JEE Main 2025 (Online) 22nd January Evening Shift | Chemical Kinetics and Nuclear Chemistry Question 24 | Chemistry

JEE Main 2025 (Online) 22nd January Evening Shift

MCQ (Single Correct Answer)

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Consider the given figure and choose the correct option :

JEE Main 2025 (Online) 22nd January Evening Shift Chemistry - Chemical Kinetics and Nuclear Chemistry Question 20 English

Activation energy of both forward and backward reaction is $E_1+E_2$ and reactant is more stable than product.

Activation energy of forward reaction is $E_1+E_2$ and product is more stable than reactant.

Activation energy of backward reaction is $\mathrm{E}_1$ and product is more stable than reactant.

Activation energy of forward reaction is $E_1+E_2$ and product is less stable than reactant.

JEE Main 2025 (Online) 22nd January Morning Shift

MCQ (Single Correct Answer)

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Out of Syllabus

Which of the following statement is not true for radioactive decay?

Half life is $\ln 2$ times of $\frac{1}{\text { rate constant }}$.

Decay constant increases with increase in temperature.

Decay constant does not depend upon temperature.

Amount of radioactive substance remained after three half lives is $\frac{1}{8}$ th of original amount.

JEE Main 2024 (Online) 8th April Evening Shift

MCQ (Single Correct Answer)

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For a reaction $$A \xrightarrow{\mathrm{K}_1} \mathrm{~B} \xrightarrow{\mathrm{K}_2} \mathrm{C}$$ If the rate of formation of B is set to be zero then the concentration of B is given by :

$$(\mathrm{K}_1-\mathrm{K}_2)[\mathrm{A}]$$

$$(\mathrm{K}_1+\mathrm{K}_2)[\mathrm{A}]$$

$$\mathrm{K}_1 \mathrm{K}_2[\mathrm{~A}]$$

$$(\mathrm{K}_1 / \mathrm{K}_2)[\mathrm{A}]$$

JEE Main 2024 (Online) 31st January Morning Shift

MCQ (Single Correct Answer)

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Integrated rate law equation for a first order gas phase reaction is given by (where $$\mathrm{P}_{\mathrm{i}}$$ is initial pressure and $$\mathrm{P}_{\mathrm{t}}$$ is total pressure at time $$t$$)

$$k=\frac{2.303}{t} \times \log \frac{P_i}{\left(2 P_i-P_t\right)}$$

$$\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \log \frac{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}{\mathrm{P}_{\mathrm{i}}}$$

$$k=\frac{2.303}{t} \times \frac{P_i}{\left(2 P_i-P_t\right)}$$

$$\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \log \frac{2 \mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}$$

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