Review Session-13

Review Session-13 - REVIEW SESSION- 13 Chemical...

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REVIEW SESSION- 13 Chemical Thermodynamics
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For each of the following pairs, predict which will have a higher standard entropy at 25 o C. (a) CuO( s ) or Cu 2 O( s ) (b) 1 mol N 2 O 4 ( g ) or 2 mol NO 2 ( g ) (c) CH 3 OH ( g ) or CH 3 OH( l )
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(a) CuO( s ) or Cu 2 O( s ) S 0 usually increases, for compounds in the same phase, when the molar mass increases and/or the number of atoms increases. From Appendix C, the S 0 values are 42.59 vs. 92.36 J/K-mol.
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(b) 1 mol N 2 O 4 ( g ) or 2 mol NO 2 ( g ) Although N 2 O 4 (g) is a more complex molecule, the 2 nd sample has twice the number of molecules and the associated, increased number of positional states will “trump” the molecular complexity. From Appendix C, the S 0 values are (1 mol)(304.3 J/K-mol) = 304.3 J/K and (2 mol)(240.45 J/K-mol) = 480.90 J/ mol.
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c) CH 3 OH ( g ) or CH 3 OH( l ) Methanol in the gas phase will have the higher entropy. From Appendix C, S 0 = 237.6 vs. 126.8 J/K-mol.
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Calculate Δ S 0 values for the following reactions by using Appendix C. In each case, explain the value of Δ S 0 . (a) N 2 H 4 ( g ) + H 2 ( g ) 2 NH 3 ( g ) (b) 2 Al( s ) + 3 Cl 2 ( g ) 2 AlCl 3 ( s ) (c) Mg(OH) 2 ( s ) + 2 HCl( g ) MgCl 2 ( s ) + 2 H 2 O( l ) (d) 2 CH 4 ( g ) C 2 H 6 ( g ) + H 2 ( g )
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(a) N 2 H 4 ( g ) + H 2 ( g ) 2 NH 3 ( g ) Δ S 0 = 2 S 0 [NH 3 ( g )] – S 0 [N 2 H 4 ( g )] – S 0 [H 2 ( g )] Δ S 0 = [2(192.5) – 238.5 – 130.58] J/K-mol = 15.9 J/K-mol The entropy change is very low. There is no change in the total number of gas molecules. It must be that the internal “degrees of freedom” of 2 NH 3 molecules are greater than that of the sum of one N 2 H 4 and one H 2 molecule. Nonetheless, the entropy change is very small.
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(b) 2 Al( s ) + 3 Cl 2 ( g ) 2 AlCl 3 ( s ) Δ S 0 = 2 S 0 [AlCl 3 ( s )] – 2 S 0 [Al( s )] – 3 S 0 [Cl 2 ( g )] Δ S 0 = [2(109.3) – 2(28.32) – 3(222.96)] J/K-mol Δ S 0 = -506.9 J/K-mol The removal of the gaseous species Cl 2 ( g ) causes a large entropy decrease. In addition, the reaction leads to a smaller number of mols of products relative to reactants.
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(c) Mg(OH) 2 ( s ) + 2 HCl( g ) MgCl 2 ( s ) + 2 H 2 O( l ) Δ S 0 = S 0 [MgCl 2 ( s )] + 2 S 0 [H 2 O( l )] – S 0 [Mg(OH) 2 ( s )] – 2 S 0 [HCl( g )] Δ S 0 = [89.6 + 2 (69.91) – 63.24 – 2 (186.69)] J/K-mol Δ S 0 = -207.2 J/K-mol The total number of mols of reactants equals the total number of mols of products. The main effect is HCl( g ) to H 2 O( l ), which carries a negative entropy.
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(d) 2 CH 4 ( g ) C 2 H 6 ( g ) + H 2 ( g ) Δ S 0 = S 0 [C 2 H 6 ( g )] + S 0 [H 2 ( g )] – 2 S 0 [CH 4 ( g )] Δ S 0 = [229.5 + 130.58 – 2 (186.3)] J/K-mol Δ S 0 = -12.5 J/K-mol The entropy change is very low. There is no change in the total number of gas molecules. It must be that the internal “degrees of freedom” of two CH 4 molecules are greater than that of the sum of one C 2 H 6 and one H 2 molecule. Nonetheless, the entropy change is very small.
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What is the difference between Δ G 0 and Δ G?
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Review Session-13 - REVIEW SESSION- 13 Chemical...

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