320364
What is the value of rate constant of first order reaction, if it takes 15 minutes for consumption of \(20 \%\) of reactants?
1 \(1.38 \times 10^{-2} \mathrm{~min}^{-1}\)
2 \(1.48 \times 10^{-2} \mathrm{~min}^{-1}\)
3 \(1.07 \times 10^{-2} \mathrm{~min}^{-1}\)
4 \(1.84 \times 10^{-2} \mathrm{~min}^{-1}\)
Explanation:
Given, \(t=15 \mathrm{~min}\) for \(20 \%\) of reactant to react. For first order reaction, Rate constant \((\mathrm{k})=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\left[\mathrm{R}_{0}\right]}{[\mathrm{R}]}\) where, \(\left[\mathrm{R}_{0}\right]=\) original amount of reactant \([R]=\) reactant remaining unreacted So, \(\left[\mathrm{R}_{0}\right]=100\) \([R]=80(20 \%\) reacted \()\) \(\mathrm{k}=\dfrac{2.303}{15 \min } \log \left[\dfrac{100}{80}\right]\) \(=1.48 \times 10^{-2} \mathrm{~min}^{-1}\)
CHXII04:CHEMICAL KINETICS
320365
For the first order reaction half-life is \(14 \mathrm{sec}\), the time required for the initial concentration to reduce to \(1 / 8\) of its value is:
320366
A first order reaction has a specific reaction rate of \({\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{se}}{{\rm{c}}^{{\rm{ - 1}}}}\). How much time it take for \({\rm{20\;g}}\) of the reactant to reduce to \({\rm{5\;g}}\) ?
1 \({\rm{138}}{\rm{.6}}\,\,{\rm{sec}}\)
2 \({\rm{346}}{\rm{.5}}\,\,{\rm{sec}}\)
3 \({\rm{693}}{\rm{.0}}\,\,{\rm{sec}}\)
4 \({\rm{238}}{\rm{.6}}\,\,{\rm{sec}}\)
Explanation:
Half life of first order reaction, \(\mathrm{t}_{1 / 2}=\dfrac{0.693}{\mathrm{k}}=\dfrac{0.693}{10^{-2}}=69.3 \mathrm{sec}\) Method - 1 Total time \(=2 \mathrm{t}_{1 / 2}=2 \times 69.3=138.6 \mathrm{sec}\) Method - 2 \(\begin{aligned}& \mathrm{t}=\dfrac{2.303}{\mathrm{k}} \log \dfrac{[\mathrm{A}]_{o}}{[\mathrm{~A}]_{\mathrm{t}}} \\& \mathrm{t}=\dfrac{2.303}{10^{-2}} \log \dfrac{20}{5} \Rightarrow \mathrm{t}=138.6 \mathrm{sec}\end{aligned}\)
CHXII04:CHEMICAL KINETICS
320367
Half life for a first order reaction is 6.93 hour. What is the time required for \(80 \%\) completion of the reaction?
1 12 hours
2 18 hours
3 6 hours
4 16 hours
Explanation:
For first order reaction, \(\mathrm{k}=\dfrac{0.693}{\mathrm{t}_{\frac{1}{2}}}=\dfrac{0.693}{6.93}=0.1\) hour \(^{-1}\) Here, \([\mathrm{A}]_{0}=100,[\mathrm{~A}]_{\mathrm{t}}=100-80=20\) \(\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \log _{10} \dfrac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{\mathrm{t}}}\) \(\therefore \mathrm{t}=\dfrac{2.303}{0.1} \log _{10} \dfrac{100}{20}=23.03 \times \log _{10} 5\) \(=23.03 \times 0.699=16.10\) hours
MHTCET - 2021
CHXII04:CHEMICAL KINETICS
320368
If the rate constant for a first order reaction is \(\mathrm{k}\), the time \((\mathrm{t})\) required for the completion of \(99 \%\) of the reaction is givne by
320364
What is the value of rate constant of first order reaction, if it takes 15 minutes for consumption of \(20 \%\) of reactants?
