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2 answers:
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Answer:
Explanation:
Given that:
number of moles of super cooled liquid water = 1
Melting enthalpy of ice = 6020 J/mol
Freezing point =0 °C = (0 + 273 K)= 273 K
The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K
Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)
Δ S = -1.4 J/K
Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K
Entropy change of universe = entropy change of the system+ entropy change of the surrounding
According to the second law of thermodynamics
Entropy change of universe >0
SO,
Entropy change of the system + entropy change in the surrounding > 0
Entropy change in the surrounding > - entropy change of the system
Entropy change in the surrounding > - (- 23.53 J/K)
Entropy change in the surrounding > 23.53 J/K
b) Make some comments on entropy changes from the obtained data.
From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.
Explanation :
The balanced chemical reaction will be,
By the stiochiometry, 3 moles of solid copper(II)oxide react with 2 moles of ammonia gas to give 3 moles of copper metal, 1 mole of nitrogen gas and 3 moles of liquid water.
The states of matter of each elements and compound is,
Copper(II)oxide is in solid state
Ammonia is in gaseous state
Copper metal is in solid state
Nitrogen is in gaseous state
Water is in liquid state