Question

The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperature, the chemical reactions associated with the cooking of food take place at a greater rate. (a) Some food cooks fully in 7.00 min in a pressure cooker at 113.0°C and in 49.0 minutes in an open pot at 100.0°C. Calculate the average activation energy for the reactions associated with the cooking of this food. kJ mol-1 (b) How long will the same food take to cook in an open pot of boiling water at an altitude of 10000 feet, where the boiling point of water is 89.8 °C? min

Answer 1

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

$T_1 = 100^0 C = 100+273 = 373 \ K \\ \\ T_2 = 113^0 C = 113 + 273 = 386 \ K$

$R_1 = \dfrac{1}{7}$

$R_2 = \dfrac{1}{49}$

Thus; $\dfrac{R_2}{R_1} = 7$

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

$In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})$

1.9459 = $\dfrac{Ea}{8.314}* 9.0292 *10^{-5}$

$1.9459*8.314 = Ea * 9.0292*10^{-5}$

$16.1782126= Ea * 9.0292*10^{-5}$

$Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}$

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

b)

here;

$T_2 = 386 \ K \\ \\T_1 = (89.8 + 273)K = 362.8 \ K$

$In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})$

$In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})$

$In (\dfrac{R_2}{R_1}) = 0.00357$

$\dfrac{R_2}{R_1}= e^{0.00357}$

$\dfrac{R_2}{R_1}= 1.0035$

where ;

$R_2 = \dfrac{1}7{}$

$R_1 = \dfrac{1}{t}$

Now;

$\dfrac{t}{7}= 1.0035$

t = 7.0245 mins

Therefore; it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.