2.303}{t}\ln\dfrac{p_A^0}{(2p_A^0 - p_t)}$ (C) $k = \dfrac{2.303}{t}\log\dfrac{(2p_A^0 - p_t)}{p_A^0}$ (D) $k = \dfrac{1}{t}\ln\dfrac{p_A^0}{(3p_A^0 - p_t)}$

   ›Reveal solutionSolution

   Tracking total pressure for $A\rightarrow B+C$ shows the pressure of remaining A at time $t$ is $2p_A^0-p_t$; substituting into the first-order rate law gives option (A).

   Concept and Intuition

   When a gas-phase reaction changes the number of moles, we can follow its progress via total pressure instead of concentration, since pressure is proportional to moles at constant volume/temperature. For $A(g) \rightarrow B(g) + C(g)$, every mole of A consumed produces one mole each of B and C — so the total moles (and hence total pressure) increase as the reaction proceeds, and this pressure change directly tells us how much A has reacted.

   Step-by-Step Solution

   1. At $t=0$: only A present, pressure $= p_A^0$.
   2. At time $t$: let the pressure of A remaining be $p$. Then pressure of A reacted $= p_A^0 - p$, and by stoichiometry, pressure of B formed $= p_A^0-p$ and pressure of C formed $=p_A^0-p$ (1:1:1 ratio).
   3. Total pressure: $p_t = p + (p_A^0-p) + (p_A^0-p) = 2p_A^0 - p$
   4. Solve for $p$ (pressure of A remaining): $p = 2p_A^0 - p_t$ …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Half-life of a first order reaction is 10 minutes. What is the rate of reaction after 20 minutes, if the initial concentration of the reactant is 10 M? (A) $1.73 \times 10^{-1}\ M\ min^{-1}$ (B) $1.73 \times 10^{-2}\ M\ min^{-1}$ (C) $3.46 \times 10^{-1}\ M\ min^{-1}$ (D) $4.19 \times 10^{-2}\ M\ min^{-1}$

   ›Reveal solutionSolution

   After 2 half-lives the concentration drops to 2.5 M; the instantaneous first-order rate at that point is $k[A] = 1.73\times10^{-1}$ M min⁻¹.

   Concept and Intuition

   For a first-order reaction, the rate constant is fixed by the half-life: $k = 0.693/t_{1/2}$, independent of concentration. The instantaneous rate at any time is $\text{rate} = k[A]_t$, so we just need $[A]$ at $t=20$ min.

   Since 20 minutes equals exactly 2 half-lives (10 min each), the concentration halves twice.

   Step-by-Step Solution

   1. $k = 0.693/10 = 0.0693\ \text{min}^{-1}$.
   2. Number of half-lives elapsed in 20 min $= 20/10 = 2$.
   3. $[A]_{20} = \dfrac{[A]_0}{2^2} = \dfrac{10}{4} = 2.5$ M. …

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4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Hydrolysis of benzene diazonium chloride follows first order kinetics. The time taken for its decomposition to $\frac{1}{8}$ and $\frac{1}{10}$ of its original concentration are $[t_{1/8}]$ and $[t_{1/10}]$ respectively. What is the ratio of $[t_{1/8}]$ to $[t_{1/10}]$? [log 2 = 0.30, log 3 = 0.48, log 4 = 0.60] (A) 9 : 10 (B) 10 : 9 (C) 3 : 5 (D) 5 : 3

   ›Reveal solutionSolution

   For a first-order reaction, time to reach a given fraction remaining is proportional to $\log(C_0/C)$; comparing $\log 8$ and $\log 10$ gives the required ratio. The answer is 9 : 10.

   Concept and Intuition

   For a first-order reaction, $k = \dfrac{2.303}{t}\log\dfrac{C_0}{C}$, so $t = \dfrac{2.303}{k}\log\dfrac{C_0}{C}$. Since $k$ (the rate constant) is the same throughout for a given reaction at a given temperature, the time taken to reach any particular fraction of the original concentration is directly proportional to $\log(C_0/C)$ for that fraction. This lets us compare two different "fraction remaining" times without ever needing to know $k$.

