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Recent questions in Chemistry
0
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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
MavisIei1672
(
370
points)
0
votes
1
answer
32
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
MarioY134267
(
270
points)
0
votes
1
answer
33
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
Ethel50G7107
(
330
points)
0
votes
1
answer
50
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
Samual94V215
(
310
points)
0
votes
1
answer
30
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
LeonaSpivey
(
330
points)
0
votes
1
answer
46
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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
AliciaB88894
(
470
points)
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votes
1
answer
42
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
MckenzieRxm
(
490
points)
0
votes
1
answer
41
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
Jon40C315633
(
380
points)
0
votes
1
answer
45
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
ShawneeSalmo
(
170
points)
0
votes
1
answer
49
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
SoilaThibode
(
370
points)
0
votes
1
answer
44
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
OnitaLerner
(
410
points)
0
votes
1
answer
46
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
SantoQuinton
(
330
points)
0
votes
1
answer
54
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
MeriLathrop0
(
310
points)
0
votes
1
answer
41
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
LorrineUlm46
(
350
points)
0
votes
1
answer
43
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
ShaunEarnest
(
710
points)
0
votes
1
answer
36
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
GidgetAuster
(
330
points)
0
votes
1
answer
47
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
BarryBabbidg
(
310
points)
0
votes
1
answer
42
views
Calculate the standard enthalpy change for the reaction: CH3OH (l) + 3/2 O2 (g) -> CO2 (g) + 2H2O (l) given the standard enthalpies of formation for CH3OH (l), CO2 (g) and H2O (l) as -238.6 kJ/mol, -393.5 kJ/mol and -285.8 kJ/mol respectively.
asked
Feb 3
in
Chemical thermodynamics
by
VelmaBiscoe4
(
330
points)
0
votes
1
answer
45
views
Calculate the standard enthalpy change for the reaction: C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l)Given the following standard enthalpies of formation: ΔHf(C6H12O6 (s)) = -1273 kJ/molΔHf(CO2 (g)) = -393.5 kJ/molΔHf(H2O (l)) = -285.8 kJ/mol
asked
Feb 3
in
Chemical thermodynamics
by
ShelaGabriel
(
490
points)
0
votes
1
answer
47
views
Calculate the standard enthalpy change for the reaction: 2NaOH (s) + H2SO4 (aq) → Na2SO4 (aq) + 2H2O (l)given that the standard enthalpy of formation of Na2SO4 (aq) is -1385 kJ/mol, and the standard enthalpies of formation of NaOH (s), H2SO4 (aq) and H2O (l) are -425 kJ/mol, -814 kJ/mol and -286 kJ/mol respectively.
asked
Feb 3
in
Chemical thermodynamics
by
AshleeHarvil
(
370
points)
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