Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

At $\mathrm{30^{\circ} \mathrm{C},}$ the half life for the decomposition of $\mathrm{\mathrm{AB}_{2}}$ is $\mathrm{200 \mathrm{~s}}$ and is independent of the initial concentration of $\mathrm{\mathrm{AB}_{2}}$ . The time required for $\mathrm{ 80 \%}$ of the $\mathrm{ \mathrm{AB}_{2}}$ to decompose is

$\mathrm{ Given: \log 2=0.30 \log 3=0.48 }$
Option: 1
$\mathrm{200 \mathrm{~s}}$

Option: 2
$\mathrm{323 \mathrm{~s}}$

Option: 3
$\mathrm{467 \mathrm{~s}}$

Option: 4
$\mathrm{532 \mathrm{~s}}$

Answers (1)

best_answer

The half life is 200 S and it is independent of the initial concentration of $\text{AB}_{2}$

So, it is first order reaction.

$\mathrm{\begin{aligned} &\mathrm{t v_2}=\frac{\ln 2}{\text{k}}=200\; \mathrm{sec} \\ &\Rightarrow \quad \text{k}=\frac{\ln 2}{200} \sec ^{-1} \\ \end{aligned}}$

Now, $\mathrm{k=\frac{2.303}{t} \log \frac{a}{a-x}}$

80% completion , So, $\mathrm{x=0.89\Rightarrow a-x=a-0.89=0.29}$

Then, $\mathrm{k=\frac{2.303}{t_{80}}\log\left ( \frac{9}{0.29} \right )}$

$\mathrm{\frac{\ln 2}{200}=\frac{2.303}{t_{80}} \log \left(\frac{10}{2}\right)}$

$\begin{aligned} &\operatorname{t}_{80}=\frac{\ln 10-\ln 2}{\ln 2} \times 200 \\ \end{aligned}$

$\mathrm{t_{80}=\frac{\log 10-\log^{2}}{\log^{2}}\times 200}$

$\begin{aligned} &\text {t}_{80}=\left(\frac{1}{\log 2}-1\right) \times 200=\left(\frac{1}{0.3}-1\right) \times 200 \\ &\operatorname{t}_{80}=466.6 \approx 467 \mathrm{sec} . \end{aligned}$

Hence, Option (3) is correct.

Posted by

rishi.raj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

5 g of Na₂SO₄ was dissolved in x g of H₂O. The change in freezing point was found to be 3.82⁰C. If Na₂SO₄ is 81.5% ionised, the value of x (K

A capacitor is made of two square plates each of side 'a' making a very small angle $\alpha$

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO₃ to give fraction A. The leftover organic phase was extracted with d

Trending Articles/News

anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

anam-khan

Bihar Board extends BSEB Super 50 free coaching application last date to March 26

anam-khan

JEE Main City Intimation Slip 2025 Live: NTA session 2 exam city slips soon at jeemain.nta.nic.in

Latest Question