The molar heat capacity of C6H6(ℓ) is 136 J/mol ·◦C and of C6H6(g) is 81.6 J/mol · ◦C.
The molar heat of fusion for benzene is 9.92 kJ/mol and its molar heat of vaporization is
30.8 kJ/mol. The melting point of benzene is 5.5 ◦C, its boiling point is 80.1 ◦C, and its
molecular weight 78.0 g/mol. How much heat would be required to convert 624 g of solid
benzene (C6H6(s)) at 5.5 ◦C into benzene vapor (C6H6(g)) at 100.0 ◦C?
bruh just do the math for the ut quest
To solve this problem, we need to consider the energy required for each step in the process: heating the solid benzene to its melting point, melting the benzene, heating the liquid benzene to its boiling point, and vaporizing the benzene.
1. Heating the solid benzene to its melting point:
- We know the specific heat capacity (C) of benzene in its liquid state (136 J/mol · ◦C).
- The change in temperature (ΔT) is the melting point (5.5 ◦C) minus the initial temperature (0 ◦C).
- We have the mass of solid benzene (624 g), so calculate the number of moles using the molecular weight (78.0 g/mol).
- The equation for the energy required is: Q = n * C * ΔT, where Q is the heat energy, n is the number of moles, C is the specific heat capacity, and ΔT is the change in temperature.
- Calculate Q using the given values.
2. Melting the benzene:
- We know the molar heat of fusion for benzene (9.92 kJ/mol).
- The equation for the energy required is: Q = n * ΔH_fusion, where Q is the heat energy, n is the number of moles, and ΔH_fusion is the molar heat of fusion.
- Calculate Q using the given values.
3. Heating the liquid benzene to its boiling point:
- We know the specific heat capacity of benzene in its liquid state (136 J/mol · ◦C).
- The change in temperature is the boiling point (80.1 ◦C) minus the melting point (5.5 ◦C).
- The equation for the energy required is: Q = n * C * ΔT, where Q is the heat energy, n is the number of moles, C is the specific heat capacity, and ΔT is the change in temperature.
- Calculate Q using the given values.
4. Vaporizing the benzene:
- We know the molar heat of vaporization for benzene (30.8 kJ/mol).
- The equation for the energy required is: Q = n * ΔH_vaporization, where Q is the heat energy, n is the number of moles, and ΔH_vaporization is the molar heat of vaporization.
- Calculate Q using the given values.
Finally, add up the energy required for each step to get the total heat energy required to convert the solid benzene at 5.5 ◦C into benzene vapor at 100.0 ◦C.
You do all of the problems the same way.
First, since the heat capacity is listed in J/mol or kJ/mol, convert 624 g benzene to moles. mol = grams/molar mass.Then
within a phase, q is
mass x specific heat in that phase x (Tfinal-Tinitial).
Converting a phase (solid to liquid or reverse, liquid to vapor or reverse), q is
mass x heat fusion at melting point or
mass x heat vaporization at boiling point.
Then add all of the qs together.