Question

7.5 The following statement appears on the side of a milk carton: The contents will spoil in 8 days if stored at 5°C. The contents will spoil in 12 hours if stored at 30°C. If the mechanism and definition of spoilage remain constant in this tem- perature range, how long will the milk remain unspoiled at 20°C? 1. ---nhua nakımer fiber is being treated to form polymer crystals. You

Answer 1

7.5.

Rate of reaction $\propto$   

Time

Or, rate (R)  $\propto$ Rate constant (K)

At , T₁ = 5⁰c = 278 K, t₁ = 8 days = 8 × 24 = 192 hours,

rate constant = K₁

At T_{2 ⁼} 30⁰c = 303 K, t₂ = 12 hours , rate constant = K₂ .

Then, _{$\frac{K_1}{K_2}$} = _{$\frac{t_2}{t_1}$} = (12/196) = 0.0625

By , Arrhenious equation.

ln ( _{$\frac{t_2}{t_1}$})  = $\frac{Ea}{R}$ × [  $\frac{1}{T_2}$ -  $\frac{1}{T_1}$ ]

Or, ln (0.0625) = $\frac{Ea}{R}$ × ( $\frac{1}{303}$ - $\frac{1}{278}$ )

Or, - 2.77 = $\frac{Ea}{R}$ × (- 0.000297)

Or, Ea/R = (2.77/0.000297)

Or, Ea/R = 9333 K.

Now, let at 20^oc ( 273+20 = 293 K ) , time required to spoil is t₃

Then,

ln ($\frac{t_2}{t_3}$ )  = $\frac{Ea}{R}$ × [ $\frac{1}{T_2}$ - $\frac{1}{T_3}$ ]

Or, ln ($\frac{12}{t_3}$) = 9333 × [ $\frac{1}{303}$ - $\frac{1}{293}$ ]

Or, ln ($\frac{12}{t_3}$) = 9333 × (-0.000113)

Or, ln($\frac{12}{t_3}$) = -1.05

Or, ($\frac{12}{t_3}$) = e^-1.05

Or, ($\frac{12}{t_3}$) = 0.35

Or, t₃ = (12/0.35) = 34 hours (2 significant figures) .

So, the milk remain unspoiled upto 34 hours at 20⁰ c .