What is the Standard Enthalpy Change for the Reaction 4NH3 + 3O2 --> 2N2 + 6H2O?

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In summary, the standard enthalpy change for the reaction 4NH3 + 3O2 --> 2N2 + 6H2O is -1531 kJ/mol, calculated by subtracting the standard heat of formation of the reactants from the products. The standard heat of formation of gaseous water is -241.826 kJ/mol, which may explain the discrepancy in the given value of -285.83 kJ/mol. It is unclear whether the problem specified gaseous or liquid water, which could affect the calculation. There is also some ambiguity regarding whether the resulting value should be divided by the number of moles of each reactant or product.
  • #1
jumbogala
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Homework Statement


Determine the standard enthalpy change for the reaction below.

4NH3 + 3O2 --> 2N2 + 6H2O

Given:
Standard heat of formation of NH3: -45.9 kJ/mol

Standard heat of formation of H2O: -285.83 kJ/mol


Homework Equations





The Attempt at a Solution


I had this question on a quiz and got points taken off, but I can't figure out why. Here's my solution:

{ (6 x -285.83 kJ/mol) + (2 x 0) } - { (3 x 0) + (4 x -45.9 kJ/mol)} = -1531 kJ/mol
 
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  • #2
Because the standard heat of formation of gaseous water is -241.826 kJ/mol.
I'm not sure where you got the -285.83 from, but I believe it's the liquid value.
 
  • #3
Reaction (as entered) doesn't state whether it should be gaseous or liquid water. My thermo is somewhat rusty, but isn't standard state of water liquid?
 
  • #4
Borek said:
Reaction (as entered) doesn't state whether it should be gaseous or liquid water. My thermo is somewhat rusty, but isn't standard state of water liquid?

It is. But it wouldn't be wrong then! :)

I assumed the gas-phase states since, well, it's a gas-phase reaction.
If the problem was stated without specifying H2O (l) or H2O (g) then either would be correct, depending on whether you mean the 'standard enthalpy' in the strict sense, or the 'reaction enthalpy at standard pressure and temperature'.
 
  • #5
There is always a slight possibility that there was a mistake in the test :smile:
 
  • #6
Just to clarify, the reaction was wriiten without states in the problem, and there were no states specified in the heat of formation given either.

Was I supposed to divide the value I got by four?
 
  • #7
That would be enthalpy change for a mole of ammonia, not for the reaction as written. I am not sure if there exeist some precise convention, I have seen it discussed (without any decisive conclusions) by chemistry teachers.
 
  • #8
Yeah, I always thought you didn't divide by anything when it was for an actual reaction. It wouldn't make sense here because it doesn't specify per mole of ammonia vs. per mole of oxygen.

Anyways, thanks for your help.
 

1. What is enthalpy in chemistry?

Enthalpy is a thermodynamic property that measures the total energy of a system. In chemistry, it refers to the energy change that occurs during a chemical reaction or process.

2. How is enthalpy related to heat?

Enthalpy is often referred to as "heat content" because it is a measure of the amount of heat energy in a system. It takes into account the internal energy of a system as well as the pressure and volume of the system.

3. What is the difference between enthalpy and entropy?

Enthalpy and entropy are both thermodynamic properties, but they measure different things. Enthalpy measures the total energy of a system, while entropy measures the amount of disorder or randomness in a system.

4. How is enthalpy calculated?

Enthalpy is typically calculated using the formula H = U + PV, where H is enthalpy, U is internal energy, P is pressure, and V is volume. It can also be calculated by measuring the heat absorbed or released during a reaction.

5. What are some real-world applications of enthalpy in chemistry?

Enthalpy is used in various industries such as pharmaceuticals, food and beverage, and petroleum refining. It is also important in understanding and predicting chemical reactions, as well as in the development of new materials and technologies.

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