4 edited body
source | link

My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$

(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)

For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:

$$\Delta G^\circ_{rxn}=-RT~ln~K$$

I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C: $$\ce{N2 + 3H2 <=> 2NH3}$$ From standard tables, at 25$^\circ$C (all kJ, mole, and K):

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$$${N_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$

$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$

For this reaction:

$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$

$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$

$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$

First approach (how I think it is):

1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$

2. Calculate K (assuming kJ, mole, and K for R and other values):

$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$

Second approach (also commonly seen):

1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:

$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$

Edit: I want to also add that this reaction is more favorable at lower temperatures, so Approach 1 makes sense and Approach 2 does not, even though many treat problems this way. Also, I get $K=10^{5.77}$ at 25$^\circ$C, so the decreasing K matches up.

Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?

And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:

$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~Q$$

My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$

(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)

For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:

$$\Delta G^\circ_{rxn}=-RT~ln~K$$

I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C: $$\ce{N2 + 3H2 <=> 2NH3}$$ From standard tables, at 25$^\circ$C (all kJ, mole, and K):

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$

$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$

For this reaction:

$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$

$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$

$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$

First approach (how I think it is):

1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$

2. Calculate K (assuming kJ, mole, and K for R and other values):

$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$

Second approach (also commonly seen):

1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:

$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$

Edit: I want to also add that this reaction is more favorable at lower temperatures, so Approach 1 makes sense and Approach 2 does not, even though many treat problems this way. Also, I get $K=10^{5.77}$ at 25$^\circ$C, so the decreasing K matches up.

Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?

And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:

$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~Q$$

My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$

(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)

For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:

$$\Delta G^\circ_{rxn}=-RT~ln~K$$

I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C: $$\ce{N2 + 3H2 <=> 2NH3}$$ From standard tables, at 25$^\circ$C (all kJ, mole, and K):

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$

$${N_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$

$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$

For this reaction:

$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$

$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$

$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$

First approach (how I think it is):

1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$

2. Calculate K (assuming kJ, mole, and K for R and other values):

$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$

Second approach (also commonly seen):

1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:

$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$

Edit: I want to also add that this reaction is more favorable at lower temperatures, so Approach 1 makes sense and Approach 2 does not, even though many treat problems this way. Also, I get $K=10^{5.77}$ at 25$^\circ$C, so the decreasing K matches up.

Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?

And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:

$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~Q$$

3 edited body
source | link

My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$

(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)

For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:

$$\Delta G^\circ_{rxn}=-RT~ln~K$$

I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C: $$\ce{N2 + 3H2 <=> 2NH3}$$ From standard tables, at 25$^\circ$C (all kJ, mole, and K):

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$

$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$

For this reaction:

$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$

$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$

$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$

First approach (how I think it is):

1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$

2. Calculate K (assuming kJ, mole, and K for R and other values):

$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$

Second approach (also commonly seen):

1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:

$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$

Edit: I want to also add that this reaction is more favorable at lower temperatures, so Approach 1 makes sense and Approach 2 does not, even though many treat problems this way. Also, I get $K=10^{5.77}$ at 25$^\circ$C, so the decreasing K matches up.

Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?

And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:

$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~K$$$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~Q$$

My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$

(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)

For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:

$$\Delta G^\circ_{rxn}=-RT~ln~K$$

I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C: $$\ce{N2 + 3H2 <=> 2NH3}$$ From standard tables, at 25$^\circ$C (all kJ, mole, and K):

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$

$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$

For this reaction:

$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$

$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$

$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$

First approach (how I think it is):

1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$

2. Calculate K (assuming kJ, mole, and K for R and other values):

$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$

Second approach (also commonly seen):

1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:

$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$

Edit: I want to also add that this reaction is more favorable at lower temperatures, so Approach 1 makes sense and Approach 2 does not, even though many treat problems this way. Also, I get $K=10^{5.77}$ at 25$^\circ$C, so the decreasing K matches up.

Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?

And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:

$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~K$$

My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$

(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)

For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:

$$\Delta G^\circ_{rxn}=-RT~ln~K$$

I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C: $$\ce{N2 + 3H2 <=> 2NH3}$$ From standard tables, at 25$^\circ$C (all kJ, mole, and K):

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$

$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$

For this reaction:

$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$

$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$

$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$

First approach (how I think it is):

1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$

2. Calculate K (assuming kJ, mole, and K for R and other values):

$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$

Second approach (also commonly seen):

1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:

$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$

Edit: I want to also add that this reaction is more favorable at lower temperatures, so Approach 1 makes sense and Approach 2 does not, even though many treat problems this way. Also, I get $K=10^{5.77}$ at 25$^\circ$C, so the decreasing K matches up.

Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?

And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:

$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~Q$$

2 added 257 characters in body
source | link

My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$

(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)

For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:

$$\Delta G^\circ_{rxn}=-RT~ln~K$$

I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C: $$\ce{N2 + 3H2 <=> 2NH3}$$ From standard tables, at 25$^\circ$C (all kJ, mole, and K):

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$

$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$

For this reaction:

$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$

$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$

$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$

First approach (how I think it is):

1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$

2. Calculate K (assuming kJ, mole, and K for R and other values):

$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$

Second approach (also commonly seen):

1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:

$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$

Edit: I want to also add that this reaction is more favorable at lower temperatures, so Approach 1 makes sense and Approach 2 does not, even though many treat problems this way. Also, I get $K=10^{5.77}$ at 25$^\circ$C, so the decreasing K matches up.

Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?

And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:

$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~K$$

My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$

(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)

For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:

$$\Delta G^\circ_{rxn}=-RT~ln~K$$

I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C: $$\ce{N2 + 3H2 <=> 2NH3}$$ From standard tables, at 25$^\circ$C (all kJ, mole, and K):

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$

$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$

For this reaction:

$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$

$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$

$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$

First approach (how I think it is):

1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$

2. Calculate K (assuming kJ, mole, and K for R and other values):

$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$

Second approach (also commonly seen):

1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:

$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$

Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?

And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:

$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~K$$

My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$

(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)

For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:

$$\Delta G^\circ_{rxn}=-RT~ln~K$$

I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C: $$\ce{N2 + 3H2 <=> 2NH3}$$ From standard tables, at 25$^\circ$C (all kJ, mole, and K):

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$

$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$

For this reaction:

$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$

$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$

$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$

First approach (how I think it is):

1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$

2. Calculate K (assuming kJ, mole, and K for R and other values):

$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$

Second approach (also commonly seen):

1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:

$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$

Edit: I want to also add that this reaction is more favorable at lower temperatures, so Approach 1 makes sense and Approach 2 does not, even though many treat problems this way. Also, I get $K=10^{5.77}$ at 25$^\circ$C, so the decreasing K matches up.

Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?

And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:

$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~K$$

1
source | link