JEE Main 2024 — Chemistry Physical Chemistry

2024integereasy

Consider the following data for the given reaction

$2{\mathrm{HI}}_{(g)}\rightarrow {H}_{2(g)}+{I}_{2(g)}$

$\begin{matrix}\mathrm{HI}({\mathrm{molL}}^{-1}) & 0.005 & 0.01 & 0.02\end{matrix}$

Rate $({\mathrm{molL}}^{-1}{s}^{-1})7.5\times {10}^{-4}3.0\times {10}^{-3}1.2\times {10}^{-2}$

The order of the reaction is __________.

Official previous-year question

Held on 27 Jan 2024 · Verified 6 Jul 2026.

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Solution

The initial concentrations of $\mathrm{HI}$ as $0.005,0.01,\text{and}0.02$ mol/L, respectively, and the corresponding initial rates as given:

$\begin{matrix}\mathrm{HI}({\mathrm{molL}}^{-1}) & 0.005 & 0.01 & 0.02\end{matrix}$

Rate $({\mathrm{molL}}^{-1}{s}^{-1})7.5\times {10}^{-4}3.0\times {10}^{-3}1.2\times {10}^{-2}$

For the reaction, rate law is

$\mathrm{rate}=k{[\mathrm{HI}]}^{P}$

$P$ is the order of the reaction.

From the given data

$7.5\times {10}^{-4}=k{[0.05]}^{P}$ -------(1)

$3\times {10}^{-3}=k{[0.01]}^{P}$ --------(2)

$1.2\times {10}^{-2}=k{[0.02]}^{P}$ ---------(3)

On dividing Equation (3) by equation (2)

$\frac{12\times {10}^{-3}}{3\times {10}^{-3}}=\frac{k{[0.02]}^{P}}{k{[0.01]}^{P}}$

$\Rightarrow {2}^{P}={2}^{2}$

$\Rightarrow P=2$

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JEE Main Physical Chemistry in other years

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JEE Main 2024 — Chemistry Physical Chemistry

2024integereasy

Consider the following data for the given reaction

$2{\mathrm{HI}}_{(g)}\rightarrow {H}_{2(g)}+{I}_{2(g)}$

$\begin{matrix}\mathrm{HI}({\mathrm{molL}}^{-1}) & 0.005 & 0.01 & 0.02\end{matrix}$

Rate $({\mathrm{molL}}^{-1}{s}^{-1})7.5\times {10}^{-4}3.0\times {10}^{-3}1.2\times {10}^{-2}$

The order of the reaction is __________.

Official previous-year question

Held on 27 Jan 2024 · Verified 6 Jul 2026.

Did you get this right?

Solution

The initial concentrations of $\mathrm{HI}$ as $0.005,0.01,\text{and}0.02$ mol/L, respectively, and the corresponding initial rates as given:

$\begin{matrix}\mathrm{HI}({\mathrm{molL}}^{-1}) & 0.005 & 0.01 & 0.02\end{matrix}$

Rate $({\mathrm{molL}}^{-1}{s}^{-1})7.5\times {10}^{-4}3.0\times {10}^{-3}1.2\times {10}^{-2}$

For the reaction, rate law is

$\mathrm{rate}=k{[\mathrm{HI}]}^{P}$

$P$ is the order of the reaction.

From the given data

$7.5\times {10}^{-4}=k{[0.05]}^{P}$ -------(1)

$3\times {10}^{-3}=k{[0.01]}^{P}$ --------(2)

$1.2\times {10}^{-2}=k{[0.02]}^{P}$ ---------(3)

On dividing Equation (3) by equation (2)

$\frac{12\times {10}^{-3}}{3\times {10}^{-3}}=\frac{k{[0.02]}^{P}}{k{[0.01]}^{P}}$

$\Rightarrow {2}^{P}={2}^{2}$

$\Rightarrow P=2$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

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

Browse all Chemistry topics

JEE Main 2024 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

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Organic Chemistry

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JEE Main 2024 — 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