True Balance forces can not change the motion or direction of an object.
If there is a close container with some water, the following procedures take place.
Initially, the system contains only liquid, and air above it. As evaporation starts (the rate of evaporation is constant for the specific temperature of the water), the molecules from the surface of the liquid escape into vapour state, in the confined space above. Therefore, the level of liquid falls.
Then starts the process of condensation. This is the conversion of vapour into liquid. Initially, escaped molecules (from liquid state) move randomly in all directions and collide with one another. As more and more molecules enter the confined space, some slow-moving molecules are pushed back. They collide with the surface of the liquid to reconvert into liquid.
In the initial stages, the rate of evaporation (constant) is more than the rate of condensation because only small number of molecules are present in the gaseous state. The rate of condensation thereafter gradually increases as the number of molecules in the gaseous phase increases. Finally, a stage is reached when the rate of the two opposing processes is the same.
The state where the rate of evaporation becomes equal to the rate of condensation is called a state of dynamic equilibrium. In such a state, although the amount of liquid level in the container does not change, evaporation has not stopped and the system is not at rest. In fact, the number of molecules, which escape from the liquid to the gaseous phase (due to evaporation), becomes equal to the number of vapour molecules that return to the liquid
Answer:
+523 kJ.
Explanation:
The following data will be used to calculate the average C-S bond energy in CS2(l).
S(s) ---> S(g)
ΔH = 223 kJ/mol
C(s) ---> C(g)
ΔH = 715 kJ/mol
Enthalpy of formation of CS2(l)
ΔH = 88 kJ/mol
CS2(l) ---> CS2(g)
ΔH = 27 kJ/mol
CS2(g) --> C(g) + 2S(g)
So we must construct it stepwise.
1: C(s) ---> C(g) ΔH = 715 kJ
2: 2S(s) ---> 2S(g) ΔH = 446 kJ
adding 1 + 2 = 3
ΔH = 715 + 446
= 1161 kJ
3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ
4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ
adding (reversed 3) from 4 = 5
ΔH = -1161 + 88
= -1073 kJ
5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ
6: CS2(l) ---> CS2(g) ΔH = 27 kJ
adding 5 + 6 = 7
ΔH = -1073 + 27
= -1046 kJ
7. C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ
Reverse and divide by 2 for C-S bond enthalpy
= -(-1046)/2
= +523 kJ.
Its C, Outermost Electrons.
