What is the standard Gibbs free energy for the transformation of diamond to graphite at 298 K? Cdiamond?Cgraphite Exp...

Question

Question

What is the standard Gibbs free energy for the transformation of diamond to graphite at 298 K? Cdiamond?Cgraphite Exp...

What is the standard Gibbs free energy for the transformation of diamond to graphite at 298 K?

Cdiamond?Cgraphite

Express your answer to three significant figures and include the appropriate units.

Gibbs free energy is a measure of the spontaneity of a chemical reaction. It is the chemical potential for a reaction, which is minimized at equilibrium. It is defined as

G=H?TS

Elemental carbon usually exists in one of two forms: graphite or diamond. It is generally believed that diamonds last forever. Here are the standard enthalpy of formation (?H?f) and the standard molar entropy (S?) values for diamond and graphite.

Substance	?H?f (kJ/mol)	S? (J/mol?K)
Cgraphite	0	5.740
Cdiamond	1.897	2.38

science chemistry

Add a comment Improve this question Transcribed image text

Answer 1

✔ Recommended Answer

Answer #1

Concepts and reason

• Gibbs free energy is explained by using the parameters entropy, enthalpy and temperature.

• Gibbs free energy is explained as the difference of enthalpy to the product of temperature and entropy.

• Gibbs energy is termed as a state function, since the function is depending only on initial and final states of the system.

Fundamentals

Gibbs free energy: In Gibbs free energy calculation, three parameters are used; Temperature, entropy and enthalpy.

G=H-TS
where,
G=Gibbs free energy.
H=Enthalpy of the system.
T= Absolute temperature.
S=Entropy of the system.

Since Gibbs free energy is a state function, a change in Gibbs free energy exists as follows.

AG=AH-TAS
i) if AG <0, reaction is spontaneous.
ii) if AG>0, reaction is non spontaneous.
iii) if AG=0, reaction is at equili

The given transformation is shown below as a reaction.

diamond(s) —>graphite(s)

The given enthalpy and entropy values for this transformation reaction are shown below.

Substance	Enthalpy (kJ/mol)	Entropy (J/mol K)
C_graphite	0	5.740
C_diamond	1.897	2.38

Change in enthalpy calculation for the given transformation is different between the enthalpies of products and reactants.

ΔΗ. =ΣΔΗεις - ΣΔΗ.

In this transformation reaction:

ΔΗ. = ΔΗρδίες -ΔΗ.κοσκά
=(Imol x0 kJ/mol )-(Imol x1.897 kJ/mol)
=-1.897kJ

Now,

Change in entropy calculation for the given transformation is different between the entropies of products and reactants.

AS n = E AS products - E AS Catants

In this transformation reaction:

ASR=ASCraphite - Aiamond
=1 mol 5.740J/mol K-1 mol x 2.38J/mol. K
=3.36 J/K
=0.00336kJ/K

AH and AS values for the given transformation are written below.

ΔΗ. =.1.897 kJ

ASX=0.00336 kJ/K

Transformation occurs at 298K temperature, so the change in free energy is as follows.

AG=AH-TAS
=(-1.897kJ)-[(298 K)x(0.00336 kJ/K)
=(-1.897 kJ)-(1.00128 kJ)
=-2.898 kJ

Therefore, the standard Gibbs free energy for the given transformation is সি 8687- .

Ans:

The standard Gibbs free energy for the given transformation is.

Add a comment

Answer 2

Similar Homework Help Questions

What is the standard Gibbs free energy for the transformation of diamond to graphite at 298...

What is the standard Gibbs free energy for the transformation of diamond to graphite at 298 K? Cdiamond+Cgraphite Express your answer to three significant figures and include the appropriate units. ► View Available Hint(s) kJ AGran = - 1.90 mol
What is the standard Gibbs energy for the transformation of diamond to graphite at 298 K?...

