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Answer:
Kp = 0.81666
Explanation:
Pressure of PCl₅ = 0.500 atm
Considering the ICE table for the equilibrium as:
PCl₅ (g) ⇔ PCl₃ (g) + Cl₂ (g)
t = o 0.500
t = eq -x x x
--------------------------------------------- --------------------------
Moles at eq: 0.500-x x x
Given the pressure of PCl₅ at equilibrium = 0.150 atm
Thus, 0.500 - x = 0.150
x = 0.350 atm
The expression for the equilibrium constant is:
So,
x = 0.350 atm
Thus,
<u>Thus, Kp = 0.81666</u>
Explanation:
The given data is as follows.
Volume of water = 0.25
Density of water = 1000
Therefore, mass of water = Density × Volume
=
= 250 kg
Initial Temperature of water () =
Final temperature of water =
Heat of vaporization of water () at
is 2133 kJ/kg
Specific heat capacity of water = 4.184 kJ/kg/K
As 25% of water got evaporated at its boiling point () 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)
= 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 = = 37 kJ/s or kW
The power rating of electric heating element is 37 kW.
Hence, heat required to raise the temperature from to
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
=
= 3392 sec
Time required to raise the temperature from to
of 0.25
water is calculated as follows.
= 56 min
Thus, we can conclude that the time required to raise the temperature is 56 min.