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Marizza181 [45]
3 years ago
9

A sample of CaCO3(s) is introduced into a sealed container of volume 0.638 L and heated to 1000 K until equilibrium is reached.

The Kp for the reaction CaCO3(s)⇌CaO(s)+CO2(g) is 3.8×10−2 at this temperature. Calculate the mass of CaO(s) that is present at equilibrium.
Chemistry
1 answer:
JulsSmile [24]3 years ago
7 0

Answer:

the mass of CaO present at equilibrium is, 0.01652g

Explanation:

K_p = [CO_2] = 3.8×10⁻²

Now we have to calculate the moles of CO₂

Using ideal gas equation,

PV =nRT

P = pressure of gas = 3.8×10⁻²

T = temperature of gas = 1000 K

V = volume of gas = 0.638 L

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mole.k

\frac{3.8 * 10^2 * 0.638}{0.0821 * 1000} \\= 2.95 * 10^-^4

Now we have to calculate the mass of CaO

mass = 2.95 * 10 ⁻⁴ × 56

= 0.01652g

Therefore,

the mass of CaO present at equilibrium is, 0.01652g

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The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

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