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Www.soran.edu.iq Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Calculating the work done in gas expansion.  Enthalpy and the first law of.

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Presentation on theme: "Www.soran.edu.iq Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Calculating the work done in gas expansion.  Enthalpy and the first law of."— Presentation transcript:

1 Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Calculating the work done in gas expansion.  Enthalpy and the first law of thermodynamics-  calculating the internal energy change of a gaseous reaction.  Free energy. 1

2  Calculating the work done in gas expansion:  A useful example of mechanical work is the expansion of a gas  As the gas expands, it pushes the piston upward against a constant, opposing, external atmospheric pressure (P). the gas ( system) is doing work on the surroundings. The work can be calculated as follows: w = - PΔV ………… (1) where: P is the external pressure ΔV is the change in volume (V f – V i ).

3  For gas expansion, (ΔV > 0), so ( - PΔV) is a negative quantity. When a gas expands, it's doing work on the surroundings; the internal energy of the system decreases.  For gas compression, (ΔV < 0), so (- PΔV) is a positive quantity. When a gas is compressed, the surroundings are doing work on the system, increasing the internal energy. Example: A gas initially at a pressure of (10.0 atm) and occupying a volume of (5.0L) is allowed to expand at constant temperature against a constant external pressure of ( 4.0 atm). After expansion, the gas occupies a volume of (12.5 L). calculate the work done by the gas on the surroundings.

4 Solution: Step (1): we are given the external pressure in the problem, but we must calculate (ΔV). ΔV = V f – V i = 12.5 L – 5.0 L = 7.5 L. Step( 2): substitute (P) and (ΔV) into equation (1) to solve for work. w = - P ΔV = - ( 4.0 atm ) ( 7.5 L) = x 10 1 L.atm.  it would be more convenient to express (w) in units of joules. The following conversion factor can be obtained: 1 L.atm = J. w = x 10 1 L.atm x = x 10 3 J

5  Enthalpy and the first law of thermodynamics-calculating the internal energy change of a gaseous reaction:  Let's return to the following form of the first law of thermodynamics. ΔE sys = q +w  Under constant-pressure conditions we can write: ΔE = q P +w  Recall that the heat evolved or absorbed (q) by a reaction carried out under constant-pressure conditions is equal to the enthalpy change of the system, ΔH. Thus, ΔE = ΔH +w ……………..(2)

6  Also, we know that for gas expansion or compression under a constant external pressure ( w = - PΔV). Substituting into equation (2) we have: ΔE = ΔH – PΔV …………....(3)  Also, for an ideal gas at constant pressure. PΔV =Δ( nRT) ΔV =

7  Substituting into equation (3) we can obtain: ΔE = ΔH – Δ(nRT)  Finally at constant temperature: ΔE = ΔH – RTΔn ………………(4)  Where (Δn) is defined as: Δn = number of moles of products gases – number of moles of reactant gases.

8 Example: Calculate the change in internal energy when ( 1mole) of (H 2 ) and (1 /2 mole) of (O 2 ) are converted to (1 mole) of (H 2 O) at ( 1 atm) and (25°). For the following reaction: H 2 (g) + 1/ 2 O 2 (g) → H 2 O (l) ΔH° = kJ. Solution: Step(1): calculate the total change in the number of moles of gas. Note that the product is a liquid. Δn = (0 mol ) – ( 1 mol + 1/2 mol) = mol.

9 Step (2): substitute the values for (ΔH°) and (Δn) into equation (4) T = 25° + 273° = 298K. ΔE° = ΔH° – RTΔn. ΔE° = kJ - ( 298 K)( mol) = kJ.

10  Entropy:  Entropy (S) is a direct measure of the randomness or disorder of a system.  The greater the disorder of a system, the greater its entropy. Conversely, the more ordered a system the lower its entropy.  Predicting entropy change: For any substance, the particles in the solid are more ordered than those in the liquid state, which in turn are more ordered than those in the gaseous state. S solid < S liquid < S gas.

