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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 GRXColleen50 (210 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 BernieceOFla (370 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 TajDown22661 (390 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 EmoryMoorhou (350 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 MellisaDurgi (450 points)

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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 LatoyaBertra (330 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 KarineWinnin (370 points)

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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 DewayneDto26 (590 points)

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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 JimmyBecker9 (390 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 VirgilioGree (530 points)

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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 CliftonS8245 (410 points)

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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 DanielleCand (370 points)

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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 HermelindaPe (430 points)

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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 MaryPhilips4 (450 points)

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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 BethanyMcRae (370 points)

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1 answer 32 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 KBKJerri587 (250 points)

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1 answer 30 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 TyreeHayward (330 points)

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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 SonHusk8627 (490 points)

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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 ReeceHodges (750 points)

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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 Zack25G18861 (390 points)

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