1 \(1.38 \times 10^{-2} \mathrm{~min}^{-1}\)
2 \(1.48 \times 10^{-2} \mathrm{~min}^{-1}\)
3 \(1.07 \times 10^{-2} \mathrm{~min}^{-1}\)
4 \(1.84 \times 10^{-2} \mathrm{~min}^{-1}\)
Explanation:
Given, \(t=15 \mathrm{~min}\) for \(20 \%\) of reactant to react. For first order reaction, Rate constant \((\mathrm{k})=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\left[\mathrm{R}_{0}\right]}{[\mathrm{R}]}\) where, \(\left[\mathrm{R}_{0}\right]=\) original amount of reactant \([R]=\) reactant remaining unreacted So, \(\left[\mathrm{R}_{0}\right]=100\) \([R]=80(20 \%\) reacted \()\) \(\mathrm{k}=\dfrac{2.303}{15 \min } \log \left[\dfrac{100}{80}\right]\) \(=1.48 \times 10^{-2} \mathrm{~min}^{-1}\)
CHXII04:CHEMICAL KINETICS
320365
For the first order reaction half-life is \(14 \mathrm{sec}\), the time required for the initial concentration to reduce to \(1 / 8\) of its value is:
320366
A first order reaction has a specific reaction rate of \({\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{se}}{{\rm{c}}^{{\rm{ - 1}}}}\). How much time it take for \({\rm{20\;g}}\) of the reactant to reduce to \({\rm{5\;g}}\) ?
1 \({\rm{138}}{\rm{.6}}\,\,{\rm{sec}}\)
2 \({\rm{346}}{\rm{.5}}\,\,{\rm{sec}}\)
3 \({\rm{693}}{\rm{.0}}\,\,{\rm{sec}}\)
4 \({\rm{238}}{\rm{.6}}\,\,{\rm{sec}}\)
Explanation:
Half life of first order reaction, \(\mathrm{t}_{1 / 2}=\dfrac{0.693}{\mathrm{k}}=\dfrac{0.693}{10^{-2}}=69.3 \mathrm{sec}\) Method - 1 Total time \(=2 \mathrm{t}_{1 / 2}=2 \times 69.3=138.6 \mathrm{sec}\) Method - 2 \(\begin{aligned}& \mathrm{t}=\dfrac{2.303}{\mathrm{k}} \log \dfrac{[\mathrm{A}]_{o}}{[\mathrm{~A}]_{\mathrm{t}}} \\& \mathrm{t}=\dfrac{2.303}{10^{-2}} \log \dfrac{20}{5} \Rightarrow \mathrm{t}=138.6 \mathrm{sec}\end{aligned}\)
CHXII04:CHEMICAL KINETICS
320367
Half life for a first order reaction is 6.93 hour. What is the time required for \(80 \%\) completion of the reaction?
1 12 hours
2 18 hours
3 6 hours
4 16 hours
Explanation:
For first order reaction, \(\mathrm{k}=\dfrac{0.693}{\mathrm{t}_{\frac{1}{2}}}=\dfrac{0.693}{6.93}=0.1\) hour \(^{-1}\) Here, \([\mathrm{A}]_{0}=100,[\mathrm{~A}]_{\mathrm{t}}=100-80=20\) \(\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \log _{10} \dfrac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{\mathrm{t}}}\) \(\therefore \mathrm{t}=\dfrac{2.303}{0.1} \log _{10} \dfrac{100}{20}=23.03 \times \log _{10} 5\) \(=23.03 \times 0.699=16.10\) hours
MHTCET - 2021
CHXII04:CHEMICAL KINETICS
320368
If the rate constant for a first order reaction is \(\mathrm{k}\), the time \((\mathrm{t})\) required for the completion of \(99 \%\) of the reaction is givne by
320364
What is the value of rate constant of first order reaction, if it takes 15 minutes for consumption of \(20 \%\) of reactants?
