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.

Answer 2

a). The activation energy given by the reactions related to the cooking of food in the pressure cooker would be:

$Ea = 179.176 kJ/mol$

b). The time duration that is taken by the same food to cook in an open vessel would be:

$7.0245 mins$

Activation Energy

a). Given that,

Temperature $1$  $= 100$° C

Temperature $2$ $= 113$° C

In Kelvin,

Temperature  $1$ $= 100 + 273$

$= 373 K$

Temperature  $2$ $= 113 + 273$

$= 386 K$

$R_{1} = 1/7\\R_{2} = 1/49$

∵ $R_{2}/R_{1} = 49/7 = 7$

It is given that at  $113$ rate $=$ $7$ × $100$°C

Therefore,

$Ea/8.314 (1/373 - 1/386) =$ $In(7)$

so,

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

∵ Activation energy $= 179.176 kJ/mol$

b). As we know,

$T_{2}$ $= 386 K$

$T_{1}$ $= (89. 8 + 273)$

$= 362.8 K$

by employing the formulae,

$In(\frac{R_{2} }{R_{1} }) = \frac{Ea}{R} (1/T_{1} - 1/T_{2})$

$In(\frac{R_{2} }{R_{1} }) = 179.176/8.314 (1/362.8 - 1/386)$

By solving this, we get

$R_{2}/R_{1} = 1.0035$

Thus,

$R_{2} = 1/7$

$R_{1} = 1/t$

∵ t $= 7.0245 min$

Thus, the time duration would be $7.0245 minutes$.

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