contestada

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.

Respuesta :

superman1987

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.

Otras preguntas