Integrated Rate Equations
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CHXII04:CHEMICAL KINETICS

320343 In a first order reaction the concentration of reactant decreases from \(800 \mathrm{M}\) to \(50 \mathrm{M}\) in \(2 \times 10^{2} \mathrm{sec}\). The rate constant of reaction in \(\sec ^{-1}\) is

1 \(2 \times 10^{4}\)
2 \(3.45 \times 10^{-5}\)
3 \(1.386 \times 10^{-2}\)
4 \(2 \times 10^{-4}\)
CHXII04:CHEMICAL KINETICS

320344 In a first order reaction, it takes 40.5 minutes for the reactant to be \(24 \%\) decomposed. Find the rate of the reaction.

1 \(9.4 \times 10^{-3} \mathrm{~min}^{-1}\)
2 \(7.0 \times 10^{-3} \mathrm{~min}^{-1}\)
3 \(25.2 \times 10^{-3} \mathrm{~min}^{-1}\)
4 \(10.5 \times 10^{-3} \mathrm{~min}^{-1}\)
CHXII04:CHEMICAL KINETICS

320345 Identify the formula which is applicable to the conversion of \(20 \%\) of the initial concentration of the reactant to the product in a first order reaction. (Rate constant = k )

1 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{5}{\rm{log}}\frac{{100}}{{20}}\)
2 \({{\text{t}}_{{\text{20\% }}}} = \frac{{2.303}}{{20}}\,\,{\text{log}}\,\,\frac{{100}}{{\text{k}}}\)
3 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{{\rm{k}}}{\rm{log}}\frac{5}{4}\)
4 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{{100}}{\rm{log}}\frac{{\rm{k}}}{{80}}\)
CHXII04:CHEMICAL KINETICS

320346 A first order reaction has a rate constant \(0.00813 \mathrm{~min}^{-1}\). How long will it take for \(60 \%\) completion?

1 \(56.35 \mathrm{~min}\)
2 \(98.7 \mathrm{~min}\)
3 \(62.77 \mathrm{~min}\)
4 \(112.7 \mathrm{~min}\)
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CHXII04:CHEMICAL KINETICS

320343 In a first order reaction the concentration of reactant decreases from \(800 \mathrm{M}\) to \(50 \mathrm{M}\) in \(2 \times 10^{2} \mathrm{sec}\). The rate constant of reaction in \(\sec ^{-1}\) is

1 \(2 \times 10^{4}\)
2 \(3.45 \times 10^{-5}\)
3 \(1.386 \times 10^{-2}\)
4 \(2 \times 10^{-4}\)
CHXII04:CHEMICAL KINETICS

320344 In a first order reaction, it takes 40.5 minutes for the reactant to be \(24 \%\) decomposed. Find the rate of the reaction.

1 \(9.4 \times 10^{-3} \mathrm{~min}^{-1}\)
2 \(7.0 \times 10^{-3} \mathrm{~min}^{-1}\)
3 \(25.2 \times 10^{-3} \mathrm{~min}^{-1}\)
4 \(10.5 \times 10^{-3} \mathrm{~min}^{-1}\)
CHXII04:CHEMICAL KINETICS

320345 Identify the formula which is applicable to the conversion of \(20 \%\) of the initial concentration of the reactant to the product in a first order reaction. (Rate constant = k )

1 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{5}{\rm{log}}\frac{{100}}{{20}}\)
2 \({{\text{t}}_{{\text{20\% }}}} = \frac{{2.303}}{{20}}\,\,{\text{log}}\,\,\frac{{100}}{{\text{k}}}\)
3 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{{\rm{k}}}{\rm{log}}\frac{5}{4}\)
4 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{{100}}{\rm{log}}\frac{{\rm{k}}}{{80}}\)
CHXII04:CHEMICAL KINETICS

320346 A first order reaction has a rate constant \(0.00813 \mathrm{~min}^{-1}\). How long will it take for \(60 \%\) completion?

