An experiment produces evidence that the evaporation of 4.00 g of liquid butane C4H10 (l) requires a gain in enthalpy of 1.67 kJ. Find the molar enthalpy of vaporization in kJ/mol for butane from this evidence.

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superman1987 superman1987 · Answer 1 · 2019-03-13T20:31:00+01:00

Answer:

The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.

Explanation:

Firstly, we need to calculate the no. of moles of 4.00 g of liquid butane C₄H₁₀:

n = mass/molar mass = (4.0 g)/(58.12 g/mol) = 0.0688 mol.

∴ 0.0688 mol of butane requires a gain in enthalpy of 1.67 kJ to be evaporized.

Know using cross multiplication:

0.0688 mol of butane to be vaporized requires → 1.67 kJ.

1.0 mol of butane to be vaporized requires → ??? kJ.

∴ 1.0 mol of butane to be vaporized requires = (1.0 mol)(1.67 kJ)/(0.0688 mol) = 24.265 kJ.

∴ The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.