   Step-by-Step Solution

   1. $t_{1/8}$ is the time for concentration to fall to $1/8$ of $C_0$, i.e., $C_0/C = 8$: $t_{1/8} = \dfrac{2.303}{k}\log 8$.
   2. $\log 8 = \log(2^3) = 3\log 2 = 3 \times 0.30 = 0.90$. …

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5. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.The time taken for 60% completion of a first order reaction is 13.22 min. What is its half-life ($t_{1/2}$) in min? ($\log(2.5) = 0.398$) (A) 11 (B) 9 (C) 8 (D) 10

   ›Reveal solutionSolution

   Using the first-order rate law with the given time for 60% completion, the half-life works out to 10 minutes.

   Concept and Intuition

   For first-order kinetics, the rate constant $k$ can be found from any fraction reacted using $k=\frac{2.303}{t}\log\frac{[A]_0}{[A]_t}$, and the half-life $t_{1/2}=\frac{0.693}{k}$ is independent of initial concentration — it only depends on $k$.

   Step-by-Step Solution

   1. At 60% completion, 40% of the reactant remains: $\dfrac{[A]_0}{[A]_t} = \dfrac{100}{40} = 2.5$.
   2. $k = \dfrac{2.303}{t}\log(2.5) = \dfrac{2.303\times0.398}{13.22}$.
   3. Numerator $= 2.303\times0.398 = 0.9166$; so $k = \dfrac{0.9166}{13.22} = 0.0693\ \text{min}^{-1}$. …

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6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.A → products, is a first order reaction. The following data is obtained for this reaction at T(K). The value of x : y is Rate (mol L$^{-1}$ min$^{-1}$) : [A] 0.2 : 0.02 M 0.4 : x M 1.0 : y M (A) 1 : 5 (B) 2 : 3 (C) 5 : 2 (D) 2 : 5

   ›Reveal solutionSolution

   Tests the defining rate law of a first-order reaction, Rate $=k[A]$; the answer is (D) 2 : 5.

   Concept and Intuition

   For a first-order reaction $A \to \text{products}$, the rate law is

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

   This means rate is directly proportional to concentration — double the concentration, double the rate. Since $k$ is a constant at a fixed temperature $T$, every (Rate, [A]) pair in the table must give the same $k$. This lets us solve for the unknown concentrations $x$ and $y$ from the first data row.

   Step-by-Step Solution

   1. Use the first row to find $k$: $0.2 = k(0.02) \Rightarrow k = \dfrac{0.2}{0.02} = 10\ \text{min}^{-1}$.
   2. Apply the same $k$ to row 2: $0.4 = k\cdot x = 10x \Rightarrow x = 0.04$ M.
   3. Apply the same $k$ to row 3: $1.0 = k\cdot y = 10y \Rightarrow y = 0.10$ M. …

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7. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.In a first order reaction, the concentration of the reactant is reduced to 1/8 of the initial concentration in 75 minutes. The $t_{1/2}$ of the reaction (in minutes) is ($\log 2 = 0.30$, $\log 3 = 0.47$, $\log 4 = 0.60$) (A) 60.2 (B) 50.2 (C) 25.1 (D) 75.1

   ›Reveal solutionSolution

   Falling to $1/8$ of the initial concentration is exactly 3 half-lives (since $1/8 = (1/2)^3$), so $t_{1/2} = 75/3 = 25$ min; using the given log values precisely gives 25.1 min. The answer is (C) 25.1 min.

   Concept and Intuition

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

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

   A defining feature of first-order kinetics is that the half-life $t_{1/2} = 0.693/k$ is independent of concentration — each half-life always halves whatever concentration remains. So if the concentration falls to $\tfrac18$ of the original, that is exactly $\left(\tfrac12\right)^3$, meaning exactly 3 half-lives have elapsed.

   Step-by-Step Solution

   1. Concentration ratio: $[A]_0/[A] = 8$.
   2. Since $8 = 2^3$, note $1/8$ of initial concentration corresponds to exactly 3 half-lives.
   3. Total time for 3 half-lives = 75 minutes, so one half-life $t_{1/2} = 75/3 = 25$ minutes (quick estimate).
   4. To match the precision implied by the given log values, compute via the rate constant: …

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8. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.$A \to P$ is a first order reaction. At T(K), the concentration of reactant (A) after 10 min of the reaction is $x\ mol\,L^{-1}$. After 20 min of the reaction, the concentration of A was $y\ mol\,L^{-1}$. What is its rate constant (in $min^{-1}$) ? (A) $0.2303\log\dfrac{x}{y}$ (B) $2.303\log\dfrac{x}{y}$ (C) $2.303\log\dfrac{y}{x}$ (D) $0.2303\log\dfrac{y}{x}$

   ›Reveal solutionSolution

   Using the first-order integrated rate law between $t=10$ min ($[A]=x$) and $t=20$ min ($[A]=y$), with $\Delta t=10$ min, gives $k = 0.2303\log(x/y)$.

