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A food substance kept at 0°C becomes rotten (as determined by a good quantitative test) in 8.3 days. The same food rots in 10.6 hours at 30°C. Assuming the kinetics of the microorganisms enzymatic action is responsible for the rate of decay, what is the activation energy for the decomposition process? Hint: Rate varies INVERSELY with time; a faster rate produces a shorter decomposition time. 1.67.2 kJ/mol 2.2.34 kJ/mol 3.23.4 kJ/mol 4.0.45 kJ/mol

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Answer:

1.   67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

[tex]In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})[/tex]

Given that:

time [tex]t_1[/tex] = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time [tex]t_2[/tex] = 10.6 hours

Temperature [tex]T_1[/tex] = 0° C = (0+273 )K = 273 K

Temperature [tex]T_2[/tex] = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since [tex](\frac{k_2}{k_1}=\frac{t_2}{t_1})[/tex]

Then we can rewrite the above expression as:

[tex]In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})[/tex]

[tex]In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})[/tex]

[tex]2.934 = \frac{E_a}{8.314}(\frac{30}{82719})[/tex]

[tex]2.934 = \frac{30E_a}{687725.766}[/tex]

[tex]30E_a = 2.934 *687725.766[/tex]

[tex]E_a = \frac{2.934 *687725.766}{30}[/tex]

[tex]E_a =67255.58 \ J/mol[/tex]

[tex]E_a =67.2 \ kJ/mol[/tex]

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