Calculate the enthalpy change for the isomerization of n-butane to iso-butane if the standard enthalpy of formation for n-butane is -125.7 kJ/mol and that for iso-butane is -147.4 kJ/mol. The isomerization reaction occurs at a constant pressure of 1 atm and at a temperature of 298 K.

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Calculate the enthalpy change for the isomerization of n-butane to iso-butane if the standard enthalpy of formation for n-butane is -125.7 kJ/mol and that for iso-butane is -147.4 kJ/mol. The isomerization reaction occurs at a constant pressure of 1 atm and at a temperature of 298 K.

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HelenaGodoy5 · Answer 1 · 2025-01-23T09:30:19+0000

To calculate the enthalpy change for the isomerization of n-butane to iso-butane, we can use the following equation:H_reaction = H_f products - H_f reactants For the isomerization reaction, the reactant is n-butane and the product is iso-butane. The given standard enthalpies of formation are:H_f n-butane = -125.7 kJ/molH_f iso-butane = -147.4 kJ/molNow, we can plug these values into the equation:H_reaction = [H_f iso-butane ] - [H_f n-butane ]H_reaction = -147.4 kJ/mol - -125.7 kJ/mol H_reaction = -147.4 + 125.7H_reaction = -21.7 kJ/molThe enthalpy change for the isomerization of n-butane to iso-butane is -21.7 kJ/mol.

Calculate the enthalpy change for the isomerization of n-butane to iso-butane if the standard enthalpy of formation for n-butane is -125.7 kJ/mol and that for iso-butane is -147.4 kJ/mol. The isomerization reaction occurs at a constant pressure of 1 atm and at a temperature of 298 K.

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