1 \(1.38 \times 10^{-2} \mathrm{~min}^{-1}\)
2 \(1.48 \times 10^{-2} \mathrm{~min}^{-1}\)
3 \(1.07 \times 10^{-2} \mathrm{~min}^{-1}\)
4 \(1.84 \times 10^{-2} \mathrm{~min}^{-1}\)
Explanation:
Given, \(t=15 \mathrm{~min}\) for \(20 \%\) of reactant to react. For first order reaction, Rate constant \((\mathrm{k})=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\left[\mathrm{R}_{0}\right]}{[\mathrm{R}]}\) where, \(\left[\mathrm{R}_{0}\right]=\) original amount of reactant \([R]=\) reactant remaining unreacted So, \(\left[\mathrm{R}_{0}\right]=100\) \([R]=80(20 \%\) reacted \()\) \(\mathrm{k}=\dfrac{2.303}{15 \min } \log \left[\dfrac{100}{80}\right]\) \(=1.48 \times 10^{-2} \mathrm{~min}^{-1}\)
CHXII04:CHEMICAL KINETICS
320365
For the first order reaction half-life is \(14 \mathrm{sec}\), the time required for the initial concentration to reduce to \(1 / 8\) of its value is:
320366
A first order reaction has a specific reaction rate of \({\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{se}}{{\rm{c}}^{{\rm{ - 1}}}}\). How much time it take for \({\rm{20\;g}}\) of the reactant to reduce to \({\rm{5\;g}}\) ?
1 \({\rm{138}}{\rm{.6}}\,\,{\rm{sec}}\)
2 \({\rm{346}}{\rm{.5}}\,\,{\rm{sec}}\)
3 \({\rm{693}}{\rm{.0}}\,\,{\rm{sec}}\)
4 \({\rm{238}}{\rm{.6}}\,\,{\rm{sec}}\)
Explanation:
Half life of first order reaction, \(\mathrm{t}_{1 / 2}=\dfrac{0.693}{\mathrm{k}}=\dfrac{0.693}{10^{-2}}=69.3 \mathrm{sec}\) Method - 1 Total time \(=2 \mathrm{t}_{1 / 2}=2 \times 69.3=138.6 \mathrm{sec}\) Method - 2 \(\begin{aligned}& \mathrm{t}=\dfrac{2.303}{\mathrm{k}} \log \dfrac{[\mathrm{A}]_{o}}{[\mathrm{~A}]_{\mathrm{t}}} \\& \mathrm{t}=\dfrac{2.303}{10^{-2}} \log \dfrac{20}{5} \Rightarrow \mathrm{t}=138.6 \mathrm{sec}\end{aligned}\)
CHXII04:CHEMICAL KINETICS
320367
Half life for a first order reaction is 6.93 hour. What is the time required for \(80 \%\) completion of the reaction?
1 12 hours
2 18 hours
3 6 hours
4 16 hours
Explanation:
For first order reaction, \(\mathrm{k}=\dfrac{0.693}{\mathrm{t}_{\frac{1}{2}}}=\dfrac{0.693}{6.93}=0.1\) hour \(^{-1}\) Here, \([\mathrm{A}]_{0}=100,[\mathrm{~A}]_{\mathrm{t}}=100-80=20\) \(\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \log _{10} \dfrac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{\mathrm{t}}}\) \(\therefore \mathrm{t}=\dfrac{2.303}{0.1} \log _{10} \dfrac{100}{20}=23.03 \times \log _{10} 5\) \(=23.03 \times 0.699=16.10\) hours
MHTCET - 2021
CHXII04:CHEMICAL KINETICS
320368
If the rate constant for a first order reaction is \(\mathrm{k}\), the time \((\mathrm{t})\) required for the completion of \(99 \%\) of the reaction is givne by
320364
What is the value of rate constant of first order reaction, if it takes 15 minutes for consumption of \(20 \%\) of reactants?
