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Alla [95]
3 years ago
9

The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur

e, 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
Chemistry
1 answer:
vazorg [7]3 years ago
4 0

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.

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