Q.Find the half life period of first order reaction whose rate constant K = 4.93 × 10⁻⁴ S⁻¹.

Showing the 12 most recent of 18 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.Calculate the half-life period of a first order reaction whose rate constant is 200 s^-1.

   ›Reveal solutionSolution

   For a first-order reaction, the half-life is a constant, related to the rate constant by t(1/2) = 0.693/k, independent of the starting concentration.

   For a first-order reaction, the half-life formula is derived from the integrated rate law ln([R]0/[R]) = kt, setting [R] = [R]0/2:

   t(1/2) = ln(2)/k = 0.693/k

   …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 markMCQ
   Q.The half life period for first order reaction is ________ of its initial concentration.

   (a) Dependent

   (b) Independent

   (c) Property

   (d) None of these

   ›Reveal solutionSolution

   For a first order reaction, $t_{1/2} = 0.693/k$, which contains no concentration term, so the half-life is independent of the initial concentration.

   For a first order reaction, integrating the rate law gives $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$. Setting $[A] = [A]_0/2$ at $t = t_{1/2}$ gives $t_{1/2} = \dfrac{0.693}{k}$. Since this expression depends only on the rate constant $k$ (which is fixed at a given temperature) and not on $[A]_0$, the half-life of a first order reaction is the same regardless of how much r …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The half life period of a radioactive element is 140 days. After 560 days 1 g of element will be reduced to :

   (a) $\left(\dfrac{1}{8}\right)g$

   (b) $\left(\dfrac{1}{2}\right)g$

   (c) $\left(\dfrac{1}{16}\right)g$

   (d) $\left(\dfrac{1}{4}\right)g$

   ›Reveal solutionSolution

   Radioactive decay is a first-order process, so the amount remaining after any whole number of half-lives is simply the initial amount halved that many times; 560 days is exactly 4 half-lives of this 140-day element.

   Step 1 — number of half-lives elapsed: $n = \dfrac{\text{total time}}{t_{1/2}} = \dfrac{560\ \text{days}}{140\ \text{days}} = 4$

   …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The half-life of a first-order reaction depends on :

   (a) Initial concentration

   (b) Temperature only

   (c) Rate constant

   (d) Both

   (a) and (c)

   ›Reveal solutionSolution

   First-order half-life t(1/2) = 0.693/k depends only on k, not on initial concentration. Answer: (c) Rate constant.

   For a first-order reaction k = (2.303/t) log([A]0/[A]). Setting [A] = [A]0/2 gives t(1/2) = 0.693/k.

   • This expression contains only k, so the half-life is independent of the initial concentration. …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2026Set ANNUAL1 markMCQ
   Q.A reaction is found to be 50% complete in 20 minutes and 75% complete in 30 minutes. The order of the reaction is –

   (a) 0

   (b) 0.5

   (c) 1

   (d) 2

   ›Reveal solutionSolution

   Checking the data against the rate laws: 50% in 20 min and 75% in 30 min are consistent with a constant zero-order rate constant, so the order is 0.

   We test which order fits both data points.

   Zero order: $[A]_0 - [A] = k t$, i.e. amount reacted $\propto t$.

   • 50% reacted in 20 min: $k(20) = 0.50[A]_0 \Rightarrow k = 0.025[A]_0\ \text{min}^{-1}$.
   • 75% reacted in 30 min: $k(30) = 0.75[A]_0 \Rightarrow k = 0.025[A]_0\ \text{min}^{-1}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set ANNUAL1 mark
   Q.A first-order reaction is found to have a rate constant, $k = 5.5\times10^{-14}\ \text{s}^{-1}$. Find the half-life of the reaction.

   ›Reveal solutionSolution

   The half-life of a first-order reaction is obtained directly from $t_{1/2}=0.693/k$.

