Question

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

Answer 1

1. 67.2 kJ/mol

Step-by-step explanation:

Using the derived expression from Arrhenius Equation

`In \ ((k_2)/(k_1)) = (E_a)/(R)((T_2-T_1)/(T_2*T_1))`

Given that:

time
`t_1` = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time
`t_2` = 10.6 hours

Temperature
`T_1` = 0° C = (0+273 )K = 273 K

Temperature
`T_2` = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since
`((k_2)/(k_1)=(t_2)/(t_1))`

Then we can rewrite the above expression as:

`In \ ((t_2)/(t_1)) = (E_a)/(R)((T_2-T_1)/(T_2*T_1))`

`In \ ((199.2)/(10.6)) = (E_a)/(8.314)((303-273)/(273*303))`

`2.934 = (E_a)/(8.314)((30)/(82719))`

`2.934 = (30E_a)/(687725.766)`

`30E_a = 2.934 *687725.766`

`E_a = (2.934 *687725.766)/(30)`

`E_a =67255.58 \ J/mol`

`E_a =67.2 \ kJ/mol`