Standard enthelpy of formation water

In summary: Yes, using Hess' Law, the standard enthalpy of formation for water vapor would be higher than the standard enthalpy of formation for liquid water. This is because the heat released during the reaction in post #5 is less than the heat required to condense one mole of water vapor to form liquid water in post #6. Therefore, the answer to the question in post #1 is (A) -480.7 kJ/mol.In summary, using Hess' Law, it can be determined that the standard enthalpy of formation for water vapor is higher than the standard enthalpy of formation for liquid water. This is due to the fact that more heat is released when reacting hydrogen and oxygen to form water compared to the
  • #1
brake4country
216
7

Homework Statement


The standard enthalpy of formation for liquid water is:
H2 = 1/2O2→H2O ΔHf=-285.8 kJ/mol
Which of the following could be the standard enthalpy of formation for water vapor?
(A) -480.7 kJ/mol
(B) -285.8 kJ/mol
(C) -241.8 kJ/mol
(D) +224.6 kJ/mol

Homework Equations


NA

The Attempt at a Solution


The equation above shows that condensation occurs, which is an exothermic process giving a negative ΔH. Perhaps I am not understanding the question because how can the standard enthalpy of formation of water vapor be determined from the above equation?
 
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  • #2
brake4country said:

Homework Statement


The standard enthalpy of formation for liquid water is:
H2 = 1/2O2→H2O ΔHf=-285.8 kJ/mol
Which of the following could be the standard enthalpy of formation for water vapor?
(A) -480.7 kJ/mol
(B) -285.8 kJ/mol
(C) -241.8 kJ/mol
(D) +224.6 kJ/mol

Homework Equations


NA

The Attempt at a Solution


The equation above shows that condensation occurs, which is an exothermic process giving a negative ΔH. Perhaps I am not understanding the question because how can the standard enthalpy of formation of water vapor be determined from the above equation?
Is the enthalpy of water vapor higher or lower than the enthalpy of liquid water? Is the heat of vaporization of water higher or lower than the heat given off when H2 and O2 react to form water?

Chet
 
  • #3
Ok, so the enthalpy of fusion is lower in energy than the enthalpy of water vapor. ΔHvap is higher than ΔHfus. When I look at a heating curve, this true. Well, wouldn't we expect the standard enthalpy of formation of water vapor to be higher than the standard enthalpy of formation of liquid water based on a heating curve?
 
  • #4
brake4country said:
Ok, so the enthalpy of fusion is lower in energy than the enthalpy of water vapor. ΔHvap is higher than ΔHfus. When I look at a heating curve, this true. Well, wouldn't we expect the standard enthalpy of formation of water vapor to be higher than the standard enthalpy of formation of liquid water based on a heating curve?
Who said anything about fusion? How did ice get into the discussion? The standard heat of formation of either liquid water or water vapor is the result of forming chemical bonds between the hydrogen atoms and the oxygen atoms.

Chet
 
  • #5
Alright, this is where I think my confusion is. My book does not clearly define these different ΔH's that well. We discussed standard heat of formation in a previous thread which defines as the change in enthalpy for a reaction that creates one mole of that compound from its raw elements in their standard state. We know that if the ΔHf is negative, then the products are more stable than the reactants. To see this more clearly, I wrote down the equations for what is known and what is being asked:

H2(g) + 1/2O2(g) →H2O (l) ΔHf = -285.8 kJ/mol

H2(g) + 1/2O2(g) →H2O(g) ΔHf = ?

For the first equation, condensation occurs, which releases heat (this makes sense). But the second equation describes a gas going to a gas, so how can the sign for the ΔHf be determined? Would we assume that it is lower since it is not technically going through a phase change?
 
  • #6
brake4country said:
Alright, this is where I think my confusion is. My book does not clearly define these different ΔH's that well. We discussed standard heat of formation in a previous thread which defines as the change in enthalpy for a reaction that creates one mole of that compound from its raw elements in their standard state. We know that if the ΔHf is negative, then the products are more stable than the reactants. To see this more clearly, I wrote down the equations for what is known and what is being asked:

H2(g) + 1/2O2(g) →H2O (l) ΔHf = -285.8 kJ/mol

H2(g) + 1/2O2(g) →H2O(g) ΔHf = ?

For the first equation, condensation occurs, which releases heat (this makes sense). But the second equation describes a gas going to a gas, so how can the sign for the ΔHf be determined? Would we assume that it is lower since it is not technically going through a phase change?
The easiest way to do this is to use Hess' Law. What is ΔH for:

H2O(l)-->H2O(v)

Chet
 
  • #7
Well, I would write it down, but the question does not provide that information. That's why I'm so confused! They just give multiple choice answer to choose from so I assume that they want you to try to estimate the ΔHf.
 
  • #8
brake4country said:
Well, I would write it down, but the question does not provide that information. That's why I'm so confused! They just give multiple choice answer to choose from so I assume that they want you to try to estimate the ΔHf.
Do you think that more heat is given off when you react hydrogen and oxygen to form one mole of water, or when you condense one mole of water vapor to form one mole of liquid water? At least tell me what you think the sign of ΔH in post #6 is.
 
  • #9
Response for post #6: Condensation releases heat so ΔH <0 and vaporization requires energy so ΔH>0.
 
  • #10
Response to post #8: I would predict that condensation would be less since the water vapor is already in the H2O molecular structure.
 
  • #11
brake4country said:
Response to post #8: I would predict that condensation would be less since the water vapor is already in the H2O molecular structure.
Both this and your previous post are correct. Now, can you apply Hess' law to the first reaction equation in post #5 and the phase change reaction in post #6 to determine the answer to the question posed in post #1?

Chet
 

Related to Standard enthelpy of formation water

What is standard enthalpy of formation water?

The standard enthalpy of formation of water is the enthalpy change that occurs when one mole of water is formed from its elements, oxygen and hydrogen, under standard conditions of temperature and pressure.

Why is standard enthalpy of formation water important?

The standard enthalpy of formation water is an important thermodynamic property that allows scientists to determine the energy change associated with a chemical reaction involving water. It is also used to calculate the standard enthalpy of reaction for various chemical processes.

How is standard enthalpy of formation water measured?

The standard enthalpy of formation water is measured using calorimetry, which involves measuring the heat exchange between a reaction and its surroundings. The enthalpy change can then be calculated using the heat capacity of the system and the temperature change.

What factors affect the standard enthalpy of formation water?

The standard enthalpy of formation water is affected by factors such as temperature, pressure, and the physical state of the water (solid, liquid, or gas). Changes in any of these factors can alter the energy required to form water and thus affect its enthalpy of formation.

How does standard enthalpy of formation water relate to other thermodynamic properties?

The standard enthalpy of formation water is related to other thermodynamic properties such as standard enthalpy of reaction, standard enthalpy of formation of other compounds, and heat capacity. These properties can be used together to calculate the enthalpy change for a chemical reaction involving water.

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