Chem390_solution_5

Chem390_solution_5 - Solution to PS 5 Problem 1(a Standard...

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Unformatted text preview: Solution to PS 5 Problem 1 (a) Standard molar entropy change is given by the total molar entropy for the products minus the entropy for the reactants. In[53]:= S HBRg = 198.7 J ê mol ê K; S Br2g = 245.5 J ê mol ê K; S H2g = 130.7 J ê mol ê K; S HBRg − 0.5 ∗ S Br2g − 0.5 ∗ S H2g Out[56]= 10.6 J K mol (b) The entropy of Br2(g) is larger than Br2(l) at the same T and p by the amount of D vap H/ T vap . The large entropy increase in this case is consistentwith the loss of condensed material H Br2 H l LL to produce gaseous HBr. In[57]:= S Br2l = 152.2 J ê mol ê K; S HBRg − 0.5 ∗ S Br2l − 0.5 ∗ S H2g Out[58]= 57.25 J K mol (c) D S(p,T)= D S(1bar,T)+ Ÿ 1 bar p I ∑ p ' S M T „ p ' =D S(1bar,T)- Ÿ 1 bar p H ∑ T V L p ' „ p '. Assuming ideal gas, the second term becomes - Ÿ 1 bar p H R ê p ' L „ p ' = - Rln H p ê 1 bar L . At 5 bar, the entropy increase is the same for the product H total 1 mole L and the reactants H total 1 mole of gas L . Hence the molar entropy change for 5 bar is the same as at 1 bar. Answer : 10.6 J ê K ê mol Problem 2 In[121]:= S trans = Log B 2 π M kB T h 2 3 ê 2 R T 1 bar Æ 5 ê 2 ë NA F ê . constJSI S rot = Log @ T Æ ê H σ Θ rot LD ; S vib = − Log A 1 − Æ −Θ vib ë T E + Θ vib ê T Æ Θ vib ë T − 1 ; S el = Log @ gel D ; (1) The formula above are given in units of R. Below all entropy values are given in units of R.units of R....
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Chem390_solution_5 - Solution to PS 5 Problem 1(a Standard...

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