Q.Calculate the half-life of a first order reaction from their rate constants given below:

Showing the 12 most recent of 15 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.The initial pressure of the system before decomposition for a first order gas phase reaction $A_{(g)} \rightarrow B_{(g)}+C_{(g)}$ was $\mathrm{P}_{\mathrm{i}}$. After lapse of time ' t ', total pressure of the system increased by ' x ' units and became $\mathrm{P}_{\mathrm{t}}$. The rate constant ' k ' for the reaction is given as: (A) $k=\frac{2.303}{t} \log \frac{P_i}{P_i+x}$ (B) $k=\frac{2.303}{t} \log \frac{P_i}{2 P_i-P_t}$ (C) $k=\frac{2.303}{t} \log \frac{P_i}{2 P_i+P_t}$ (D) $k=\frac{2.303}{t} \log \frac{P_i}{P_i-P_t}$

   ›Reveal solutionSolution

   For a first-order gas-phase reaction where one molecule splits into two, the total pressure increases as the reaction proceeds. The key is to relate the partial pressure of the reactant at time $t$ to the total pressure, then substitute into the first-order rate law. The correct expression is $k = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}$, which corresponds to option (B).

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   Concept & Intuition

   In a first-order reaction, the rate depends only on the concentration (or partial pressure) of one reactant. Here, $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$ starts with pure A at initial pressure $P_i$. As A decomposes, it produces two gas molecules, so the total number of moles increases — and therefore the total pressure rises. The increase in total pressure tells us how much A has reacted. If we can express the remaining pressure of A in terms of the measured total pressure $P_t$, we can plug that into the first-order integrated rate law.

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   Step-by-step derivation

   1. Set up the initial and changing conditions Initially, only A is present at pressure $P_i$. Let the partial pressure of A that has reacted by time $t$ be $p$. Then:

   $\begin{array}{c|ccc} & A & B & C \\ \hline \text{Initial} & P_i & 0 & 0 \\ \text{At time } t & P_i - p & p & p \end{array}$

   The total pressure at time $t$ is:

   $P_t = (P_i - p) + p + p = P_i + p$

   So the increase in total pressure, $x = P_t - P_i$, equals $p$, the amount of A that has reacted.

   2. Express the remaining pressure of A The partial pressure of A at time $t$ is:

   $P_A = P_i - p = P_i - (P_t - P_i) = 2P_i - P_t$

   This is the key relation: the reactant’s pressure at time $t$ is $2P_i - P_t$.

   3. Apply the first-order rate law …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.$60 \%$ of a first order reaction was completed in 60 min , then $50 \%$ of the same reaction can be completed in: $[\log 4=0.60, \log 5=0.69]$ (A) 60 min (B) 65 min (C) 50 min (D) 45 min

   ›Reveal solutionSolution

   For a first‑order reaction, the time to reach a given fraction depends only on the rate constant. Using the 60% completion data we find $k$, then compute the half‑life. The answer is 45 min, option (D).

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   Concept & Intuition

   First‑order reactions have a constant half‑life: the time to go from any concentration to half of that is always the same. The integrated rate law is

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

   so the time to reach a certain fraction $f$ completed (i.e., $[A] = (1-f)[A]_0$) is

   $t = \frac{1}{k}\,\ln\frac{1}{1-f}.$

   We are given that $f=0.60$ takes 60 min. That lets us find $k$. Then we want the time for $f=0.50$ (the half‑life). No need to compute $k$ explicitly — we can take a ratio.

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   Step‑by‑step reasoning

   1. Write the time for 60% completion For a first‑order reaction,

   $t_{60} = \frac{1}{k}\,\ln\frac{1}{1-0.60} = \frac{1}{k}\,\ln\frac{1}{0.40} = \frac{1}{k}\,\ln(2.5).$

   Given $t_{60}=60$ min, we have

   $60 = \frac{1}{k}\,\ln(2.5).$

   2. Express the half‑life The half‑life $t_{1/2}$ corresponds to $f=0.50$:

   $t_{1/2} = \frac{1}{k}\,\ln\frac{1}{0.50} = \frac{1}{k}\,\ln 2.$

   3. Take the ratio to eliminate $k$

   $\frac{t_{1/2}}{60} = \frac{\ln 2}{\ln(2.5)}.$

   So

   $t_{1/2} = 60 \times \frac{\ln 2}{\ln(2.5)}.$

   4. Convert to base‑10 logs using given values Recall $\ln x = 2.303 \log_{10} x$. The factor $2.303$ cancels in the ratio: $\frac{\ln 2}{\ln(2.5)} = \frac{\log 2}{\log(2.5)}.$ …

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.The time needed for completion of $80 \%$ is $y$ times the half-life period of a first order reaction. What is the value of $y$ ? (A) 0.648 (B) 3.46 (C) 2.32 (D) 0.322

   ›Reveal solutionSolution

   For a first‑order reaction, the time to reach 80% completion is about 2.32 times the half‑life, so the answer is (C).

