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

Answer:

1.   67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

$In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})$

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 $(\frac{k_2}{k_1}=\frac{t_2}{t_1})$

Then we can rewrite the above expression as:

$In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})$

$In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})$

$2.934 = \frac{E_a}{8.314}(\frac{30}{82719})$

$2.934 = \frac{30E_a}{687725.766}$

$30E_a = 2.934 *687725.766$

$E_a = \frac{2.934 *687725.766}{30}$

$E_a =67255.58 \ J/mol$

$E_a =67.2 \ kJ/mol$