1 \(1.38 \times 10^{-2} \mathrm{~min}^{-1}\)
2 \(1.48 \times 10^{-2} \mathrm{~min}^{-1}\)
3 \(1.07 \times 10^{-2} \mathrm{~min}^{-1}\)
4 \(1.84 \times 10^{-2} \mathrm{~min}^{-1}\)
Explanation:
Given, \(t=15 \mathrm{~min}\) for \(20 \%\) of reactant to react. For first order reaction, Rate constant \((\mathrm{k})=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\left[\mathrm{R}_{0}\right]}{[\mathrm{R}]}\) where, \(\left[\mathrm{R}_{0}\right]=\) original amount of reactant \([R]=\) reactant remaining unreacted So, \(\left[\mathrm{R}_{0}\right]=100\) \([R]=80(20 \%\) reacted \()\) \(\mathrm{k}=\dfrac{2.303}{15 \min } \log \left[\dfrac{100}{80}\right]\) \(=1.48 \times 10^{-2} \mathrm{~min}^{-1}\)
CHXII04:CHEMICAL KINETICS
320365
For the first order reaction half-life is \(14 \mathrm{sec}\), the time required for the initial concentration to reduce to \(1 / 8\) of its value is:
320366
A first order reaction has a specific reaction rate of \({\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{se}}{{\rm{c}}^{{\rm{ - 1}}}}\). How much time it take for \({\rm{20\;g}}\) of the reactant to reduce to \({\rm{5\;g}}\) ?
1 \({\rm{138}}{\rm{.6}}\,\,{\rm{sec}}\)
2 \({\rm{346}}{\rm{.5}}\,\,{\rm{sec}}\)
3 \({\rm{693}}{\rm{.0}}\,\,{\rm{sec}}\)
4 \({\rm{238}}{\rm{.6}}\,\,{\rm{sec}}\)
Explanation:
Half life of first order reaction, \(\mathrm{t}_{1 / 2}=\dfrac{0.693}{\mathrm{k}}=\dfrac{0.693}{10^{-2}}=69.3 \mathrm{sec}\) Method - 1 Total time \(=2 \mathrm{t}_{1 / 2}=2 \times 69.3=138.6 \mathrm{sec}\) Method - 2 \(\begin{aligned}& \mathrm{t}=\dfrac{2.303}{\mathrm{k}} \log \dfrac{[\mathrm{A}]_{o}}{[\mathrm{~A}]_{\mathrm{t}}} \\& \mathrm{t}=\dfrac{2.303}{10^{-2}} \log \dfrac{20}{5} \Rightarrow \mathrm{t}=138.6 \mathrm{sec}\end{aligned}\)
CHXII04:CHEMICAL KINETICS
320367
Half life for a first order reaction is 6.93 hour. What is the time required for \(80 \%\) completion of the reaction?
1 12 hours
2 18 hours
3 6 hours
4 16 hours
Explanation:
For first order reaction, \(\mathrm{k}=\dfrac{0.693}{\mathrm{t}_{\frac{1}{2}}}=\dfrac{0.693}{6.93}=0.1\) hour \(^{-1}\) Here, \([\mathrm{A}]_{0}=100,[\mathrm{~A}]_{\mathrm{t}}=100-80=20\) \(\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \log _{10} \dfrac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{\mathrm{t}}}\) \(\therefore \mathrm{t}=\dfrac{2.303}{0.1} \log _{10} \dfrac{100}{20}=23.03 \times \log _{10} 5\) \(=23.03 \times 0.699=16.10\) hours
MHTCET - 2021
CHXII04:CHEMICAL KINETICS
320368
If the rate constant for a first order reaction is \(\mathrm{k}\), the time \((\mathrm{t})\) required for the completion of \(99 \%\) of the reaction is givne by
320364
What is the value of rate constant of first order reaction, if it takes 15 minutes for consumption of \(20 \%\) of reactants?
