Respuesta :
Explanation:
The given data is as follows.
[tex](t_{\frac{1}{2}})_{1}[/tex] = 24 ms,
[tex](t_{\frac{1}{2}})_{2}[/tex] = 39 ms,
where, [tex](t_{\frac{1}{2}})_{1}[/tex] = half-life for the forward reaction
[tex](t_{\frac{1}{2}})_{2}[/tex] = half-life for the backward reaction
It is known that the formula for first-order reaction is as follows.
[tex]t_{\frac{1}{2}} = \frac{0.693}{K}[/tex]
Therefore,
[tex]K_{1} = \frac{0.693}{(t_{\frac{1}{2}})_{1}}[/tex]
= [tex]\frac{0.693}{24 ms}[/tex]
= 0.0289 [tex]ms^{-1}[/tex]
[tex]K_{2} = \frac{0.693}{(t_{\frac{1}{2}})_{2}}[/tex]
= [tex]\frac{0.693}{39 ms}[/tex]
= 0.0178 [tex]ms^{-1}[/tex]
Hence, formula for the relaxation time is as follows.
[tex]\tau = \frac{1}{K_{1} + k_{2}}[/tex]
= [tex]\frac{1}{(0.0289 + 0.0178) ms^{-1}}[/tex]
= 21.41 ms
Thus, we can conclude that the corresponding relaxation time(s) for the return to equilibrium after a temperature jump is 21.41 ms.