11  There are a number of other factors that you need to consider to predict entropy change: Heating increases the entropy of a system because it increases the random motion of atoms and molecules. If a reaction produces more gas molecules than it consumes, the entropy of the system increases. If the total number of gas molecules diminishes during a reaction, the entropy of the system decreases.

12 If there is no net change in the total number of gas molecules in a reaction, then the entropy of the system may increase or decrease, but it will be a relatively small change. Example: Predict whether the entropy increases, decreases or remains essentially unchanged for the following reactions. (a) H 2 O 2 (l) → H 2 O (l) + 1/2 O 2 (g) (b) H + (aq) + OH - (aq) → H 2 O (l) (c ) Ca(OH) 2 (s) + CO 2 (g) → CaCO 3 (s) + H 2 O (g) Solution:

13 (a) The number of moles of gaseous compounds in the products is greater than in the reactant. Entropy increases during this reaction. (b) Two reactants combine into one product in this reaction. Order is increased and so entropy decreases. ( c) The number of moles of gas phase products is the same as the reactants. The entropy remains essentially unchanged.  Calculating entropy changes of a system:  The universe is made up of the system and the surroundings. The entropy change in the universe (ΔS univ. ) for any process is the sum of the entropy changes in the system (ΔS sys ) and in the surrounding (ΔS surr ).

14 ΔS univ. = ΔS sys + ΔS surr  For a spontaneous process, the entropy of the universe increases: ΔS univ. = ΔS sys + ΔS surr > 0.  In chemistry, we typically focus on the entropy of the system. Let's suppose that the system is represented by the following reaction: aA + bB → cC + dD  As is the case for the enthalpy of a reaction, the standard entropy change (ΔS°) is given by:

15 ΔS° rxn = [ c S° ( C ) + d S° ( D)] – [ a S° ( A) + b S° ( B)]  Or, using ( Σ) to represent summation and (m) and (n) for the stiochiometric coefficients in the reaction. ΔS° rxn = Σ n S° (products) – Σ m S° ( reactants).  We can calculate ( ΔS° rxn ), which corresponds to (ΔS sys ). Example: Use standard entropy values to calculate the standard entropy change (ΔS° rxn ) For the following reaction: H 2 (g) + 1 /2 O 2 (g) → H 2 O (l)

16 S° of H 2 O (l), H 2 (g), and O 2 (g) are 69.9, 131.0, and J / mol.K, respectively. Solution: The standard entropy change is given by: ΔS° rxn = S° [H 2 O (l) ] - { S° [ H 2 (g) ] + (1/2)S° [O 2 (g) ]} ΔS° rxn = (1 mol) – [(1 mol) + (1/2 mol) = 69.6 J/K – J/K = J /K.  Tip: this reaction is known to be spontaneous, so you would think that (ΔS) should be positive. Remember, that the value of (ΔS° rxn ) applies only to the system; (ΔS° univ ) will have a positive value.

17  Free energy: The second law of thermodynamic tells us that a spontaneous reaction increases the entropy of the universe, that is, (ΔS° univ > 0). calculate (ΔS° univ ) we must calculate both (ΔS surr ) and (ΔS sys ). However, it can be difficult to calculate (ΔS surr ), and typically we are only concerned with that happens in a particular system.  function, called (Gibbs free energy) (G). Or simply free energy is given by the following equation:

18 G = H – TS Where: H is the enthalpy S is the entropy T is the temperature ( in K)  The change in free energy (ΔG) of a system for a reaction at constant temperature is : ΔG = ΔH – ΔTS o The sign of (ΔG) will allow us to predict whether a reaction is spontaneous at equilibrium. at constant temperature and pressure, we can summarize the following condition in terms of (ΔG).