   Concept and Intuition

   For a first-order reaction $A\to P$, the integrated rate law between any two times $t_1$ and $t_2$ (with concentrations $C_1$ and $C_2$) is

   $k = \frac{2.303}{t_2-t_1}\log\frac{C_1}{C_2}$

   This works between any two time points, not just from $t=0$, as long as the concentrations at those two times are known.

   Step-by-Step Solution

   1. At $t_1=10$ min, $[A] = x$. At $t_2=20$ min, $[A] = y$.
   2. Time interval: $\Delta t = t_2-t_1 = 10$ min.
   3. $k = \dfrac{2.303}{\Delta t}\log\dfrac{C_1}{C_2} = \dfrac{2.303}{10}\log\dfrac{x}{y}$.
   4. $\dfrac{2.303}{10} = 0.2303$.
   5. So $k = 0.2303\log\dfrac{x}{y}$ (in $\text{min}^{-1}$).

   Common Mistakes …

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9. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.$A \rightarrow P$ is a first order reaction. The reaction was started at 10.00 AM. At 10.10 AM, the concentration of A was x mol L$^{-1}$. At 10.20 AM, the concentration of A was y mol L$^{-1}$. The half life (in min) of the reaction is equal to (A) $\dfrac{2.303}{\log(x/y)}$ (B) $\dfrac{\log(x/y)}{3.01}$ (C) $\dfrac{3.01}{\log(y/x)}$ (D) $\dfrac{3.01}{\log(x/y)}$

   ›Reveal solutionSolution

   Since $x$ and $y$ are concentrations 10 minutes apart in a first-order reaction, the rate constant follows from the integrated rate law over that interval, giving $t_{1/2} = \dfrac{3.01}{\log(x/y)}$.

   Concept and Intuition

   For a first-order reaction, the integrated rate law between any two times separated by $\Delta t$ is

   $k = \frac{2.303}{\Delta t}\log\frac{[A]_{t_1}}{[A]_{t_2}}$

   regardless of what the initial concentration was — because first-order kinetics only cares about the ratio of concentrations over the elapsed time, not the absolute starting point. This is why concentrations "10 minutes into the reaction" and "20 minutes into the reaction" can be treated as a self-contained 10-minute window.

   Step-by-Step Solution

   1. From 10.00 AM to 10.10 AM to 10.20 AM, the interval between the two given concentrations is $\Delta t = 10$ min.
   2. Apply the first-order integrated law over this window: $k = \dfrac{2.303}{10}\log\dfrac{x}{y}$.
   3. Half-life: $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693 \times 10}{2.303\log(x/y)}$. …

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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.For a first order reaction, the ratio between the time taken to complete $\dfrac{3}{4}$th of the reaction and time taken to complete half of the reaction is (A) 2 (B) 3 (C) 1.5 (D) 2.5

    ›Reveal solutionSolution

    This tests the first-order rate law's independence from concentration; the ratio $t_{3/4}/t_{1/2}=2$ exactly.

    Concept and Intuition

    For a first-order reaction, the time to reach any fixed fraction of completion is a fixed multiple of the half-life, regardless of the starting concentration — this is the defining signature of first-order kinetics. Doubling the 'half-life count' needed (from one half-life to two half-lives) to go from 50% to 75% completion is exactly why the ratio comes out to a clean integer.

