Answer
Verified
28.5k+ views
Hint: Obtain the order of the reaction from the unit of the rate constant.For the first order reaction, the unit of the rate constant is the reciprocal of time.
Complete step by step answer:
The rate constant for the reaction \[{{\rm{N}}_2}{{\rm{O}}_5}\left( g \right) \to 2{\rm{N}}{{\rm{O}}_2}\left( g \right) + \dfrac{1}{2}{{\rm{O}}_2}\] is \[2.3 \times {10^{ - 2}}{\rm{ se}}{{\rm{c}}^{ - 1}}\].
This suggests that the reaction follows first order kinetics. For the first order reaction, the unit of the rate constant is the reciprocal of time.
For the first order reaction, the rate of the reaction is directly proportional to the first power of the concentration of the reaction.
\[\begin{array}{l}
{\rm{Rate}} \propto \left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]\\
{\rm{Rate}} = k \times \left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]
\end{array}\]
But the rate of the reaction is \[{\rm{Rate}} = - \dfrac{{d\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}}{{dt}}\].
Hence \[{\rm{Rate}} = - \dfrac{{d\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}}{{dt}} = k \times \left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]\] … …(1)
When the above equation is integrated, the following relationship is obtained.
\[\begin{array}{l}
{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0}{e^{ - kt}}\\
{\rm{Rearrange above expression}}\\
{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t}{e^{kt}}
\end{array}\]
This is the same as in option A. Hence, option A is correct.
Upon integration of equation (1), the following reactions are also obtained.
\[\begin{array}{l}
\ln \dfrac{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_0}}}{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_t}}} = kt\\
\log \dfrac{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_0}}}{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_t}}} = \dfrac{{kt}}{{2.303}}
\end{array}\]
Hence, the option B) is incorrect.
Upon integration of equation (1), the following reactions are also obtained.
\[\begin{array}{l}
\ln {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = \ln {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} - kt\\
2.303{\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = 2.303{\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} - kt\\
{\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} - \dfrac{{kt}}{{2.303}}
\end{array}\]
Hence, the option C) is incorrect.
The relation \[{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} + kt\] is incorrect as it shows a linear relationship between concentration and time, however, in first order reaction, the concentration shows exponential decay.
Hence, the option C) is incorrect.
Hence, only option A) is the correct option.
Note:
Please take care of positive and negative signs used in the formula. Also remember that when natural logarithm is converted to logarithm to base 10, it is multiplied with 2.303.
Complete step by step answer:
The rate constant for the reaction \[{{\rm{N}}_2}{{\rm{O}}_5}\left( g \right) \to 2{\rm{N}}{{\rm{O}}_2}\left( g \right) + \dfrac{1}{2}{{\rm{O}}_2}\] is \[2.3 \times {10^{ - 2}}{\rm{ se}}{{\rm{c}}^{ - 1}}\].
This suggests that the reaction follows first order kinetics. For the first order reaction, the unit of the rate constant is the reciprocal of time.
For the first order reaction, the rate of the reaction is directly proportional to the first power of the concentration of the reaction.
\[\begin{array}{l}
{\rm{Rate}} \propto \left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]\\
{\rm{Rate}} = k \times \left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]
\end{array}\]
But the rate of the reaction is \[{\rm{Rate}} = - \dfrac{{d\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}}{{dt}}\].
Hence \[{\rm{Rate}} = - \dfrac{{d\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}}{{dt}} = k \times \left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]\] … …(1)
When the above equation is integrated, the following relationship is obtained.
\[\begin{array}{l}
{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0}{e^{ - kt}}\\
{\rm{Rearrange above expression}}\\
{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t}{e^{kt}}
\end{array}\]
This is the same as in option A. Hence, option A is correct.
Upon integration of equation (1), the following reactions are also obtained.
\[\begin{array}{l}
\ln \dfrac{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_0}}}{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_t}}} = kt\\
\log \dfrac{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_0}}}{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_t}}} = \dfrac{{kt}}{{2.303}}
\end{array}\]
Hence, the option B) is incorrect.
Upon integration of equation (1), the following reactions are also obtained.
\[\begin{array}{l}
\ln {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = \ln {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} - kt\\
2.303{\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = 2.303{\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} - kt\\
{\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} - \dfrac{{kt}}{{2.303}}
\end{array}\]
Hence, the option C) is incorrect.
The relation \[{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} + kt\] is incorrect as it shows a linear relationship between concentration and time, however, in first order reaction, the concentration shows exponential decay.
Hence, the option C) is incorrect.
Hence, only option A) is the correct option.
Note:
Please take care of positive and negative signs used in the formula. Also remember that when natural logarithm is converted to logarithm to base 10, it is multiplied with 2.303.
Recently Updated Pages
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
What is the difference between solvation and hydra class 11 chemistry JEE_Main
IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main
Sodium chloride is purified by passing hydrogen chloride class 11 chemistry JEE_Main
Consider the following oxyanions PO43P2O62SO42MnO4CrO4S2O52S2O72 class 11 chemistry JEE_Main
Other Pages
Which of the following compounds has zero dipole moment class 11 chemistry JEE_Main
Differentiate between mass and inertia class 11 physics JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main
An air capacitor of capacity C10 mu F is connected class 12 physics JEE_Main
According to classical free electron theory A There class 11 physics JEE_Main
Mulliken scale of electronegativity uses the concept class 11 chemistry JEE_Main