= Solution
Subtract the new equilibrium <concentration> and divide by the initial displacement:
$$
\frac{[\mathrm{trans}](t)-[\mathrm{trans}]_{\rm eq2}}{[\mathrm{trans}]_{\rm eq1}-[\mathrm{trans}]_{\rm eq2}}
=e^{-(k'_1+k'_{-1})t}.
$$
At the halfway <concentration> this ratio is $1/2$, so
$$
\boxed{t_{1/2}=\frac{\ln2}{k'_1+k'_{-1}}=465\,\mathrm s\simeq7.75\,\mathrm{min}.}
$$
This is the half-time of the displacement from equilibrium. It is not the <time> for the absolute trans <concentration> to fall to half its initial value; that value lies far below the attainable new equilibrium plateau. The corresponding <chemical relaxation time> is $671$ s.
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