Chemical Kinetics and Nuclear ChemistryHard
Question
To very good approximations, the cooling of a hot body to room temperature follows first-order kinetics (in this case, however, the unit that is changing is temperature (in kelvin), not molarity). If the rate constant for a body is 0.04 s−1, then [ln 2 = 0.7, ln(323/25) = 2.6]
Options
A.the time taken for that body to go from 323°C to 25°C is 17.5 s.
B.the time taken for that body to go from 1192 K to 298 K is 35 s.
C.the time taken for that body to go from 323°C to 25°C is 65 s.
D.the time taken for that body to go from 1192 K to 298 K is 130 s.
Solution
$- \frac{d\theta}{dt} = K.\theta \Rightarrow Kt = \ln\frac{\theta_{o}}{\theta}$
$(a)t = \frac{1}{K}.\ln\frac{\theta_{o}}{\theta} = \frac{1}{0.04}.\ln\frac{596}{298} = 17.5\text{ sec}$
$(b)t = \frac{1}{0.04}.\ln\frac{1192}{298} = 35\text{ sec}$
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