Are all exothermic processes exergonic?


by MathewsMD
Tags: exergonic, exothermic, processes
MathewsMD
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Jan15-14, 06:18 PM
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Are all exothermic processes exergonic?

What is exactly meant by the term of standard enthalpy of fusion? For example, if you have water that has a melting point at 273.15 K, what exactly is meant/measured by this enthalpy when it is already fully melted at 298.15 K?

Just had these questions that were slightly confusing me,
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Quote Quote by MathewsMD View Post
What is exactly meant by the term of standard enthalpy of fusion? For example, if you have water that has a melting point at 273.15 K, what exactly is meant/measured by this enthalpy when it is already fully melted at 298.15 K?
Enthalpy of fusion describes process of fusion - no fusion, no fusion enthalpy, no problem.

That being said, if you change the pressure you will move the melting point - then, enthalpy of fusion will describe the melting at the temperature it happens. But it will be a slightly different value (numerically).
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Quote Quote by Borek View Post
Enthalpy of fusion describes process of fusion - no fusion, no fusion enthalpy, no problem.

That being said, if you change the pressure you will move the melting point - then, enthalpy of fusion will describe the melting at the temperature it happens. But it will be a slightly different value (numerically).
Just to add on, how exactly is the standard enthalpy of vapourization of water found experimentally? Since water boils at 373.15 K and standard conditions require 298.15K, how is this value measured? I am just slightly confused on what is meant by standard enthalpy (of any reaction) when the reaction taking place only occurs at a temperature and pressure besides 298.15 K and 101.325 kPa.

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Are all exothermic processes exergonic?


Quote Quote by MathewsMD View Post
Just to add on, how exactly is the standard enthalpy of vapourization of water found experimentally? Since water boils at 373.15 K and standard conditions require 298.15K, how is this value measured? I am just slightly confused on what is meant by standard enthalpy (of any reaction) when the reaction taking place only occurs at a temperature and pressure besides 298.15 K and 101.325 kPa.
Water will also boil at 298.15 K if water vapor is the only component present in the gas phase, and the pressure is maintained at the equilibrium vapor pressure of water at 298.15 K.

The standard enthalpy of a reaction is the amount of heat that has to be added starting with pure reactants at the standard temperature and pressure, and ending up with the pure products at the
standard temperature and pressure. To get the heat of reaction at other temperatures, you need to calculate the changes in the enthalpy of the pure products and reactants to take them from 298.15 to the desired temperature. Look up Hess's Law. For ideal gases, the heat of reaction is not a function of pressure.
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Jan23-14, 09:11 AM
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Quote Quote by Chestermiller View Post
Water will also boil at 298.15 K if water vapor is the only component present in the gas phase, and the pressure is maintained at the equilibrium vapor pressure of water at 298.15 K.

The standard enthalpy of a reaction is the amount of heat that has to be added starting with pure reactants at the standard temperature and pressure, and ending up with the pure products at the
standard temperature and pressure. To get the heat of reaction at other temperatures, you need to calculate the changes in the enthalpy of the pure products and reactants to take them from 298.15 to the desired temperature. Look up Hess's Law. For ideal gases, the heat of reaction is not a function of pressure.
Okay, but what if the pure products cannot be isolated at 298.15 K and 1 atm? For example, if you have water and want to figure out the energy required to freeze it...the temperature would have to go lower than 273.15 K for ice to be produced. Would we just bring a piece of ice to room temperature and quickly put it in a calorimeter. Then just measure the enthalpy change for it to melt and reverse the sign of this enthalpy to find the enthalpy change for a reaction in which the original reactant was liquid water that was going to form ice?

With regards to your last point, ΔH = q + pΔV so if volume is kept constant, then the second term cancels. Is this what you meant by "For ideal gases, the heat of reaction is not a function of pressure"?
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Quote Quote by MathewsMD View Post
Okay, but what if the pure products cannot be isolated at 298.15 K and 1 atm? For example, if you have water and want to figure out the energy required to freeze it...the temperature would have to go lower than 273.15 K for ice to be produced. Would we just bring a piece of ice to room temperature and quickly put it in a calorimeter. Then just measure the enthalpy change for it to melt and reverse the sign of this enthalpy to find the enthalpy change for a reaction in which the original reactant was liquid water that was going to form ice?
No. Ice can't exist at thermodynamic equilibrium at 298.15. So you shouldn't find a value of the enthalpy of ice at 298.15 in your tables (unless, they do what they sometimes do and extrapolate to 298.15 using the heat capacity of ice at 273). If the enthalpy of liquid water at 298 is taken as 0, then the enthalpy at 273 is -15 cal/gm, and the enthalpy of ice at 273 is -93 cal/gm (because the heat of fusion is 80 cal/gm). The enthalpy of ice at temperatures below 273 is less than -93. You can also get the enthalpy of water vapor at 298 by taking the heat of vaporization at 298. Vapor can also exist at lower temperatures, and you can get the enthalpy at the lower temperatures by multiplying the temperature difference by the heat capacity, and subtracting. You can also use this to get the heat of vaporization of ice.
With regards to your last point, ΔH = q + pΔV so if volume is kept constant, then the second term cancels. Is this what you meant by "For ideal gases, the heat of reaction is not a function of pressure"?
No. This is not what I meant. First of all, at constant pressure in a closed system, ΔH = q. If you measure the heat of reaction for ideal gases reacting (from the pure reactants to the pure products) at constant system temperature and constant pressure, you get the same value for q at all low pressures, irrespective of the pressure level, and it doesn't even matter if the final pressure is equal to the initial pressure.

