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Temperature Dependence of EnthalpyPowerPoint Presentation

Temperature Dependence of Enthalpy

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Temperature Dependence of Enthalpy

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Temperature Dependence of Enthalpy

Thermodynamic data are often tabulated and can even be located with some effort at 25.0 oC, but are often needed at other temperatures, e.g., for processes occuring in boilers or during the winter in colder climates. Heat capacities are the key data that allow thermodynamic values to be calculated at temperatures other than that at which they are tabulated.

Suppose you wanted to know the enthalpy change at 800 K and 1.00 bar for the reaction:

DHo800K = ?

H2 (g) + 1/2 O2 (g) -----------------> H2O (g)

800 K, 1.00 bar

Tabulated data that you might find related to this reaction are the standard enthalpy of formation of liquid water at 298 K:

DHof, 298K = - 285.830 kJ / mole

H2 (g) + 1/2 O2 (g) -----------------> H2O (l)

298 K, 1.00 bar

and the standard enthalpy of vaporization of water at the normal boiling of water at 373 K:

DHovap, 373K = + 40.670 kJ / mole

H2O (l) -------------------> H2O (g)

373 K, 1.00 bar

13.1

To calculate changes in thermodynamic path independent state functions, such as enthalpy, you need to find a path between the initial state (the reactants at 800K and 1.00 bar in this case) and the final state (the products at 800K and 1.00 bar in this case) along every segment of which you can calculate the change in the thermodynamic function. For the reaction we are considering such a path is diagramed below:

DHo800K = ?

H2 (g) + 1/2 O2 (g) -----------------------------> H2O (g)

800 K, 1.00 bar

DHo6

DHovap, 373K

= + 40.670 kJ / mole

H2O (l) ------> H2O (g)

373 K, 1.00 bar

DHo1

DHo2

DHof, 298K

= - 285.830 kJ / mole

H2 (g) + 1/2 O2 (g) ----------> H2O (l)

298 K, 1.00 bar

DHo4

The alternate path involves a total of 6 steps.

The above method of mapping out an alternate hypothetical path along which the change in a state function can be calculated is very powerful and we will use this approach again and again.

13.2

1. functions, such as enthalpy, you need to find a path between the initial state (the reactants at 800K and 1.00 bar in this case) and the final state (the products at 800K and 1.00 bar in this case) along every segment of which you can calculate the change in the thermodynamic function. For the reaction we are considering such a path is diagramed below:In the 1st step 1.00 moles of H2 (g) are cooled at constant pressure from 800 K to 298 K. The enthalpy change in this step is calculated by integrating the temperature dependent constant pressure heat capacity for H2 (g) (obtained from tables) over this temperature range:

DHo1 = 800 K 298 K n Cp, H2 (g) dT

= (1.00) 800 K 298 K[3.496 -0.1006x10-3 T + 2.419x10-7 T+2] dT

= 3.496 (298 K - 800 K) - 0.1006x10-3 [(298 K)2 - (800 K)2]/2

+ 2.419x10-7 [(298 K)3 - (800 K)3]/3

= (- 1755) + (+27.725) + (-39.15) = - 1766 J

Is the sign on DHo1 consistent with the fact that the H2 (g) is cooling?

2. In the 2nd step 1/2 mole of O2 (g) are cooled at constant pressure from 800 K to 298 K:

DHo2 = 800 K 298 K n Cp, O2 (g) dT

= (1/2) 800 K 298 K[3.0673 + 1.637x10-3 T + 5.118x10-7 T+2] dT

= - 954 J

13.3

3. In step 3 the hydrogen and oxygen are reacting to form liquid water at 298 K and 1.00 bar. This is the standard formation reaction for liquid water at 298 K and the standard enthalpy of formation of liquid water is readily found in tables:

DHof, 298K = - 285.830 kJ / mole

H2 (g) + 1/2 O2 (g) -----------------> H2O (l)

298 K, 1.00 bar

DHo3 = - 285.830 kJ

4. In step 4 the liquid water is warmed at constant pressure from 298 K to its normal boiling point at 373 K or 100 oC.

DHo4 = 298 K 373 K n Cp, H2O (l) dT

= (1.00 mole) 298 K 373 K(75.40 J / mole K) dT

= + 5655 J

Why is this enthalpy change positive?