   Concept & Intuition

   First‑order kinetics follow an exponential decay: the fraction remaining after time $t$ is $e^{-kt}$. The half‑life $t_{1/2}$ is the time when half remains, giving $k = \frac{\ln 2}{t_{1/2}}$. For 80% completion, 20% remains. We set up the decay equation, solve for $t_{80\%}$, and then find the ratio $y = t_{80\%} / t_{1/2}$. The key is that the ratio depends only on the logarithms of the fractions, not on the actual rate constant.

   Step‑by‑step reasoning

   1. Write the first‑order decay law For a first‑order reaction, the amount remaining after time $t$ is

   $[A] = [A]_0 e^{-kt}$

   where $k$ is the rate constant.

   2. Relate the half‑life to $k$ At $t = t_{1/2}$, $[A] = \frac{1}{2}[A]_0$. So

   $\frac{1}{2} = e^{-k t_{1/2}} \quad\Rightarrow\quad k t_{1/2} = \ln 2 \quad\Rightarrow\quad k = \frac{\ln 2}{t_{1/2}}.$

   3. Set up the condition for 80% completion 80% completion means 20% remains:

   $\frac{[A]}{[A]_0} = 0.20 = e^{-k t_{80\%}}.$

   4. Solve for $t_{80\%}$ Take natural logs:

   $\ln(0.20) = -k t_{80\%} \quad\Rightarrow\quad t_{80\%} = -\frac{\ln(0.20)}{k}.$

   Substitute $k = \frac{\ln 2}{t_{1/2}}$:

   $t_{80\%} = -\frac{\ln(0.20)}{\ln 2} \cdot t_{1/2}.$

   5. Compute the ratio $y = t_{80\%} / t_{1/2}$ $y = -\frac{\ln(0.20)}{\ln 2}.$ …

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4. COMEDK 2025Set 2025-E1 markMCQ
   Q.The decomposition of $\mathrm{PH}_3$ follows first order kinetics. The time required for $3 / 4^{\text {th }}$ of $\mathrm{PH}_3$ to decompose is 75.76 s . Calculate the fraction of original amount of $\mathrm{PH}_3$ which will remain after 2.0 minutes. (A) 0.354 (B) 0.111 (C) 0.954 (D) 0.898

   ›Reveal solutionSolution

   For a first‑order reaction, the fraction remaining after time $t$ is $e^{-kt}$. Using the given time for $3/4$ decomposition to find $k$, then evaluating at $t = 120\ \text{s}$ gives fraction $\approx 0.111$, which corresponds to option (B).

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   Concept & Intuition

   First‑order kinetics means the rate of decomposition is proportional to the amount present. This leads to an exponential decay:

   $N(t) = N_0 e^{-kt}$

   where $N_0$ is the initial amount, $N(t)$ the amount remaining after time $t$, and $k$ the rate constant.

   The key idea: if we know how long it takes for a certain fraction to decompose, we can solve for $k$. Then we can predict the fraction remaining after any other time.

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   Step‑by‑step solution

   1. Interpret the given data “Time required for $3/4$ of $\mathrm{PH}_3$ to decompose” means that after 75.76 s, only $1/4$ of the original amount remains. So:

   $\frac{N(75.76)}{N_0} = \frac{1}{4}$

   2. Write the first‑order decay equation

   $\frac{N(t)}{N_0} = e^{-kt}$

   For $t = 75.76\ \text{s}$:

   $e^{-k \cdot 75.76} = \frac{1}{4}$

   3. Solve for the rate constant $k$ Take natural logs:

   $-k \cdot 75.76 = \ln\!\left(\frac{1}{4}\right) = -\ln 4$

   So:

   $k = \frac{\ln 4}{75.76}$

   Numerically, $\ln 4 \approx 1.38629$, thus:

   $k \approx \frac{1.38629}{75.76} \approx 0.01830\ \text{s}^{-1}$

   4. Find the fraction remaining after 2.0 minutes 2.0 minutes = 120 s.

   $\frac{N(120)}{N_0} = e^{-k \cdot 120} = e^{-0.01830 \times 120}$

   Compute exponent: $0.01830 \times 120 = 2.196$.

