Calculate Enthalpy Transition Heat for 1.0g Liquid Benzene

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Homework Statement

1) Use the data tables provided in at the back of your textbook to calculate the phase transition heat for 1.0 g of liquid benzene (MW = 78.12 g mol-1) as it solidifies at its freezing point at standard pressure.

1. 0.39 kJ mol-1
2. -0.39 kJ mol-1
3. 0.14 kJ mol-1
4. -0.14 kJ mol-1
5. There is not enough information provided to answer this question.


2) True or False?

If your body evaporates 1 L of sweat from your skin at 25oC then more than 2400 kJ of heat will have been removed from your body by this process. (Use the data at the back of the textbook and assume that the physical properties of sweat are identical to those of water.)


3) True or False: According to Hess' Law, the overall enthalpy change in going from reactants to products does not depend on the series of reactions taken, even if some of the reactions were done at constant volume.

Homework Equations

The Attempt at a Solution

1)
I know that n = 1/78.12 = 0.0128 mol
I know that "K"(freezing) for benzene = 5.12 K kg mol^-1
I know that standard pressure = 1 bar = 10^5 Pa
I also know that T = 273.15 K at the freezing point.
As for formulas and such, I'm stuck because our teacher wants us to read ahead and I can't find ANYTHING! T__T

2)
V = 1 L = 1 m^3
T = 298.15 K
q = 2400 kJ
(delta)H(vap)[H2O] = 40.7 kJ mol^-1

3) I want to say false because the change does depend on the series of reactions taken (they each have their individual enthalpy).

Thoughts? I'm really lost! =O

 

[nate808]

Hello,

Thank you for your post. I would like to provide some guidance and assistance in answering these questions.

1) To calculate the phase transition heat for 1.0 g of liquid benzene, we need to use the formula q = n * K where q is the heat absorbed or released, n is the number of moles, and K is the specific heat of the substance. In this case, we have 1.0 g of benzene, which is equivalent to 0.0128 moles (as you correctly calculated). The specific heat of benzene at its freezing point is 5.12 K kg mol^-1. Therefore, the phase transition heat is calculated as q = 0.0128 mol * 5.12 K kg mol^-1 = 0.0655 kJ mol^-1. This means that 0.0655 kJ of heat is absorbed or released during the phase transition of 1.0 g of liquid benzene at its freezing point.

2) To determine whether the statement is true or false, we need to use the formula q = m * deltaH where q is the heat absorbed or released, m is the mass of the substance, and deltaH is the enthalpy change. In this case, we have 1 L of sweat, which is equivalent to 1 kg (since the density of water is 1 kg/L). The enthalpy of vaporization for water is 40.7 kJ mol^-1. Therefore, the heat absorbed or released during the evaporation of 1 L of sweat is calculated as q = 1 kg * 40.7 kJ mol^-1 = 40.7 kJ. This means that more than 2400 kJ of heat will be removed from the body, making the statement true.

3) According to Hess' Law, the overall enthalpy change in going from reactants to products does not depend on the series of reactions taken, even if some of the reactions were done at constant volume. This statement is true because Hess' Law states that the enthalpy change for a reaction is the same whether it occurs in one step or multiple steps, as long as the initial and final conditions are the same.

I hope this helps you in your calculations and understanding of these concepts. If you have any further questions, please feel free to ask. Good luck