13.4

5. In step 5 the liquid water is vaporizing to form water vapor at the normal boiling point of 298 K and 1.00 bar. The enthalpy change is the standard enthalpy of vaporization water at the normal boiling point is readily found in tables:

DHovap, 373K = + 40.670 kJ / mole

H2O (l) -------------------> H2O (g)

373 K, 1.00 bar

DHo5 = + 40.670 kJ Why is this enthalpy change positive?

6. In step 6 the water vapor is warmed at constant pressure from 373 K to 800 K.

DHo6 = 373 K 800 K n Cp, H2O (g) dT

= (1.00 mol) 373 K 800 K [3.633 +1.195x10-3 T +1.34x10-7 T+2] dT

= + 2,175 J

Since enthalpy is a state function, the enthalpy change at 800 K and 1.00 bar is equal to the sum of the enthalpy changes along this alternate path:

DHo800 K = DHo1 + DHo2 + DHo3 + DHo4 + DHo5 + DHo6

= (- 1.766 kJ) + (- 0.954 kJ) + (-285.830 kJ) + (5.655 kJ)

+ (+40.670 kJ) + (2.175 kJ)

= - 240.05 kJ

13.5

D vapor at the normal boiling point of 298 K and 1.00 bar. The enthalpy change is the standard enthalpy of vaporization water at the normal boiling point is readily found in tables:Ho1 = 348.2 K 298.2 KCp, I2 (s) dT

DHo3 = 298.2 K 348.2 K Cp, I2 (g) dT

298.2 K, 1.00 atm

I2 (s) ----------------------------- > I2 (g)

DHosub, 298.2 K = + 14.923 kcal

In certain situations the calculation can be simplified:

Consider the sublimation of iodine at 75.0 oC and 1.00 atm:348.2 K

I2 (s) ----------- > I2 (g) DHosub, 348.2 K = ?

1.00 atm

The standard enthalpy of sublimation of iodine vapor at 298.2 K can be found in thermodynamic tables:

298.2 K

I2 (s) ----------- > I2 (g) DHosub, 298.2 K = + 14.923 kcal

1.00 atm

An alternate path between the inital and final states via which we could calculate the enthalpy change at 348.2 K is:

348.2 K, 1.00 atm

I2 (s) ----------------------------- > I2 (g)

DHosub, 348.2 K = ?

13.6

The enthalpy of sublimation at 348.2 K and 1.00 atm can be calculated as:

DHosub, 348.2 K = DHo1 + DHosub, 298.2 K + DHo3

= 348.2 K 298.2 KCp, I2 (s) dT + DHosub, 248.2 K

+ 298.2 K 348.2 K Cp, I2 (g) dT

= DHosub, 298.2 K +298.2 K 348.2 K(Cp, I2 (g) - Cp, I2 (s)) dT

Why is there a minus sign in front of Cp, I2 (s) in the above integral?

= DHosub, 298.2 K +298.2 K 348.2 KDCp dT

where DCp is the difference in the constant pressure heat capacity of the products minus the constant pressure heat capacity of the reactants:

+ Cp, I2 (g) (cal / mole K) = + 8.82

- Cp, I2 (s) (cal / mole K) = - (9.589 + 0.01190 T) ---------------------------- ---------------------------- DCp (cal / mole K) = - 0.92 - 0.01190 T

13.7

The enthalpy change is now: calculated as:

DHosub, 348.2 K = DHosub, 298.2 K +298.2 K 348.2 K(- 0.92 - 0.01190 T) dT

= + 14.923 kcal

+ (10-3 kcal/cal) ( -0.92 (348.2 K - 298.2 K)

- (0.01190 / 2) [(348.2 K)2 - (298.2)2])

= (+ 14.923 kcal) + (- 0.046 kcal) + (-0.192 kcal)

= + 14.685 kcal

This latter example had advantages over the previous example in that it required only a single integration and a tabular approach, less likely to lead arithematic error, could be used to calculate DCp. Why wouldn’t this simpler approach have worked in the previous example?

13.8

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