Suppose we redefine the standard state as P=2 atm. Find the new standard ΔGf^∘ values of each su

Suppose we redefine the standard state as P=2 atm. Find the...

Key Concepts

Chemical Potential and Pressure Correction

For an ideal gas, the chemical potential depends on pressure through the relation ? = ?° + RT ln(P/P°). Changing the standard state pressure requires adding RT ln(P_new/P_standard) to the chemical potential. This correction leads to modified ?G values because the free energy is a sum over the chemical potentials of the reactants or products involved in the formation reaction.

Standard State Definition

The standard state is a set of reference conditions, such as pressure and temperature, against which thermodynamic properties are measured. Redefining the standard state (for example, from 1 atm to 2 atm) means that the baseline for reporting free energies changes, and adjustments must be made to account for this new reference state.

Relation Between Gibbs Free Energy and Entropy

Gibbs free energy is defined by the equation G = H – TS, linking energy, temperature, and entropy. When the standard state is altered, the contribution of entropy becomes important, particularly for gases where entropy is sensitive to pressure. A change in pressure changes the degree of molecular disorder (entropy), which in turn influences the free energy of formation of the substance.

Formation Reactions and Stoichiometry

In formation reactions, elements in their standard states combine to form compounds, and the number of moles of gaseous species can differ between reactants and products. When pressure is redefined for gaseous species, the effect on ?G_f° depends on the reaction stoichiometry. For instance, if the formation reaction produces fewer gas molecules than it consumes, the pressure change can cause a more significant shift in the free energy due to the corresponding change in entropy.

Transcript

00:01 For this question, we want to find what the gibbs free energy is with a new standard atm of 2.

00:08 So starting with the first part of this question, what i'm going to do first is write out the equation fully balanced, so we can see what we're working with a little bit better.

00:30 And all of these will be in the gases form.

00:38 So once you have your balanced equation, the first thing you want to do is calculate the equilibrium concept.

00:43 And we want to calculate the equilibrium constant so that we can compare what we see here.

00:49 Given our new standard atm with what we know from any standard thermodynamic table of what we would expect for hcl at standard conditions, which you can see here.

01:02 So first, how you calculate the equal room constant is you do the pressure of the product over the pressure the first reactant, raised the how many moles we have, multiplied by the second reactant, also raised the power of how many moles.

01:27 We have.

01:29 So if we do out, you know that we have a recent question to atm to raised one half.

01:42 This is how many rules we have here.

01:45 And we'll be the same here.

01:52 And this is equivalent to one.

01:55 So since this is one, we know the gives for energy at 2 atm is equivalent to the gives for energy that we see at 1 atm.

02:14 Since these are equivalent, you can again use a standard demodynamic table to know that it gives free energy of the formation of hcl in the gaseous form is negative 95 .5 .3 kilojures since this gives free energy value is negative if this is there.

02:43 You know that this is a thermodynamically favorable reaction.

02:48 Moving on to look at b, we'll line it out that we have the balance equation to look at.

03:05 And we'll proceed in the same way that we did a by first finding people of birth constant.

03:11 So we have one product and two reactants again.

03:15 So we'll do the pressure of the product over the pressure of the first reactor, which is two.

03:23 And since there is a one here in front of the nitrogen, we will not raise it to anything but one.

03:29 Then we will also do the same for the second, but we have half here.

03:39 And this calculated out, it's equivalent to 0 .707 .1.

03:48 Since this value is not one, we know that the gives free energy at 2 atm is not the same as it gives free energy at 1 .m.

03:58 So in order to find what it would be, we know the gives free energy at 2 .8 .m is not the same...

Please give Ace some feedback