What is the standard Gibbs energy for the transformation of diamond to graphite at 298 K? Cdiamond+Cgraphite Express your answer to three significant figures and include the appropriate units. ► View Available Hint(s)
The standard Gibbs free energy for the transformation of diamond to graphite at 298 K is...

The standard Gibbs free energy for the transformation of diamond to graphite at 298 K is -2.9 kJ/mol. Why are diamonds not spontaneously transformed to graphite? C(diamond)→C(graphite) Δ?0=−2.9kJ/mol

LHA What is the standard Gibbs free energy for the transformation of diamond to graphite at...

LHA What is the standard Gibbs free energy for the transformation of diamond to graphite at 298 K? Cdiamond+Cgraphite Express your answer to three significant figures and include the appropriate units. You did not open hints for this part. ANSWER: AGTx =
1. Determine the standard Gibb's free energy change for the conversion of diamond into graphite in...

1. Determine the standard Gibb's free energy change for the conversion of diamond into graphite in units of kJ/mol and is it spontaneous, nonspontaneous, or neither. Cdiamond (s) ? Cgraphite (s) f° for diamond = 2.9 kJ/mol f° for graphite = 0.0 kJ/mol (Please be careful wtih sig figs.) 2. For the following reaction, = 68.1 kJ/mol: 2 C(s) + 2 H2(g) ? C2H4(g) Determine for the reaction in units of kJ/mol at 1800. K, a partial pressure of C2H2 of...
In Class Exercise - The Gibbs Free Energy Change, AG 1) Determining the Standard Gibbs Free...

In Class Exercise - The Gibbs Free Energy Change, AG 1) Determining the Standard Gibbs Free Energy Change (AGⓇ) for a Chemical Reaction 2) Using AGº to Determine Spontaneity Name: Date: Lab section: Show your work when there are calculations, write units, and use correct significant figures. Consider the following reaction (balanced as written) and thermodynamic data from tables in your book: CO(NH2)2(aq) + H2O(1) ► CO2(g) + 2NH3(g) Substance CO(NH3)2(aq) H00 AH(kJ/mol) 1-391.2 -285.9 -3935 -46.19 S'J/mol K) 173.8...

In Class Exercise - The Gibbs Free Energy Change, AG 1) Determining the Standard Gibbs Free...

In Class Exercise - The Gibbs Free Energy Change, AG 1) Determining the Standard Gibbs Free Energy Change (AGⓇ) for a Chemical Reaction 2) Using AGº to Determine Spontaneity Name: Date: Lab section: Show your work when there are calculations, write units, and use correct significant figures. Consider the following reaction (balanced as written) and thermodynamic data from tables in your book: CO(NH2)2(aq) + H2O(l) → CO2(g) + 2NH3(g) Substance CO(NH2)2(aq) H2O(1) CO2(g) NH AH° (kJ/mol) -391.2 -285.9 -393.5 -46.19...
A reaction is at equilibrium at 298 K. At 310 K, the Gibbs free energy for the reaction is –12.6 ...

A reaction is at equilibrium at 298 K. At 310 K, the Gibbs free energy for the reaction is –12.6 kJ/mol. Assuming that both entropy and enthalpy are independent of temperature, what are the values of the entropy and enthalpy for this reaction?
Calculate K at 298 K for the following reaction given the Gibbs free energy of formations...

Calculate K at 298 K for the following reaction given the Gibbs free energy of formations 213) substance! ΔG。 kJ/mol N2O4(g) +99.8 NO2(g)+51.3 1.13 0.32 3.1

the standard Gibbs energy of formation (in kJ-mol-) of the compound at 298 K

the standard Gibbs energy of formation (in kJ-mol-) of the compound at 298 K

What is the standard Gibbs free energy for the transformation of diamond to graphite at 298 K? Cdiamond?Cgraphite Exp...

Homework Answers

Add Answer to:
What is the standard Gibbs free energy for the transformation of diamond to graphite at 298 K? Cdiamond?Cgraphite Exp...

Post as a guest

Earn Coins