   So:

   $$\frac{N(120)}{N_0} = e^{-2.196} \approx 0.111 …$$

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.Choose the correct statement. (A) The plot of $[A]$ versus time " $t$ " for a zero order reaction is a horizontal line parallel to the $x$-axis which represents " $t$ " (B) For a first order reaction the time taken for $3 / 4^{\text {th }}$ completion is 2 times the half-life period. (C) For a second order reaction the rate of reaction with respect to reactant becomes 27 times when its concentration is tripled. (D) Molecularity of a chemical reaction is always greater than order of the reaction.

   ›Reveal solutionSolution

   The key is to recall the integrated rate laws and definitions for zero, first, and second order reactions, and the distinction between molecularity and order. Only statement (B) is correct: for a first order reaction, the time for 3/4 completion is exactly twice the half-life.

   Concept & Intuition

   Each statement tests a different core idea from chemical kinetics. We must check each against the standard mathematical forms. The half-life for a first order reaction is constant, so the time to go from start to half is one half-life; to go from half to three-quarters is another half-life — hence two half-lives total. The other statements contain common misconceptions: zero order plots are linear with negative slope, not horizontal; second order rate depends on concentration squared, so tripling gives a 9-fold increase, not 27; molecularity and order are unrelated in magnitude.

   Step-by-step verification

   1. Statement (A): For a zero order reaction, the integrated rate law is $[A] = [A]_0 - kt$. A plot of $[A]$ vs. $t$ is a straight line with slope $-k$, not a horizontal line. A horizontal line would mean $[A]$ constant, which implies no reaction. So (A) is false.

   2. Statement (B): For a first order reaction, $t = \frac{1}{k} \ln \frac{[A]_0}{[A]}$.

      • Half-life: $t_{1/2} = \frac{\ln 2}{k}$.
      • Time for $3/4$ completion means $[A] = \frac{1}{4}[A]_0$, so

   $t_{3/4} = \frac{1}{k} \ln \frac{[A]_0}{[A]_0/4} = \frac{1}{k} \ln 4 = \frac{2\ln 2}{k} = 2 \cdot \frac{\ln 2}{k} = 2 t_{1/2}.$

    Hence (B) is correct. …
   

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.$\mathrm{X} \rightarrow 2 \mathrm{Y}$ is a first order reaction where $1.0 \mathrm{~mol} / \mathrm{L}$ of the reactant yields $0.4 \mathrm{~mol} / \mathrm{L}$ of Y in 200 minutes. Calculate the half-life period of the reaction in minutes. (A) 151.24 (B) 203.69 (C) 620.96 (D) 271.34

   ›Reveal solutionSolution

   For a first‑order reaction, the half‑life is constant and can be found from the integrated rate law using the given concentration change. The half‑life works out to about 620.96 minutes, so the correct option is (C).

   Concept & Intuition

   The reaction $\mathrm{X} \rightarrow 2\mathrm{Y}$ is first order in X. That means the rate depends only on [X], and the time for half of X to react is always the same, no matter how much you start with.

   We are told: initial [X] = 1.0 mol/L, and after 200 minutes, [Y] = 0.4 mol/L. Because each X that reacts produces two Y molecules, the amount of X consumed is half the amount of Y produced. So we can find how much X is left, then use the first‑order integrated rate law to find the rate constant $k$, and finally the half‑life $t_{1/2} = \frac{\ln 2}{k}$.

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   Step‑by‑step reasoning

   1. Relate [Y] produced to [X] consumed The stoichiometry: 1 mol X → 2 mol Y. If [Y] formed = 0.4 mol/L, then moles of X that reacted = $\frac{0.4}{2} = 0.2$ mol/L. So the concentration of X remaining after 200 min is:

   $[X] = 1.0 - 0.2 = 0.8 \ \text{mol/L}.$

   2. Apply the first‑order integrated rate law For a first‑order reaction:

   $\ln \frac{[X]_0}{[X]} = k t$

   Here $[X]_0 = 1.0$, $[X] = 0.8$, $t = 200$ min.