Chet
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Jan23-14, 11:13 AM
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Quote Quote by Chestermiller View Post
No. Ice can't exist at thermodynamic equilibrium at 298.15. So you shouldn't find a value of the enthalpy of ice at 298.15 in your tables (unless, they do what they sometimes do and extrapolate to 298.15 using the heat capacity of ice at 273). If the enthalpy of liquid water at 298 is taken as 0, then the enthalpy at 273 is -15 cal/gm, and the enthalpy of ice at 273 is -93 cal/gm (because the heat of fusion is 80 cal/gm). The enthalpy of ice at temperatures below 273 is less than -93. You can also get the enthalpy of water vapor at 298 by taking the heat of vaporization at 298. Vapor can also exist at lower temperatures, and you can get the enthalpy at the lower temperatures by multiplying the temperature difference by the heat capacity, and subtracting. You can also use this to get the heat of vaporization of ice.

No. This is not what I meant. First of all, at constant pressure in a closed system, ΔH = q. If you measure the heat of reaction for ideal gases reacting (from the pure reactants to the pure products) at constant system temperature and constant pressure, you get the same value for q at all low pressures, irrespective of the pressure level, and it doesn't even matter if the final pressure is equal to the initial pressure.

Chet
I think I know what I'm thinking about incorrectly. When we say that the reaction occurs at standard temperature and pressure, that is, the temperature and pressure of the surroundings is 298.15K and 1atm. This does not specify the actual temperature of the reactants, does it? And in a calorimeter, ΔV = 0 so ΔH = q.
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Quote Quote by MathewsMD View Post
I think I know what I'm thinking about incorrectly. When we say that the reaction occurs at standard temperature and pressure, that is, the temperature and pressure of the surroundings is 298.15K and 1atm. This does not specify the actual temperature of the reactants, does it? And in a calorimeter, ΔV = 0 so ΔH = q.
No. It means that the temperature and pressure of the pure reactants and the temperature and pressure of the pure products are 298.15 and 1 atm. This does specify the actual temperature of the reactants. In a calorimeter, ΔP=0, so ΔH=q. This is the amount of heat you have to add to hold the temperature of the reactor at 298 and the pressure at 1 atm. If ΔV=0, then ΔU=q, but this does not give the standard heat of reaction. The term "heat of reaction" refers to constant pressure and temperature.
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Jan23-14, 05:01 PM
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Quote Quote by Chestermiller View Post
No. It means that the temperature and pressure of the pure reactants and the temperature and pressure of the pure products are 298.15 and 1 atm. This does specify the actual temperature of the reactants. In a calorimeter, ΔP=0, so ΔH=q. This is the amount of heat you have to add to hold the temperature of the reactor at 298 and the pressure at 1 atm. If ΔV=0, then ΔU=q, but this does not give the standard heat of reaction. The term "heat of reaction" refers to constant pressure and temperature.
ΔH = ΔE + ΔPV but since P is constant ΔH = ΔE + PΔV
ΔE = q + w = q - PΔV (at constant pressure)
So ΔH = q - PΔV + PΔV = q

So isn't ΔU = ΔH in this case, if pressure is constant?

Just wondering, if you have a bomb calorimeter and volume is kept constant, how can we say pressure is constant before and after? Do we just wait for the product(s) to settle and hopefully return to the original pressure and then calculate enthalpy changes at that time?

Sorry, I'm just having trouble understanding this concept. If you are at the melting point of water (273.15K) you have to lose enough energy to turn ALL of the liquid water to ice, correct? So despite being at a temperature a little higher (ex. 280K) there would still be ice present, but it would be in low quantities, right? It is only once you go just below 273.15K all of the water is solid. I'm confused in understanding how both the products and reactants are to be measured at 298.15K and 1atm if the other exists in only trace amounts in those conditions...wouldn't there be negligible ice in these conditions as the formation of liquid water is favoured?

