Reaction enthelpy at non-standard conditions

In summary, the conversation discussed the calculation of reaction enthalpy for a given chemical reaction at non-standard conditions. The methodology for computing the formation enthalpies at these conditions was explained, including taking into account the increase in temperature and pressure. The relevance and method for calculating the pressure correction were also discussed. However, it was determined that the pressure correction may not be significant and may not be worth the effort at a high-school level.
  • #1
giokara
9
0
Hi

I am tutoring a pupil in high-school out of necessity (no other teachers avail), while I am not so acquainted anymore with chemistry. The pupil showed me a question concerning the calculation of the reaction enthalpy which looked somewhat as follows:

1. Problem statement, all variables and given/known data

Calculate the reaction-enthalpy for the reaction:

[itex]C_3H_8 (g) + 5O_2 (g) \rightarrow 3CO_2 (g) + 4H_2O (g)[/itex]

at 30[itex]^\mathrm{o}[/itex]C and 1.08 atm.
The formation enthalpies at standard conditions and the heat capacities at constant pressure were known.

Homework Equations



The Attempt at a Solution



NOTE: I am not sure anymore about the exact reaction or the exact temperature and pressure, the above is meant as a generic example of the exercise that the pupil showed me.

As I remember, first the formation enthalpies at the non-standard conditions should be computed. This should be done in two steps:

1) Take into account the increase in temperature assuming constant pressure: use [itex]\Delta H = C_p\Delta T[/itex].
2) Take into account the increase in pressure assuming constant temperature.

I cannot remember how the second step should be computed, neither did I find some appropriate formula during a quick search.
Is the current approach the correct one? Could it be possible to give me some more information about how to proceed?

Thanks in advance
Giorgos
 
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  • #2
giokara said:
Hi

I am tutoring a pupil in high-school out of necessity (no other teachers avail), while I am not so acquainted anymore with chemistry. The pupil showed me a question concerning the calculation of the reaction enthalpy which looked somewhat as follows:

1. Problem statement, all variables and given/known data

Calculate the reaction-enthalpy for the reaction:

[itex]C_3H_8 (g) + 5O_2 (g) \rightarrow 3CO_2 (g) + 4H_2O (g)[/itex]

at 30[itex]^\mathrm{o}[/itex]C and 1.08 atm.
The formation enthalpies at standard conditions and the heat capacities at constant pressure were known.

Homework Equations



The Attempt at a Solution



NOTE: I am not sure anymore about the exact reaction or the exact temperature and pressure, the above is meant as a generic example of the exercise that the pupil showed me.

As I remember, first the formation enthalpies at the non-standard conditions should be computed. This should be done in two steps:

1) Take into account the increase in temperature assuming constant pressure: use [itex]\Delta H = C_p\Delta T[/itex].
2) Take into account the increase in pressure assuming constant temperature.

I cannot remember how the second step should be computed, neither did I find some appropriate formula during a quick search.
Is the current approach the correct one? Could it be possible to give me some more information about how to proceed?

Thanks in advance
Giorgos
Yes. Your methodology is correct. As far as the pressure effect is concerned, it is hard to imagine how the pressure correction could be significant at that pressure and with that pressure change.

If you still want to get the pressure correction, you calculate the molar enthalpy change for each of the species individually (as a pure species) using:
[tex]\frac{∂H}{∂P}=(V-T\frac{∂V}{∂T})dP[/tex]
To do this calculation, you need to know the PVT behavior of each pure species slightly beyond the ideal gas range (with an equation of state). Probably, Van Der Waals would be adequate, but you need to know the Van Der Waals parameters for each species). In my judgement, it's not worth the effort.

Chet
 
  • #3
Hi Chet,

Thanks for your answer.
The pressure change is indeed minimal, and I had difficulties understanding how this is relevant at high-school level.
But I'm glad I understand how it should be taken into account.

Giorgos
 

Related to Reaction enthelpy at non-standard conditions

What is reaction enthalpy?

Reaction enthalpy is a thermodynamic property that describes the heat released or absorbed during a chemical reaction at constant pressure. It is a measure of the change in energy of the system as reactants are converted to products.

How is reaction enthalpy calculated?

The reaction enthalpy is calculated by taking the difference between the enthalpy of the products and the enthalpy of the reactants. This can be determined experimentally by measuring the heat released or absorbed during the reaction, or it can be calculated using the enthalpy of formation values for each compound involved.

What are non-standard conditions?

Non-standard conditions refer to any conditions in which the temperature, pressure, or concentrations of reactants and products do not match the standard conditions of 298 K, 1 atm, and 1 M, respectively. This can include changes in temperature, pressure, or concentrations during a reaction, or reactions occurring under different experimental conditions.

How does reaction enthalpy change with temperature?

The reaction enthalpy is dependent on temperature and can change as the temperature is varied. In exothermic reactions, the enthalpy decreases as temperature increases, while in endothermic reactions, the enthalpy increases with temperature. This relationship can be described by the Van't Hoff equation.

What is the significance of reaction enthalpy at non-standard conditions?

Understanding reaction enthalpy at non-standard conditions is important for predicting and controlling the behavior of chemical reactions. It can help determine the feasibility and direction of a reaction, as well as the amount of heat released or absorbed. This information is crucial in industries such as chemical production, where optimizing reaction conditions can improve efficiency and yield.

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