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Calculate the standard enthalpy change for the vaporization of 1 mole of water at 100°C, given that the enthalpy of fusion of water is 6.01 kJ/mol and the enthalpy of vaporization of water is 40.7 kJ/mol.

asked Feb 3 in Chemical thermodynamics by CandraIdk507 (2.4k points)

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Calculate the standard enthalpy change for the vaporization of 1 mole of water at 100°C and 1 atm pressure. The enthalpy of formation of liquid water is -285.8 kJ/mol and the enthalpy of formation of water vapor is -241.8 kJ/mol.

asked Feb 3 in Chemical thermodynamics by DoraSturm903 (1.8k points)

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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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1 answer 111 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 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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1 answer 117 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 BYUBlake479 (2.0k points)

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1 answer 82 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 GracieGainfo (2.2k points)

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1 answer 79 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 Latonya89F54 (2.2k points)

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1 answer 122 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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1 answer 150 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 MaybellHoffm (2.2k points)

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1 answer 128 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 BeatrisSanfo (1.7k points)

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1 answer 160 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 QTHGenevieve (1.7k points)

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1 answer 123 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 YaniraRhh59 (1.9k points)

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1 answer 118 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 111 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 124 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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1 answer 131 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 JanellLunsfo (2.0k points)

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1 answer 136 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 160 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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1 answer 120 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 MacLgq130601 (2.4k points)

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