   $\ln \frac{1.0}{0.8} = k \times 200$

   $\ln(1.25) = 200k$

   $k = \frac{\ln(1.25)}{200}$

   3. Compute $k$ numerically $\ln(1.25) \approx 0.22314$

   $k \approx \frac{0.22314}{200} = 0.0011157 \ \text{min}^{-1}$

   4. Find the half‑life For a first‑order reaction:

   $t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693147}{0.0011157} \approx 621.3 \ \text{minutes}$

   More precisely, using exact values:

   $$$$

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.Given below a first order reaction in the gas phase $\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ If the initial pressure of the system is $\mathrm{P}_{\mathrm{i}}$ and the total pressure at $\mathrm{t}$ seconds is $\mathrm{P}_{\mathrm{t}}$, the rate constant $\mathrm{k}$ for the reaction is: (A) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{t}}}$ (B) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}$ (C) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}+\mathrm{x}\right)}$ (D) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_{\mathrm{i}}}{\left(\mathrm{P}_{\mathrm{i}}+\mathrm{x}\right)}$

   ›Reveal solutionSolution

   For a first-order gas-phase reaction where one molecule splits into two, the total pressure changes in a simple way; the correct rate constant expression is $k = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}$, which corresponds to option (B).

   The key idea here is that for a first-order reaction, the rate constant depends on the concentration (or partial pressure) of the reactant at time $t$, not on the total pressure directly. Since the reaction produces more gas molecules, the total pressure increases as the reaction proceeds. We need to relate the reactant’s partial pressure to the measured total pressure.

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   Why this approach works

   In a first-order reaction, the integrated rate law is:

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

   For gases at constant volume and temperature, pressure is proportional to concentration (via $P = nRT/V$). So we can replace concentrations with partial pressures of A:

   $k = \frac{2.303}{t} \log \frac{P_{A,0}}{P_{A,t}}$

   We know the initial total pressure $P_i$ is just the initial pressure of A (since no B or C yet). But at time $t$, the total pressure $P_t$ includes contributions from A, B, and C. We must express $P_{A,t}$ in terms of $P_i$ and $P_t$.

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   Step-by-step reasoning

   1. Set up the reaction and initial conditions The reaction is:

   $\mathrm{A(g)} \rightarrow \mathrm{B(g)} + \mathrm{C(g)}$

   Initially, only A is present. Let the initial pressure of A be $P_i$. So:

   $P_{A,0} = P_i, \quad P_{B,0} = 0, \quad P_{C,0} = 0$

   2. Define the extent of reaction in terms of pressure Suppose at time $t$, a fraction of A has reacted. Let the decrease in pressure of A be $x$. Then:

   $P_{A,t} = P_i - x$

   Since 1 mole of A produces 1 mole of B and 1 mole of C, the pressure of B formed is $x$ and the pressure of C formed is also $x$. So:

   $P_{B,t} = x, \quad P_{C,t} = x$

   3. Express total pressure at time $t$ The total pressure is the sum of partial pressures:

   $P_t = P_{A,t} + P_{B,t} + P_{C,t} = (P_i - x) + x + x = P_i + x$

   Therefore:

   $x = P_t - P_i$ …

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8. COMEDK 2024Set 2024-M1 markMCQ
   Q.The time required for $80 \%$ of a first order reaction is "$y$" times the half-life period of the same reaction. What is the value of "$y$"? (A) 3.22 (B) 2.32 (C) 0.322 (D) 2.96

   ›Reveal solutionSolution

   For a first-order reaction, $y = t_{80\%}/t_{1/2} = \ln5/\ln2 \approx 2.32$ — option (B).

   Step-by-step reasoning

   1. Half-life. From $\ln\dfrac{[A]_0}{[A]_t}=kt$, at $t=t_{1/2}$, $[A]_t=\tfrac12[A]_0$:

   $t_{1/2} = \frac{\ln 2}{k}$

   2. Time for 80% completion. At 80% reacted, 20% remains, $[A]_t = 0.20[A]_0$:

   $t_{80\%} = \frac{\ln(1/0.20)}{k} = \frac{\ln 5}{k}$

   3. Ratio. $y = \frac{t_{80\%}}{t_{1/2}} = \frac{\ln 5}{\ln 2} = \frac{1.6094}{0.6931} \approx 2.32$ …

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9. COMEDK 2023Set 2023-E1 markMCQ
   Q.A first order reaction proceeds to $90 \%$ completion. What will be the approximate time taken for $90 \%$ completion in relation to $t_{1 / 2}$ of the reaction? (A) 5.02 times of $\mathrm{t}_{1 / 2}$ (B) 4.54 times of $\mathrm{t}_{1 / 2}$ (C) 5.66 times of $\mathrm{t}_{1 / 2}$ (D) 3.32 times of $\mathrm{t}_{1 / 2}$

   ›Reveal solutionSolution

   So 90% completion takes about 3.32 times the half-life. (Sanity check: 90% completion is between 3 half-lives, 87.5%, and 4 half-lives, 93.75% - so ~3.3 is right.)