    Step-by-Step Solution

    1. First-order integrated law: $kt = \ln\dfrac{[A]_0}{[A]_t}$.
    2. At $t_{1/2}$: half of $[A]_0$ remains, so $kt_{1/2} = \ln\dfrac{[A]_0}{[A]_0/2} = \ln 2$.
    3. At $t_{3/4}$: three-quarters consumed means one-quarter remains, so $kt_{3/4} = \ln\dfrac{[A]_0}{[A]_0/4} = \ln 4 = 2\ln 2$. …

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11. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
    Q.$A \rightarrow P$ is a first order reaction. At 27 °C, the time taken for the completion of 20 % of the reaction is $t_1$ min. The time taken for the completion of 80 % of the reaction is $t_2$ min at the same temperature. What is the value of $\dfrac{t_2}{t_1}$? ($\log 80 = 1.9$; $\log 20 = 1.3$) (A) $\dfrac{1}{7}$ (B) 7 (C) $\dfrac{3}{7}$ (D) 14

    ›Reveal solutionSolution

    Using the first-order rate law at 20% and 80% completion, the ratio $t_2/t_1$ reduces to $\log5/\log1.25$, which evaluates to 7 using the given log values.

    Concept and Intuition

    For a first-order reaction, the time to reach a given fraction of completion depends logarithmically on the fraction of reactant remaining: $t = \dfrac{2.303}{k}\log\dfrac{a}{a-x}$. Taking the ratio of two such times eliminates the unknown rate constant $k$.

    Step-by-Step Solution

    1. At 20% completion, 80% of $a$ remains: $t_1 = \dfrac{2.303}{k}\log\dfrac{100}{80} = \dfrac{2.303}{k}\log(1.25)$.
    2. At 80% completion, 20% of $a$ remains: $t_2 = \dfrac{2.303}{k}\log\dfrac{100}{20} = \dfrac{2.303}{k}\log(5)$.
    3. Ratio: $\dfrac{t_2}{t_1} = \dfrac{\log 5}{\log 1.25}$. …

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12. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
    Q.At T(K), the decomposition of $\mathrm{N_2O_5(g)}$ is a first order reaction. The initial pressure of $\mathrm{N_2O_5(g)}$ is 'a' atm. After time, t, the total pressure of reaction is 'p' atm. The rate constant (k) of the reaction is (A) $K = \frac{1}{t}\ln\left(\frac{a}{a-2p}\right)$ (B) $K = \frac{1}{t}\ln\left(\frac{3a}{3a-2p}\right)$ (C) $K = \frac{1}{t}\ln\left(\frac{3a}{3a-p}\right)$ (D) $K = \frac{1}{2}\ln\left(\frac{3a}{5a-2p}\right)$

    ›Reveal solutionSolution

    Tracking total pressure through the stoichiometry of

    $2\mathrm{N_2O_5}\rightarrow4\mathrm{NO_2}+\mathrm{O_2}$ shows the remaining

    $\mathrm{N_2O_5}$ pressure is $\frac{5a-2p}{3}$, giving the first-order rate constant

    the form in option (D).

    Concept and Intuition

    This is the standard "gas-phase first-order decomposition tracked by total pressure"

    problem. Because the number of moles of gas changes during the reaction, the total

    pressure $p$ at time $t$ is NOT simply the pressure of unreacted $\mathrm{N_2O_5}$ — you

    must use the reaction stoichiometry to relate the increase in total pressure to how

    much $\mathrm{N_2O_5}$ has actually decomposed, then plug the leftover $\mathrm{N_2O_5}$

    pressure into the first-order integrated rate law $k=\frac1t\ln\frac{[A]_0}{[A]_t}$.

    Step-by-Step Solution

    1. Reaction: $2\mathrm{N_2O_5}(g)\rightarrow4\mathrm{NO_2}(g)+\mathrm{O_2}(g)$.
    2. Let $2\xi$ = pressure of $\mathrm{N_2O_5}$ consumed. By stoichiometry, NO$_2$ formed $=4\xi$ and O$_2$ formed $=\xi$.
    3. Species pressures at time $t$: $\mathrm{N_2O_5}=a-2\xi$, $\mathrm{NO_2}=4\xi$, $\mathrm{O_2}=\xi$.
    4. Total pressure: $p=(a-2\xi)+4\xi+\xi=a+3\xi \Rightarrow \xi=\dfrac{p-a}{3}$.
    5. Remaining $\mathrm{N_2O_5}=a-2\xi=a-\dfrac{2(p-a)}{3}=\dfrac{3a-2p+2a}{3}=\dfrac{5a-2p}{3}$. …

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