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  • For the first-order reaction: <img alt="2N_{2}O_{5 }\left ( g \right )\rightarrow4NO_{2}\left ( g \right )+O_{2}\left ( g \right )" src="http://entrancecorner.oncodecogs.com/gif.latex?2N_%7B

For the first-order reaction:
2N_{2}O_{5 }\left ( g \right )\rightarrow4NO_{2}\left ( g \right )+O_{2}\left ( g \right )

(A) The concentration of the reactant decreases exponentially with time.

(B) The half-life of the reaction decreases with increasing temperature.

(C) The half-life of the reaction depends on the initial concentration of the reactant.

(D) The reaction proceeds to 99.6% completion in eight half-life duration.

The correct statements are -

Option: 1

Only A and B


Option: 2

Only B and C


Option: 3

A, B, and D


Option: 4

A, B, C, and D


Answers (1)

best_answer

A) The concentration of reactant which is following first-order kinetics always decreases exponentially and becomes zero at infinity. 

\mathrm{A}_{\mathrm{t}}=\mathrm{A_o} \mathrm{e}^{-\mathrm{kt}}

B) The half-life of the reaction decreases with increasing temperature

As the temperature increases, the rate constant increases and the half-life decreases since half-life is inversely dependent on the rate constant.

\mathrm{t_{1/2}=\frac{ln2}{k}}

\mathrm{t_{1/2}\propto \frac{1}{k}}

K increase on increasing T.

C) The half-life of the reaction does not depend on the initial concentration of the reactant.

\mathrm{t_{1/2}= \frac{ln2}{k}}
 

D) The reaction proceeds to 99.6% completion in eight half-life duration

After eight half-lives.
\mathrm{A}=\frac{\mathrm{A}_{0}}{2^{8}}

\% \text { completion }=\frac{\mathrm{A_o}-\frac{\mathrm{A}_{\mathrm{o}}}{2^{8}}}{\mathrm{A}_{\mathrm{o}}} \times 100=99.6 \%

So, A, B, and D are correct.

Option 3 is correct.

Posted by

vishal kumar

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