   Concept: first-order kinetics, t = (2.303/k) log([A0]/[A]).

   Time for 90% completion ([A] = 0.1[A0]):

   t_90 = (2.303/k) log(10) = 2.303/k

   Half-life:

   t_1/2 = 0.693/k

   Ratio:

   t_90 / t_1/2 = 2.303 / 0.693 = 3.32 …

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10. COMEDK 2023Set 2023-E1 markMCQ
    Q.The rate constant for a First order reaction at $560 \mathrm{~K}$ is $1.5 \times 10^{-6}$ per second. If the reaction is allowed to take place for 20 hours, what percentage of the initial concentration would have converted to products? (A) 11.14 (B) 10.23 (C) 12.46 (D) 21.2

    ›Reveal solutionSolution

    First-order kinetics: the fraction converted after time $t$ is $1-e^{-kt}$. Here $kt=0.108$, giving $\approx 10.2\%$ converted.

    Time in seconds:

    $t = 20\ \text{h} \times 3600 = 72000\ \text{s}$

    Exponent:

    $kt = (1.5\times10^{-6})(72000) = 0.108$

    Fraction remaining and converted (first order, $[A]=[A]_0 e^{-kt}$): …

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11. COMEDK 2022Set 20221 markMCQ
    Q.For the reaction 2N$_2$O$_5$ $\to$ 4NO$_2$ + O$_2$. If initial pressure is 100 atm and rate constant $k$ is 3.38 $\times$ 10$^{-5}$ s$^{-1}$. after 20 min the final pressure of N$_2$O$_5$ will be (A) 96 atm (B) 50 atm (C) 70 atm (D) 60 atm

    ›Reveal solutionSolution

    (Equivalently with the log form: log(100/p) = kt/2.303 = 0.01761 -> 100/p = 1.0414 -> p = 96.0 atm.)

    Concept: First-order kinetics (the given k has units s^-1, so the decomposition of N2O5 is first order in N2O5):

    p = p0 * e^(-kt) (pressure is proportional to concentration)

    Data: p0 = 100 atm, k = 3.38 x 10^-5 s^-1, t = 20 min = 1200 s.

    k t = (3.38 x 10^-5)(1200) = 4.056 x 10^-2 = 0.04056

    p = 100 * e^(-0.04056) …

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12. KCET 2021Set B-21 markMCQ
    Q.The rate of a gaseous reaction is given by the expression $k[\mathrm{A}][\mathrm{B}]^2$. If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is (A) $\dfrac{1}{16}$ (B) $\dfrac{1}{8}$ (C) $8$ (D) $16$

    ›Reveal solutionSolution

    Compressing the vessel to half its volume doubles all concentrations; feed that into the rate law and the overall order (3) gives a factor of $2^3 = 8$.

    Step 1 — Effect of halving the volume on concentration.

    The number of moles of each gas is unchanged; only the volume shrinks:

    $[\mathrm{A}]_{\text{new}} = \frac{n_A}{V/2} = 2\,\frac{n_A}{V} = 2[\mathrm{A}], \qquad [\mathrm{B}]_{\text{new}} = 2[\mathrm{B}].$

    Step 2 — Substitute into the rate law.

    Original:

    $r_1 = k[\mathrm{A}][\mathrm{B}]^2.$

    After compression:

    $r_2 = k\,(2[\mathrm{A}])\,(2[\mathrm{B}])^2 = k \cdot 2[\mathrm{A}] \cdot 4[\mathrm{B}]^2 = 8\,k[\mathrm{A}][\mathrm{B}]^2.$

    Step 3 — Take the ratio.

    $\frac{r_2}{r_1} = \frac{8\,k[\mathrm{A}][\mathrm{B}]^2}{k[\mathrm{A}][\mathrm{B}]^2} = 8.$

    Step 4 — The shortcut worth remembering. …

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