The small bags of silica gel you often see in a new shoe box are placed there to control humidity. Despite its name, silica gel is a solid. It is a chemically inert, highly porous, amorphous form of SiO2. Water vapor readily adsorbs onto the surface of silica gel, so it acts as a desiccant. Despite not knowing mechanistic details of the adsorption of water onto silica gel, from the information provided, you should be able to make an educated guess about the thermodynamic characteristics of the process.
Predict the signs of ΔG, ΔH, and ΔS.
Sapling #10 Week 5/6
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Re: Sapling #10 Week 5/6
JaylinWangDis1L wrote:The small bags of silica gel you often see in a new shoe box are placed there to control humidity. Despite its name, silica gel is a solid. It is a chemically inert, highly porous, amorphous form of SiO2. Water vapor readily adsorbs onto the surface of silica gel, so it acts as a desiccant. Despite not knowing mechanistic details of the adsorption of water onto silica gel, from the information provided, you should be able to make an educated guess about the thermodynamic characteristics of the process.
Predict the signs of ΔG, ΔH, and ΔS.
Without knowing anything about how the desiccant works, you can say that the system is becoming more ordered as the water vapor absorbs onto the surface of the gel. Therefore, the entropy would be less than 0/decreasing.
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Re: Sapling #10 Week 5/6
In this case, water vapor is going from moving as a gas to settling on a solid, therefore losing mobility. This process is happening spontaneously which means that deltaG is negative.
Because it is going from gas to solid it is an exothermic reaction, which tells you your deltaH.
Again, because it is losing mobility, the entropy will decrease, therefore also telling you your deltaS.
Because it is going from gas to solid it is an exothermic reaction, which tells you your deltaH.
Again, because it is losing mobility, the entropy will decrease, therefore also telling you your deltaS.
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Re: Sapling #10 Week 5/6
Delta G: Since the process of water absorption happens spontaneously, we know that delta G must be negative.
Delta S: The water molecules are absorbed by the silica gel bag and so are more limited in possible positions to be in. They were originally in the gas phase and occupying a large space with more possible positions to be in. Therefore, for the system, Delta S is negative as the water molecules occupy less space and positions.
Delta H: Delta G = Delta H - (T)(Delta S). Since Delta S is negative, -(T)(Delta S) would be a positive value. In order for Delta G to be negative, Delta H must be very negative so that |Delta H|>|T(Delta S)|.
Delta S: The water molecules are absorbed by the silica gel bag and so are more limited in possible positions to be in. They were originally in the gas phase and occupying a large space with more possible positions to be in. Therefore, for the system, Delta S is negative as the water molecules occupy less space and positions.
Delta H: Delta G = Delta H - (T)(Delta S). Since Delta S is negative, -(T)(Delta S) would be a positive value. In order for Delta G to be negative, Delta H must be very negative so that |Delta H|>|T(Delta S)|.
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Re: Sapling #10 Week 5/6
For ∆G:
The water absorption on the silica gel is a spontaneous process. Therefore, ∆G is negative.
For ∆H:
Due to the absorption process, there is an exothermic reaction. For exothermic reaction ∆H is less than zero. (So ∆H is negative)
For ∆S:
There is a decrease in entropy due to the water molecules becoming immobilize. (so ∆S is negative)
The water absorption on the silica gel is a spontaneous process. Therefore, ∆G is negative.
For ∆H:
Due to the absorption process, there is an exothermic reaction. For exothermic reaction ∆H is less than zero. (So ∆H is negative)
For ∆S:
There is a decrease in entropy due to the water molecules becoming immobilize. (so ∆S is negative)
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Re: Sapling #10 Week 5/6
Since we know water vapor readily absorbs into the silica gel's surface, this process must be spontaneous (since it happens with no outside influence), so delta G must be negative.
Since the water vapor is going from being free in the air to being absorbed/bound to the solid surface, its entropy decreases, so delta S is negative.
In order for the overall reaction to still have a negative delta G if the delta S is negative, we must have a negative delta H (and it must be sufficiently negative to overpower the T delta S component). This is because delta G = delta H - T delta S, and if delta S is negative, we are actually adding the T delta S term, so delta H must be negative to make sure delta G is still negative.
Since the water vapor is going from being free in the air to being absorbed/bound to the solid surface, its entropy decreases, so delta S is negative.
In order for the overall reaction to still have a negative delta G if the delta S is negative, we must have a negative delta H (and it must be sufficiently negative to overpower the T delta S component). This is because delta G = delta H - T delta S, and if delta S is negative, we are actually adding the T delta S term, so delta H must be negative to make sure delta G is still negative.
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