The decomposition of SOCl2 is first-order in SOCl2. If the half-life for the reaction is 4.1 hr, how long would it take for the
concentration of SOCl2 to drop from 0.36 M to 0.045 M?
1 answer:
4.1 h = 14760 s
<span>t 1/2 = ln 2 / k </span>
<span>k = rate reaction = 4.97 x 10^-5 </span>
<span>ln 0.045 / 0.36 = - 4.97 x 10^-5 t </span>
<span>2.08 = 4.97 x 10^-5 t </span>
<span>t = 41839.9 s = 11 h 37 min 19 s</span>
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