Recent questions in Chemical thermodynamics

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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 MellisaOlivo (1.9k points)

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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 MammieArmyta (1.8k points)

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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 ArlenBieber1 (1.9k points)

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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 BYUBlake479 (2.0k points)

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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 GracieGainfo (2.2k points)

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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 Latonya89F54 (2.2k points)

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1 answer 70 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 DomenicAsbur (2.1k points)

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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 MaybellHoffm (2.2k points)

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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 YaniraRhh59 (1.9k points)

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1 answer 68 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 KreogMoore5 (6.4k points)

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1 answer 75 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 BroderickHom (2.1k points)

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1 answer 72 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 CarrieMadiso (1.8k points)

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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 JanellLunsfo (2.0k points)

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1 answer 81 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 BetteBoniwel (1.6k points)

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1 answer 102 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 AmeliaCuella (1.8k points)

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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 MacLgq130601 (2.4k points)

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1 answer 60 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 VickeyPeacoc (2.4k points)

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1 answer 69 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 Rosaline02Y (2.0k points)

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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 SelenaRonan (2.2k points)

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1 answer 72 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 FrancescaTor (2.0k points)

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