1 \(56.35 \mathrm{~min}\)
2 \(98.7 \mathrm{~min}\)
3 \(62.77 \mathrm{~min}\)
4 \(112.7 \mathrm{~min}\)
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CHXII04:CHEMICAL KINETICS

320343 In a first order reaction the concentration of reactant decreases from \(800 \mathrm{M}\) to \(50 \mathrm{M}\) in \(2 \times 10^{2} \mathrm{sec}\). The rate constant of reaction in \(\sec ^{-1}\) is

1 \(2 \times 10^{4}\)
2 \(3.45 \times 10^{-5}\)
3 \(1.386 \times 10^{-2}\)
4 \(2 \times 10^{-4}\)
CHXII04:CHEMICAL KINETICS

320344 In a first order reaction, it takes 40.5 minutes for the reactant to be \(24 \%\) decomposed. Find the rate of the reaction.

1 \(9.4 \times 10^{-3} \mathrm{~min}^{-1}\)
2 \(7.0 \times 10^{-3} \mathrm{~min}^{-1}\)
3 \(25.2 \times 10^{-3} \mathrm{~min}^{-1}\)
4 \(10.5 \times 10^{-3} \mathrm{~min}^{-1}\)
CHXII04:CHEMICAL KINETICS

320345 Identify the formula which is applicable to the conversion of \(20 \%\) of the initial concentration of the reactant to the product in a first order reaction. (Rate constant = k )

1 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{5}{\rm{log}}\frac{{100}}{{20}}\)
2 \({{\text{t}}_{{\text{20\% }}}} = \frac{{2.303}}{{20}}\,\,{\text{log}}\,\,\frac{{100}}{{\text{k}}}\)
3 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{{\rm{k}}}{\rm{log}}\frac{5}{4}\)
4 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{{100}}{\rm{log}}\frac{{\rm{k}}}{{80}}\)
CHXII04:CHEMICAL KINETICS

320346 A first order reaction has a rate constant \(0.00813 \mathrm{~min}^{-1}\). How long will it take for \(60 \%\) completion?

1 \(56.35 \mathrm{~min}\)
2 \(98.7 \mathrm{~min}\)
3 \(62.77 \mathrm{~min}\)
4 \(112.7 \mathrm{~min}\)
For More Updates join with Us
CHXII04:CHEMICAL KINETICS

320343 In a first order reaction the concentration of reactant decreases from \(800 \mathrm{M}\) to \(50 \mathrm{M}\) in \(2 \times 10^{2} \mathrm{sec}\). The rate constant of reaction in \(\sec ^{-1}\) is

1 \(2 \times 10^{4}\)
2 \(3.45 \times 10^{-5}\)
3 \(1.386 \times 10^{-2}\)
4 \(2 \times 10^{-4}\)
CHXII04:CHEMICAL KINETICS

320344 In a first order reaction, it takes 40.5 minutes for the reactant to be \(24 \%\) decomposed. Find the rate of the reaction.

1 \(9.4 \times 10^{-3} \mathrm{~min}^{-1}\)
2 \(7.0 \times 10^{-3} \mathrm{~min}^{-1}\)
3 \(25.2 \times 10^{-3} \mathrm{~min}^{-1}\)
4 \(10.5 \times 10^{-3} \mathrm{~min}^{-1}\)
CHXII04:CHEMICAL KINETICS

320345 Identify the formula which is applicable to the conversion of \(20 \%\) of the initial concentration of the reactant to the product in a first order reaction. (Rate constant = k )

1 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{5}{\rm{log}}\frac{{100}}{{20}}\)
2 \({{\text{t}}_{{\text{20\% }}}} = \frac{{2.303}}{{20}}\,\,{\text{log}}\,\,\frac{{100}}{{\text{k}}}\)
3 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{{\rm{k}}}{\rm{log}}\frac{5}{4}\)
4 \({{\rm{t}}_{{\rm{20\% }}}} = \frac{{2.303}}{{100}}{\rm{log}}\frac{{\rm{k}}}{{80}}\)
CHXII04:CHEMICAL KINETICS

320346 A first order reaction has a rate constant \(0.00813 \mathrm{~min}^{-1}\). How long will it take for \(60 \%\) completion?

1 \(56.35 \mathrm{~min}\)
2 \(98.7 \mathrm{~min}\)
3 \(62.77 \mathrm{~min}\)
4 \(112.7 \mathrm{~min}\)
NEET DIGITAL LIBRARY