1 \(1.38 \times 10^{-2} \mathrm{~min}^{-1}\)
2 \(1.48 \times 10^{-2} \mathrm{~min}^{-1}\)
3 \(1.07 \times 10^{-2} \mathrm{~min}^{-1}\)
4 \(1.84 \times 10^{-2} \mathrm{~min}^{-1}\)
Explanation:
Given, \(t=15 \mathrm{~min}\) for \(20 \%\) of reactant to react. For first order reaction, Rate constant \((\mathrm{k})=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\left[\mathrm{R}_{0}\right]}{[\mathrm{R}]}\) where, \(\left[\mathrm{R}_{0}\right]=\) original amount of reactant \([R]=\) reactant remaining unreacted So, \(\left[\mathrm{R}_{0}\right]=100\) \([R]=80(20 \%\) reacted \()\) \(\mathrm{k}=\dfrac{2.303}{15 \min } \log \left[\dfrac{100}{80}\right]\) \(=1.48 \times 10^{-2} \mathrm{~min}^{-1}\)
CHXII04:CHEMICAL KINETICS
320365
For the first order reaction half-life is \(14 \mathrm{sec}\), the time required for the initial concentration to reduce to \(1 / 8\) of its value is:
320366
A first order reaction has a specific reaction rate of \({\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{se}}{{\rm{c}}^{{\rm{ - 1}}}}\). How much time it take for \({\rm{20\;g}}\) of the reactant to reduce to \({\rm{5\;g}}\) ?
1 \({\rm{138}}{\rm{.6}}\,\,{\rm{sec}}\)
2 \({\rm{346}}{\rm{.5}}\,\,{\rm{sec}}\)
3 \({\rm{693}}{\rm{.0}}\,\,{\rm{sec}}\)
4 \({\rm{238}}{\rm{.6}}\,\,{\rm{sec}}\)
Explanation:
Half life of first order reaction, \(\mathrm{t}_{1 / 2}=\dfrac{0.693}{\mathrm{k}}=\dfrac{0.693}{10^{-2}}=69.3 \mathrm{sec}\) Method - 1 Total time \(=2 \mathrm{t}_{1 / 2}=2 \times 69.3=138.6 \mathrm{sec}\) Method - 2 \(\begin{aligned}& \mathrm{t}=\dfrac{2.303}{\mathrm{k}} \log \dfrac{[\mathrm{A}]_{o}}{[\mathrm{~A}]_{\mathrm{t}}} \\& \mathrm{t}=\dfrac{2.303}{10^{-2}} \log \dfrac{20}{5} \Rightarrow \mathrm{t}=138.6 \mathrm{sec}\end{aligned}\)
CHXII04:CHEMICAL KINETICS
320367
Half life for a first order reaction is 6.93 hour. What is the time required for \(80 \%\) completion of the reaction?
1 12 hours
2 18 hours
3 6 hours
4 16 hours
Explanation:
For first order reaction, \(\mathrm{k}=\dfrac{0.693}{\mathrm{t}_{\frac{1}{2}}}=\dfrac{0.693}{6.93}=0.1\) hour \(^{-1}\) Here, \([\mathrm{A}]_{0}=100,[\mathrm{~A}]_{\mathrm{t}}=100-80=20\) \(\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \log _{10} \dfrac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{\mathrm{t}}}\) \(\therefore \mathrm{t}=\dfrac{2.303}{0.1} \log _{10} \dfrac{100}{20}=23.03 \times \log _{10} 5\) \(=23.03 \times 0.699=16.10\) hours
MHTCET - 2021
CHXII04:CHEMICAL KINETICS
320368
If the rate constant for a first order reaction is \(\mathrm{k}\), the time \((\mathrm{t})\) required for the completion of \(99 \%\) of the reaction is givne by
