JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

A first order reaction has the rate constant, $k=4.6\times {10}^{-3}{s}^{-1}$ . The number of correct statement/s from the following is/are Given: $\mathrm{log}3=0.48$ .

A. Reaction completes in $1000s$ .

B. The reaction has a half-life of $500s$ .

C. The time required for $10%$ completion is 25 times the time required for 90% completion.

D. The degree of dissociation is equal to $(1-{e}^{-\mathrm{kt}})$ .

E. The rate and the rate constant have the same unit.

Official previous-year question

Held on 25 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

For a first order reaction:

${t}_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{4.6\times {10}^{-3}{s}^{-1}}=150.65s$

For a first order reaction,

${t}_{10%}=\frac{1}{K}\mathrm{ln}(\frac{a}{a-x})=\frac{1}{K}\mathrm{ln}(\frac{100}{90})$

${t}_{10%}=\frac{2.303}{K}(\mathrm{log}10-\mathrm{log}9)$

${t}_{10%}=\frac{2.093}{K}\times (0.04)$

Similarly

${t}_{90%}=\frac{1}{K}\mathrm{ln}(\frac{100}{10})$

${t}_{90%}=\frac{2.303}{K}$

$\frac{{t}_{90%}}{{t}_{10%}}=\frac{1}{0.04}=25$

We know,

${e}^{\mathrm{kt}}=\frac{a}{a-x}$

$\frac{a-x}{a}={e}^{-\mathrm{kt}}$

$x=a(1-{e}^{-\mathrm{kt}})$

$\alpha =\frac{x}{a}=(1-{e}^{-\mathrm{kt}})$

Thus, only option (D) is correct.

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JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

A first order reaction has the rate constant, $k=4.6\times {10}^{-3}{s}^{-1}$ . The number of correct statement/s from the following is/are Given: $\mathrm{log}3=0.48$ .

A. Reaction completes in $1000s$ .

B. The reaction has a half-life of $500s$ .

C. The time required for $10%$ completion is 25 times the time required for 90% completion.

D. The degree of dissociation is equal to $(1-{e}^{-\mathrm{kt}})$ .

E. The rate and the rate constant have the same unit.

Official previous-year question

Held on 25 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

For a first order reaction:

${t}_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{4.6\times {10}^{-3}{s}^{-1}}=150.65s$

For a first order reaction,

${t}_{10%}=\frac{1}{K}\mathrm{ln}(\frac{a}{a-x})=\frac{1}{K}\mathrm{ln}(\frac{100}{90})$

${t}_{10%}=\frac{2.303}{K}(\mathrm{log}10-\mathrm{log}9)$

${t}_{10%}=\frac{2.093}{K}\times (0.04)$

Similarly

${t}_{90%}=\frac{1}{K}\mathrm{ln}(\frac{100}{10})$

${t}_{90%}=\frac{2.303}{K}$

$\frac{{t}_{90%}}{{t}_{10%}}=\frac{1}{0.04}=25$

We know,

${e}^{\mathrm{kt}}=\frac{a}{a-x}$

$\frac{a-x}{a}={e}^{-\mathrm{kt}}$

$x=a(1-{e}^{-\mathrm{kt}})$

$\alpha =\frac{x}{a}=(1-{e}^{-\mathrm{kt}})$

Thus, only option (D) is correct.

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2023 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry