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votes
1
answer
38
views
Calculate the standard enthalpy change for the transition of liquid water at 100°C to steam at the same temperature, given that the enthalpy of vaporization of water is 40.7 kJ mol-1.
asked
Feb 3
in
Chemical thermodynamics
by
MarianaHerit
(
450
points)
0
votes
1
answer
26
views
Calculate the standard enthalpy change for the sublimation of iodine, given that the sublimation of 1 mole of iodine requires 62.44 kJ of energy and the standard enthalpy of formation of iodine (s) is 0 kJ/mol.
asked
Feb 3
in
Chemical thermodynamics
by
RenaUxo07372
(
310
points)
0
votes
1
answer
34
views
Calculate the standard enthalpy change for the sublimation of iodine solid to iodine gas at 298 K with the following given data: - Standard enthalpy of fusion of iodine: 15.52 kJ/mol- Standard molar entropy of iodine solid: 62.7 J/K/mol- Standard molar entropy of iodine gas: 260.6 J/K/mol
asked
Feb 3
in
Chemical thermodynamics
by
TabathaD6751
(
310
points)
0
votes
1
answer
31
views
Calculate the standard enthalpy change for the sublimation of iodine at 25°C, given that the standard enthalpy of fusion of iodine is 15.7 kJ/mol and the standard enthalpy of vaporization of iodine is 41.0 kJ/mol. (The molar mass of iodine is 126.90 g/mol)
asked
Feb 3
in
Chemical thermodynamics
by
GlennaValle0
(
230
points)
0
votes
1
answer
23
views
Calculate the standard enthalpy change for the sublimation of iodine (I2) using the given information: The standard enthalpy of fusion of iodine is 15.52 kJ/mol and the standard enthalpy of vaporization of iodine is 41.57 kJ/mol.
asked
Feb 3
in
Chemical thermodynamics
by
SherrillFred
(
570
points)
0
votes
1
answer
23
views
Calculate the standard enthalpy change for the sublimation of iodine (I2) if 1 mole of I2(s) is converted to 1 mole of I2(g) at 25°C and 1 bar pressure. Given the following information:- The standard enthalpy of fusion of I2 is 15.4 kJ/mol- The standard enthalpy of vaporization of I2 is 41.3 kJ/mol- The standard entropy change for the sublimation of I2 is 62.4 J/mol K.
asked
Feb 3
in
Chemical thermodynamics
by
AnthonyCheat
(
290
points)
0
votes
1
answer
35
views
Calculate the standard enthalpy change for the sublimation of iodine (I2) given that the enthalpy of fusion for iodine is 15.52 kJ/mol and the enthalpy of vaporization for iodine is 41.57 kJ/mol.
asked
Feb 3
in
Chemical thermodynamics
by
RosalindCust
(
510
points)
0
votes
1
answer
24
views
Calculate the standard enthalpy change for the sublimation of 5 moles of solid iodine (I2) given the following data:- Enthalpy of fusion of iodine = 15.52 kJ/mol- Enthalpy of vaporization of iodine = 41.57 kJ/mol
asked
Feb 3
in
Chemical thermodynamics
by
EmmaLeist635
(
170
points)
0
votes
1
answer
39
views
Calculate the standard enthalpy change for the reduction of iron(III) oxide to iron using the following balanced chemical equation:Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)Given that the standard enthalpy change of formation for Fe2O3(s) is -824.2 kJ/mol, the standard enthalpy change of formation for CO2(g) is -393.5 kJ/mol, and the standard enthalpy change of formation for Fe(s) is 0 kJ/mol. Assume all reactants and products are in their standard states.
asked
Feb 3
in
ThermoChemistry
by
MerleEbj441
(
370
points)
0
votes
1
answer
33
views
Calculate the standard enthalpy change for the reduction of iron (III) oxide using carbon monoxide as the reducing agent, given that the standard enthalpy of formation for iron (III) oxide is -824.2 kJ/mol and the standard enthalpy of formation for carbon monoxide is -110.5 kJ/mol.
asked
Feb 3
in
ThermoChemistry
by
ValentinaGee
(
310
points)
0
votes
1
answer
36
views
Calculate the standard enthalpy change for the reaction:Cu(H2O)62+ (aq) + 4Cl-(aq) → CuCl42- (aq) + 12H2O (l)given the following standard enthalpies of formation:Cu(H2O)62+ (aq): -1846.4 kJ/molCuCl42- (aq): -3599.5 kJ/molH2O (l): -285.8 kJ/molCl- (aq): -167.2 kJ/mol
asked
Feb 3
in
Chemical thermodynamics
by
FlorrieBourk
(
530
points)
0
votes
1
answer
33
views
Calculate the standard enthalpy change for the reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) given the following enthalpy changes: ∆Hf° for NaCl(aq) = -407.3 kJ/mol ∆Hf° for H2O(l) = -285.8 kJ/mol ∆Hf° for NaOH(aq) = -469.11 kJ/mol ∆Hf° for HCl(aq) = -167.16 kJ/mol
asked
Feb 3
in
Chemical thermodynamics
by
JustineRocho
(
390
points)
0
votes
1
answer
39
views
Calculate the standard enthalpy change for the reaction: 2Fe(s) + 3/2O2(g) → Fe2O3(s), given that the standard enthalpy of formation for Fe2O3(s) is -824.2 kJ/mol, the standard enthalpy of formation for Fe(s) is 0 kJ/mol, and the standard enthalpy of formation for O2(g) is 0 kJ/mol.
asked
Feb 3
in
Chemical thermodynamics
by
JakeGagner6
(
390
points)
0
votes
1
answer
35
views
Calculate the standard enthalpy change for the reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Given the following standard enthalpy of formation values: ΔHf° [Fe2O3(s)] = -824 kJ/mol ΔHf° [CO(g)] = -110 kJ/mol ΔHf° [CO2(g)] = -393.5 kJ/mol ΔHf° [Fe(s)] = 0 kJ/mol.
asked
Feb 3
in
Chemical thermodynamics
by
Kimberly5191
(
610
points)
0
votes
1
answer
39
views
Calculate the standard enthalpy change for the reaction: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l), given the following standard enthalpies of formation: ΔHf°[C2H5OH(l)] = -277.69 kJ/mol, ΔHf°[CO2(g)] = -393.51 kJ/mol, ΔHf°[H2O(l)] = -285.83 kJ/mol.
asked
Feb 3
in
Chemical thermodynamics
by
ErvinNeill87
(
450
points)
0
votes
1
answer
42
views
Calculate the standard enthalpy change for the reaction: 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)Given the following data:ΔHf°(NaOH) = -469.14 kJ/molΔHf°(H2SO4) = -814.00 kJ/molΔHf°(Na2SO4) = -1388.40 kJ/molΔHf°(H2O(l)) = -285.83 kJ/mol
asked
Feb 3
in
Chemical thermodynamics
by
AbrahamMcder
(
390
points)
0
votes
1
answer
31
views
Calculate the standard enthalpy change for the reaction: 2 Al(s) + 3/2 O2(g) → Al2O3(s) given the following information: ΔH°f[Al2O3(s)] = -1676.0 kJ/mol ΔH°f[Al(s)] = 0 kJ/mol ΔH°f[O2(g)] = 0 kJ/mol
asked
Feb 3
in
Chemical thermodynamics
by
HughLarge668
(
270
points)
0
votes
1
answer
29
views
Calculate the standard enthalpy change for the reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Given the following information: ΔHf° [NaOH(aq)] = -469.1 kJ/mol ΔHf° [HCl(aq)] = -167.2 kJ/mol ΔHf° [NaCl(aq)] = -407.3 kJ/mol ΔHf° [H2O(l)] = -286 kJ/mol
asked
Feb 3
in
Chemical thermodynamics
by
LinneaZox85
(
370
points)
0
votes
1
answer
23
views
Calculate the standard enthalpy change for the reaction: CuSO4 (aq) + 4NH3 (aq) -> Cu(NH3)4SO4 (aq) given that the standard enthalpy changes of formation for CuSO4 (aq) and NH3 (aq) are -771.5 kJ/mol and -46.19 kJ/mol respectively, and the standard enthalpy change of formation for Cu(NH3)4SO4 (aq) is -2130.4 kJ/mol.
asked
Feb 3
in
Chemical thermodynamics
by
Juliet26542
(
190
points)
0
votes
1
answer
38
views
Calculate the standard enthalpy change for the reaction: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) Given the necessary bond enthalpies are: C-H = 411 kJ/molO=O = 495 kJ/molO-H = 463 kJ/molC=O = 799 kJ/mol.
asked
Feb 3
in
Quantum Chemistry
by
ChristineR12
(
430
points)
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