JEE Main 2022 — Chemistry Physical Chemistry

2022integermedium

For a given chemical reaction ${\gamma }_{1}A+{\gamma }_{2}B\rightarrow {\gamma }_{3}C+{\gamma }_{4}D$ . Concentration of $C$ changes from $10\mathrm{mmol}{\mathrm{dm}}^{-3}$ to $20\mathrm{mmol}{\mathrm{dm}}^{-3}$ in $10s.$ Rate of appearance of $D$ is $1.5$ times the rate of disappearance of $B$ which is twice the rate of disappearance $A$ . The rate of appearance of $D$ has been experimentally determined to be $9\mathrm{mmol}{\mathrm{dm}}^{-3}{s}^{-1}.$ Therefore the rate of reaction is____ $\mathrm{mmol}{\mathrm{dm}}^{-3}{s}^{-1}.$ (Nearest Integer)

Official previous-year question

Held on 25 Jun 2022 · Verified 6 Jul 2026.

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Solution

$\frac{d[c]}{\mathrm{dt}}=(\frac{20-10}{10})=1\mathrm{milimole}/{\mathrm{dm}}^{3}\mathrm{sec}$

${\gamma }_{1}A+{\gamma }_{2}B\rightarrow {\gamma }_{3}C+{\gamma }_{4}D$

$\frac{d[D]}{\mathrm{dt}}=1.5(-\frac{d}{\mathrm{dt}}[B])=9\mathrm{milimole}/{\mathrm{dm}}^{3}\mathrm{sec}$

$(-\frac{d}{\mathrm{dt}}[B])=2(-\frac{d}{\mathrm{dt}}[A])$

Rate $=-\frac{1}{{\gamma }_{1}}\frac{d[A]}{\mathrm{dt}}=-\frac{1}{{\gamma }_{2}}\frac{d[B]}{\mathrm{dt}}=\frac{1}{{\gamma }_{3}}\frac{d[C]}{\mathrm{dt}}=\frac{1}{{\gamma }_{4}}\frac{d[D]}{\mathrm{dt}}$

(i) $\frac{d[D]}{\mathrm{dt}}=\frac{{\gamma }_{4}}{{\gamma }_{2}}[-\frac{d[B]}{\mathrm{dt}}]=1.5(-\frac{d}{\mathrm{dt}}[B])$

$\frac{{\gamma }_{4}}{{\gamma }_{2}}=1.5$

(ii) $\frac{{\gamma }_{2}}{{\gamma }_{1}}[-\frac{d[A]}{\mathrm{dt}}]=(-\frac{d}{\mathrm{dt}}[B])=2(-\frac{d}{\mathrm{dt}}[A])$

$\frac{{\gamma }_{2}}{{\gamma }_{1}}=2$

(iii) $\frac{d[C]}{\mathrm{dt}}=\frac{{\gamma }_{3}}{{\gamma }_{4}}[\frac{d[D]}{\mathrm{dt}}]$

$1=\frac{{\gamma }_{3}}{{\gamma }_{4}}\times 9$

$\frac{{\gamma }_{4}}{{\gamma }_{3}}=9$

So, if

${\gamma }_{3}=1\mathrm{than}{\gamma }_{4}=9 {\gamma }_{2}=6\mathrm{and}{\gamma }_{1}=3 \mathrm{Rate}\mathrm{of}\mathrm{reaction}=\frac{1}{{\gamma }_{3}}\frac{d[C]}{\mathrm{dt}}=1m\mathrm{mol}{\mathrm{dm}}^{-3}{s}^{-1}$

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JEE Main Physical Chemistry in other years

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JEE Main 2022 — Chemistry Physical Chemistry

2022integermedium

Official previous-year question

Held on 25 Jun 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

$\frac{d[c]}{\mathrm{dt}}=(\frac{20-10}{10})=1\mathrm{milimole}/{\mathrm{dm}}^{3}\mathrm{sec}$

${\gamma }_{1}A+{\gamma }_{2}B\rightarrow {\gamma }_{3}C+{\gamma }_{4}D$

$\frac{d[D]}{\mathrm{dt}}=1.5(-\frac{d}{\mathrm{dt}}[B])=9\mathrm{milimole}/{\mathrm{dm}}^{3}\mathrm{sec}$

$(-\frac{d}{\mathrm{dt}}[B])=2(-\frac{d}{\mathrm{dt}}[A])$

(i) $\frac{d[D]}{\mathrm{dt}}=\frac{{\gamma }_{4}}{{\gamma }_{2}}[-\frac{d[B]}{\mathrm{dt}}]=1.5(-\frac{d}{\mathrm{dt}}[B])$

$\frac{{\gamma }_{4}}{{\gamma }_{2}}=1.5$

(ii) $\frac{{\gamma }_{2}}{{\gamma }_{1}}[-\frac{d[A]}{\mathrm{dt}}]=(-\frac{d}{\mathrm{dt}}[B])=2(-\frac{d}{\mathrm{dt}}[A])$

$\frac{{\gamma }_{2}}{{\gamma }_{1}}=2$

(iii) $\frac{d[C]}{\mathrm{dt}}=\frac{{\gamma }_{3}}{{\gamma }_{4}}[\frac{d[D]}{\mathrm{dt}}]$

$1=\frac{{\gamma }_{3}}{{\gamma }_{4}}\times 9$

$\frac{{\gamma }_{4}}{{\gamma }_{3}}=9$

So, if

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JEE Main 2022 — Chemistry Physical Chemistry

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JEE Main Physical Chemistry in other years

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