Also, since ΔH = mcΔT, if T is not constant, then ΔT ≠ 0 and thus ΔH (in this case) is not a standard enthalpy, right?
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Quote Quote by MathewsMD View Post
ΔH = ΔE + ΔPV but since P is constant ΔH = ΔE + PΔV
ΔE = q + w = q - PΔV (at constant pressure)
So ΔH = q - PΔV + PΔV = q

So isn't ΔU = ΔH in this case, if pressure is constant?
No. That doesn't follow from your own algebra.
Just wondering, if you have a bomb calorimeter and volume is kept constant, how can we say pressure is constant before and after? Do we just wait for the product(s) to settle and hopefully return to the original pressure and then calculate enthalpy changes at that time?
No. The bomb calorimeter at constant volume does not measure what we define as "heat of reaction." However, if the reaction takes place involving ideal gases, we can calculate what the heat of reaction would be.
Sorry, I'm just having trouble understanding this concept. If you are at the melting point of water (273.15K) you have to lose enough energy to turn ALL of the liquid water to ice, correct?
Yes.
So despite being at a temperature a little higher (ex. 280K) there would still be ice present, but it would be in low quantities, right?
At 280, you can't have ice and water present simultaneously in equilibrium. Even under non-equilibrium conditions, if part of the water were at 280, the water at the interface with the ice would be at 273 (the water temperature would vary with spatial position in the system), and all the ice would be at 273 (or below).[/quote]

At equilibrium, below 273, water can't exist as a liquid. In some non-equilibrium cases, it can be cooled below 273, but the system is thermodynamically unstable, and the slightest disturbance will cause the water to solidify (provided the heat of solidification is removed).
I'm confused in understanding how both the products and reactants are to be measured at 298.15K and 1atm if the other exists in only trace amounts in those conditions...wouldn't there be negligible ice in these conditions as the formation of liquid water is favoured?
The terms products and reactants usually refer to chemical reactions. If you are referring to a phase change like this, you really can't have water at 298 change to ice at 298. Ice at 298 is a hypothetical state that can be assigned an enthalpy by extrapolating the behavior of ice from below 273. So this can still be a reference state for ice mathmatically, but the change in enthalpy from 273 to 298 will always automatically cancel out from any practical calculations.
Also, since ΔH = mcΔT, if T is not constant, then ΔT ≠ 0 and thus ΔH (in this case) is not a standard enthalpy, right?
There is no such thing as a change in enthalpy ΔH being a standard enthalpy. Here's how it works: The enthalpy of the material in some standard state is specified to have a certain value (there's a method of establishing preciesely what this value is relative to the pure atomic species making up the formula for the material, which are assigned enthalpies of zero). You can then evaluate the enthalpy of the material in some other state by evaluating the change in the enthalpy from the standard state, and then adding it to the enthalpy of the standard state.
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Jan24-14, 05:05 PM
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Quote Quote by Chestermiller View Post
No. That doesn't follow from your own algebra.

There is no such thing as a change in enthalpy ΔH being a standard enthalpy. Here's how it works: The enthalpy of the material in some standard state is specified to have a certain value (there's a method of establishing preciesely what this value is relative to the pure atomic species making up the formula for the material, which are assigned enthalpies of zero). You can then evaluate the enthalpy of the material in some other state by evaluating the change in the enthalpy from the standard state, and then adding it to the enthalpy of the standard state.
Thank you for that further clarification. The last paragraph certainly helps. So the standard enthalpy of fusion of a substance at standard conditions does not exist, it is just the enthalpy of fusion. This information is found by finding the necessary enthalpy change to go from the standard state (ex. water at 298.15 K) to the desired state (ex. ice at 273.15K).

When you say that the initial statement I showed "doesn't follow from [my] own algebra" do you mind referring me to the correct proof? Is this expression still correct, despite my faulty reasoning?

Also, on a related note: When calculating total bond enthalpy (TBE) for a substance, it is assumed that the reactants and products are gases. I don't see this explicitly stated in my workbook, but must both reactants and products behave as ideal gases for this to be simply calculated using

##ΔH_{rxn} = TBE_{reactants} - TBE_{products}## ?
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Jan24-14, 06:40 PM
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Quote Quote by MathewsMD View Post

When you say that the initial statement I showed "doesn't follow from [my] own algebra" do you mind referring me to the correct proof? Is this expression still correct, despite my faulty reasoning?
ΔE is the same thing as ΔU (if there is no change in potential energy or macroscopic kinetic energy). So ΔU=ΔH-PΔV for constant pressure.
Also, on a related note: When calculating total bond enthalpy (TBE) for a substance, it is assumed that the reactants and products are gases. I don't see this explicitly stated in my workbook, but must both reactants and products behave as ideal gases for this to be simply calculated using

##ΔH_{rxn} = TBE_{reactants} - TBE_{products}## ?
I'm not sure. My guess is yes, or else there would probably be some contribution from changes is intermolecular potential energy.


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