   Calculation

   For a first-order reaction, the half-life is independent of initial concentration and given by:

   $t_{1/2} = \frac{0.693}{k}$

   Substituting $k = 5.5\times10^{-14}\ \text{s}^{-1}$:

   $t_{1/2} = \frac{0.693}{5.5\times10^{-14}} = 0.126\times10^{14}\ \text{s} = 1.26\times10^{13}\ \text{s}$

   …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The half life of a first order reaction is 69.3 s. Its rate constant is –

   (i) 10⁻² s⁻¹

   (ii) 10⁻⁴ s⁻¹

   (iii) 10 s⁻¹

   (iv) 10² s⁻¹

   ›Reveal solutionSolution

   For a first order reaction, $k = \dfrac{0.693}{t_{1/2}}$; substituting $t_{1/2}=69.3\text{ s}$ gives $k = 10^{-2}\,\text{s}^{-1}$.

   For a first order reaction, the half-life is related to the rate constant by:

   $t_{1/2} = \dfrac{0.693}{k}$

   Rearranging for $k$: …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2024Set B1 markMCQ
   Q.A first order reaction completes 75% in 16 minutes. How much time will it take to complete 50%?

   (a) 8 minutes

   (b) 32 minutes

   (c) 24 minutes

   (d) 4 minutes

   ›Reveal solutionSolution

   For a first-order reaction, 75% completion corresponds to exactly two half-lives, so the time for 50% completion is half of 16 minutes = 8 minutes.

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

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

   When 75% reacts, 25% of the reactant is left, so $[A]_0/[A] = 100/25 = 4$:

   $k = \frac{2.303}{16}\log 4 = \frac{2.303}{16}(2\log 2)$

   When 50% reacts, $[A]_0/[A] = 2$, and the time taken is $t_{50\%}$:

   $k = \frac{2.303}{t_{50\%}}\log 2$

   Since $k$ is the same for both:

   …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2024Set ANNUAL1 markMCQ
   Q.Value of t1/2 for first order reaction is given as

   (a) t1/2 = [R0] / 2k

   (b) t1/2 = 0.693 / k

   (c) t1/2 = 6.023 / k

   (d) t1/2 = -k / 0.693

   ›Reveal solutionSolution

   The half-life of a first order reaction is derived from its integrated rate law and comes out to 0.693/k, independent of initial concentration.

   Integrated first order rate law: k = (1/t) ln([R0]/[R])

   At t = t1/2, [R] = [R0]/2:

   k = (1/t1/2) ln([R0]/([R0]/2)) = (1/t1/2) ln 2

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2023Set BZ1 markMCQ
    Q.Velocity constant for a first order reaction is $2.0\times10^{-3}\ s^{-1}$. The half-life period for this reaction is:

    (a) $3.465\times10^{3}\ s$

    (b) $3.465\times10^{2}\ s$

    (c) $3.465\times10^{-1}\ s$

    (d) $3.465\times10^{-2}\ s$

    ›Reveal solutionSolution

    For a first-order reaction $t_{1/2}=\dfrac{0.693}{k}=3.465\times10^{2}\ s$.

    Concept: The half-life of a first-order reaction is independent of the initial concentration and equals $t_{1/2}=\dfrac{0.693}{k}$.

    Calculation: …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2023Set F1 markMCQ
    Q.The half life of a first order reaction

    (a) does not depend upon rate constant

    (b) does not depend upon initial concentration

    (c) depends upon initial concentration

    (d) all of these

    ›Reveal solutionSolution

    For a first-order reaction, t1/2 = 0.693/k, so it is independent of initial concentration — option (B).

    For a first-order reaction the half-life is:

    t1/2 = 0.693 / k

    This expression contains only the rate constant k, so the half-life depends on k (hence indirectly on temperature) but is completely INDEPENDENT of the initial concentration of the reactant. Whatever the starting amount, the time to fall to half is the same. Option (a) is wrong because t1/2 does depe …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2023Set ANNUAL1 mark
    Q.Cyclopropane undergoes isomerization at $1000°C$ to propylene following first order kinetics with a rate constant $9.9\ s^{-1}$. How long would it take for the concentration of cyclopropane to decrease to 50% of its initial value ?

    ›Reveal solutionSolution

    For a first-order reaction, $t_{1/2}=0.693/k$; with $k=9.9\ s^{-1}$, this gives about 0.07 seconds.

    For a first-order reaction, the half-life is independent of the initial concentration and given by:

    $t_{1/2} = \frac{0.693}{k}$

    Substituting $k = 9.9\ s^{-1}$: …

    🔒 Start your 14-day free trial to see the full solution →