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
The standard Gibbs free energy for the transformation of diamond to graphite at 298 K is...
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...
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? Cdiamond+Cgraphite Express your answer to three significant figures and include the appropriate units. ► View Available Hint(s)
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 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...
The standard change in Gibbs free energy is Δ?°′=7.53 kJ/mol .
Calculate Δ? for this reaction at 298 K when [dihydroxyacetone
phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.00600 M .
Thank you!
For the aqueous reaction CH2OH Н— —он SO CH-0–_0 CH -0 - 0- dihydroxyacetone phosphate = glyceraldehyde-3-phosphate the standard change in Gibbs free energy is AGⓇ' = 7.53 kJ/mol. Calculate AG for this reaction at 298 K when [dihydroxyacetone phosphate) = 0.100 M and [glyceraldehyde-3-phosphate] = 0.00600 M AG = kJ/mol
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar for all speies. For the reactionC2H6(g)+H2(g)↽−−⇀2CH4(g)the standard change in Gibbs free energy is Δ𝐺∘=−32.8 kJ/mol.ΔG°=−32.8 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are 𝑃C2H6=0.400 bar,𝑃H2=0.150 bar, and 𝑃CH4=0.850 bar?𝑃CH4=0.850 bar? Δ𝐺=_____ kJ/molSk
Calculate the standard change in Gibbs free energy, Δ?∘rxnΔGrxn∘ , for the given reaction at 25.0 ∘C25.0 ∘C . Consult the table of thermodynamic properties for standard Gibbs free energy of formation values. KCl(s)↽−−⇀K+(aq)+Cl−(aq)KCl(s)↽−−⇀K+(aq)+Cl−(aq) Δ?∘rxn= Determine the concentration of K+(aq)K+(aq) if the change in Gibbs free energy, Δ?rxnΔGrxn , for the reaction is −8.31 kJ/mol−8.31 kJ/mol . [K+]=
the standard Gibbs energy of formation (in kJ-mol-) of the compound at 298 K
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?