19 ΔG < 0 a spontaneous process in the forward direction. ΔG > 0 a nonspontaneuos reaction as written. However, the reaction is spontaneous in the reverse direction. ΔG = 0 the system is at equilibrium. there is no net change in the system. Calculating standard free energy changes: Let's again suppose that the system is represented by the following reaction: aA + bB → cC + dD  as is the case for the entropy of a reaction the standard free energy change (ΔG°) is given by:

20 ΔG° rxn = [ c ΔG° f ( C) + d ΔG° f (D) ] – [ a ΔG° f (A) + b ΔG° f (B)] ΔG° rxn = Σ n ΔG° f ( products) – Σ m ΔG° f (reactants)  Where (m) and (n) are stiochiometric coefficients.  (ΔG° f ) is the standard free energy change of formation of a compound. It is the free energy change that occurs when ( 1 mole) of the compound is synthesized from its elements in their standard states, by definition, the standard free energy of formation of any element in its stable form is zero. For example: ΔG° f [ O 2 (g)] = 0 and ΔG° f [ Na(s)] = 0.

21 Example: Calculate (ΔG° rxn ) at (25°) for the following reaction given that: ΔG° f [Fe 2 O 3 (s)] = kJ / mol. ΔG° f [Al 2 O 3 (s)] = kJ / mol. 2Al (s) + Fe 2 O 3 (s) → Al 2 O 3 (s) + 2Fe (s) Solution: The standard free energy change is given by: ΔG° rxn = {ΔG° f [Al 2 O 3 (s)] +2 ΔG° f [Fe (s)]} – {2 ΔG° f [Al] (s) + ΔG° f [Fe 2 O 3 (s)]} = [(1 mol) ( kJ/mol) + 0] – [0 + (1 mol) ( kJ/mol)] = kJ kJ = kJ.

22  We can determine the sign of (ΔG) if we know the signs of both (ΔH) and (ΔS). A negative (ΔH) (an exothermic reaction) and a positive (ΔS) (increase in disorder), give a negative (ΔG)  In addition, temperature may influence the direction of a spontaneous reaction. There are four possible outcomes for the relationship.

23 1-If both (ΔH) and (ΔS) are positive, (ΔG) will be negative only when (TΔS) is greater in magnitude than (ΔH). This condition is met when (T) is large (high temperature). 2-If (ΔH) is positive and (ΔS) is negative, (ΔG) will always be positive, regardless of temperature, the reaction is nonspontaneuos. 3-If (ΔH) is negative and (ΔS) is positive, (ΔG) will always be negative, the reaction is spontaneous. 4-If (ΔH) is negative and (ΔS) is negative, (ΔG) will be negative only when (ΔH) is greater in magnitude than (TΔS). This condition is met when (T) is small (low temperature).

24  We can also calculate the value of (ΔG) if we know the values of (ΔH and ΔS), and the temperature (T). Typically, (ΔH°) is calculated from standard enthalpies of formation: ΔH° = Σ n ΔH° f (products) – Σ m ΔH° f (reactants). Also, (ΔS°) is calculated from standard entropy values. ΔS° = Σ n S°(products) – Σ m S°(reactants). Then, for reactions carried out under standard-state conditions, we can substitute (ΔH°), (ΔS°), and the temperature into the following equation to calculate (ΔG°).

25 ΔG° = ΔH° - TΔS°. Example: Calculate (ΔG°) for the following reaction at (298 K). 2H 2 (g) + CO (g) ↔ CH 3 OH (g) Given that (ΔH° = kJ) and (ΔS° = J/K) for this process. Solution: The standard free energy change is given by: ΔG° = ΔH° - TΔS°.

26 Step (1): let's convert (ΔS°) to units of kJ/K., so we have consistent units. Step (2): substitution ΔH°, ΔS°, and the temperature into the above equation to calculate (ΔG°). ΔG° = ΔH° - TΔS°. ΔG° = kJ – [(298K) ( kJ/K)] = kJ kJ = kJ


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