Bromine melts at -7.25 degree C and boils at 58.8 degree C. The enthalpy of fusion of bromine is 10.57 kJ/mol and the enthalpy of vaporization of bromine is 29.96 kJ/mol. The specific heat of liquid bromine is 0.474 J/g∙K. How much heat, in kJ, is required to convert 25.0 g of solid bromine at -7.25 degree C to the gas phase at 58.8 degree C?
From solid at the start to the melting point, heat required is
q1 = mass x specific heat solid x (Tfinal-Tinitial)
At the melting point, q for melting is
q2 = mass solid x heat fusion
From the melting point to the boiling point, q for heating the liquid phase is
q3 = mass x specific heat x (Tfinal-Tinitial)
At the boiling point, the heat to convert to vapor is
q4 = mass liquid x heat vaporization
Total heat is q1 + q2 + q3 + q4.
To answer this question, we need to consider the different stages of the phase changes involved in converting the solid bromine to the gas phase. Here's the step-by-step process to find the heat required:
1. Calculate the heat required to raise the temperature of solid bromine from -7.25°C to its melting point at 0°C:
- The specific heat capacity of bromine in the solid state is not provided, so we cannot calculate this amount directly. But we can assume it to be constant and use the average specific heat capacity between the melting and boiling points, which is:
(0.474 J/g∙K + 0.474 J/g∙K) / 2 = 0.474 J/g∙K
- The temperature change is 0°C - (-7.25°C) = 7.25°C.
- The formula to calculate the heat is: q = m × c × ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the temperature change.
- Substitute the values: q₁ = 25.0 g × 0.474 J/g∙K × 7.25°C = 85.0125 J or 0.0850 kJ (rounded to 4 decimal places).
2. Calculate the heat required to melt the solid bromine at the melting point:
- The enthalpy of fusion for bromine is given as 10.57 kJ/mol.
- Convert the mass of solid bromine (25.0 g) to moles using the molar mass of bromine (159.808 g/mol):
moles = 25.0 g / 159.808 g/mol ≈ 0.1567 mol (rounded to 4 decimal places).
- Multiply the number of moles by the enthalpy of fusion: q₂ = 0.1567 mol × 10.57 kJ/mol = 1.6518 kJ (rounded to 4 decimal places).
3. Calculate the heat required to raise the temperature of liquid bromine from its melting point at 0°C to its boiling point at 58.8°C:
- The specific heat capacity of liquid bromine is given as 0.474 J/g∙K.
- The temperature change is 58.8°C - 0°C = 58.8°C.
- Use the same formula as before: q₃ = 25.0 g × 0.474 J/g∙K × 58.8°C = 699.78 J or 0.6998 kJ (rounded to 4 decimal places).
4. Calculate the heat required to vaporize the liquid bromine at its boiling point:
- The enthalpy of vaporization for bromine is given as 29.96 kJ/mol.
- Use the same number of moles as calculated before (0.1567 mol) and multiply it by the enthalpy of vaporization: q₄ = 0.1567 mol × 29.96 kJ/mol = 4.6844 kJ (rounded to 4 decimal places).
5. Add up the heat from all the steps to get the total heat required:
q = q₁ + q₂ + q₃ + q₄ = 0.0850 kJ + 1.6518 kJ + 0.6998 kJ + 4.6844 kJ = 7.1208 kJ (rounded to 4 decimal places).
Therefore, it requires approximately 7.1208 kJ of heat to convert 25.0 g of solid bromine at -7.25°C to the gas phase at 58.8°C.