Respuesta :
The answer is - For the given second-order reaction, t = 3.63 seconds.
Second-order reaction: It is the reaction in which the rate of reaction depends on either the concentration of two reactant species or on the two times the concentration of single reactant species.
Write the integrated rate law for second-order reaction.
- The integrated rate law is-
[tex]\[\frac{{\rm{1}}}{{{\rm{[A]}}}}{\rm{ = }} & \frac{{\rm{1}}}{{{{{\rm{[A]}}}_{\rm{0}}}}}{\rm{ + kt}}\][/tex]
Where
[tex][A]_0[/tex] is the initial concentration
[tex][A][/tex] is the concentration at time t
k is the rate constant
- Now, as per the question,
[tex][NO_2}] = 0.28\ M[/tex]
[tex][NO_{2}]_0 = 0.62\ M[/tex]
k = 0.54 m-1.s-1
Thus, using the rate law of second-order reaction, solve for time, t-
[tex]\frac{1}{0.28} =\frac{1}{0.62} + (0.54\ m^-1.s^-1) .t\\ \frac{1}{0.28} -\frac{1}{0.62} =(0.54\ m^-1.s^-1) .t\\\\1.96 =(0.54\ m^-1.s^-1) .t\\\\t = \frac{1.96}{0.54} = 3.63\ s[/tex]
- Hence, it would take 3.63 seconds for the concentration of [tex]NO_2[/tex] to decrease from 0. 62 M to 0. 28 M for the given second-order reaction.
To learn more about the second-order reaction, visit:
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