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Can you please help me with this problem and show me how to do it?
Calculate the amount of heat needed to convert 54.0 grams of solid elemental bromine at -7.31 degrees Celsius to gaseous elemental bromine at 59.2 degrees Celsius. Heat of fusion for bromine is 10.57 kJ/mol. Heat of vaporization: 29.5 kJ/mol; specific heat of liquid bromine: 0.113 cal/g-K; Melting point: -7.31 degrees Celsius; boiling point: 59.2 degrees Celsius; 1 cal = 4.184 J.
1 個解答
- hcbiochemLv 75 年前
These problems require that you work in several individual steps and then add those results together.
Since you are beginning at the melting point of the solid bromine, you must first add heat to convert the solid to a liquid at that temperature. Since the heat of fusion of bromine is given in kJ/mol, first convert your mass to moles. Then multiply that by the heat of fusion to calculate that energy:
54.0 g / 159.8 g/mol = 0.338 mol Br2
q = 0.338 mol X 10.57 kJ/mol = 3.57 kJ
Next, calculate the heat required to raise the temperature of liquid Br2 to its boiling point:
q = m c (T2-T1)
q = 54.0 g (0.113 cal/gK) (59.2 - -7.31) =406 cal X 4.184 J/cal = 1698 J = 1.70 kJ
Next, find the heat required to vaporize the Br2:
q = 0.338 mol X 29.5 kJ/mol = 9.97 kJ
Now, add those values together:
heat required = 9.97 kJ + 1.70 kJ + 3.57 kJ = 15.2 kJ required
