I have answered the first question. Can you please solve just e and f for me please1.)Consider the conversion of dichloroethane (C2H4Cl2) to Cl2 and ethylene (C2H4) gases:C2H4Cl2 (g)  Cl2(g) + C2H4 (g) At 298K and standard conditions this reaction is non-spontaneous. Without doing any calculations, and just by looking at the reaction, deduce the sign of DSo.The sign of DSo is positive . I know this because (give two reasons): It has more gaseous product than the reactantThe higher number gas molecules mean higher entropy. Calculate DSo from the given values of So for the reactants and products.So (C2H4Cl2, g) = 306.5 J/Kmol ; So (Cl2,g)= 223.07 J/Kmol; So (C2H4,g) = 219.36 J/Kmol 1mol (233.07) + 1mol (219.36) – 1mol (306.5) = 135.95J/K DSo = 135.95J/K (give units) Using only the information you know so far, what must the sign of DHo be? positive Explain how you know this.Since the reaction is non-spontaneous, ∆Gorxn is positive. Also, we have found that ∆So is also positive. In the equation ∆Gorxn = ∆Horxn - T*∆So. Since ∆So is positive, then T*∆So will be part negative. So, ∆Gorxn = ∆Horxn - T*∆So will be positive, only if ∆Horxn is positive.Now calculate the value of DHo using the values of DHfo given here:DHfo(C2H4Cl2, g) = -132.2 J/Kmol ; DHfo (C2H4,g) = 52.4 J/Kmol ∆Horxn = 1 mol (0kj/ mol) + 1 mol (52.4kj/ mol) – 1 mol (-132.2kj/ mol) = 184.6 KJ DHo = 184.6 KJ (give units) e.)What is the value of Kp at 298 K? KP = Comment on the magnitude (size) of this number. What does it mean?f.)Would the spontaneity of this reaction change as temperature changes? If so, please describe what specifically happens to the spontaneity as T changes. If it does not change, say why not. If there is a “crossover temperature” calculate it.

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Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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I have answered the first question. Can you please solve just e and f for me please

1.)Consider the conversion of dichloroethane (C₂H₄Cl₂) to Cl₂ and ethylene (C₂H₄) gases:

C₂H₄Cl₂ (g)  Cl₂(g) + C₂H₄ (g)

At 298K and standard conditions this reaction is non-spontaneous.

Without doing any calculations, and just by looking at the reaction, deduce the sign of DS^o.

The sign of DS^ois positive . I know this because (give two reasons):

It has more gaseous product than the reactant
The higher number gas molecules mean higher entropy.

Calculate DS^o from the given values of S^o for the reactants and products.

S^o (C₂H₄Cl₂, g) = 306.5 J/Kmol ; S^o (Cl₂,g)= 223.07 J/Kmol; S^o (C₂H₄,g) = 219.36 J/Kmol

1mol (233.07) + 1mol (219.36) – 1mol (306.5)

= 135.95J/K

DS^o = 135.95J/K (give units)

Using only the information you know so far, what must the sign of DH^o be? positive

Explain how you know this.

Since the reaction is non-spontaneous, ∆G^o_rxn is positive. Also, we have found that ∆S^o is also positive. In the equation ∆G^o_rxn = ∆H^o_rxn - T*∆So. Since ∆S^o is positive, then T*∆So will be part negative. So, ∆G^o_rxn = ∆H^o_rxn - T*∆So will be positive, only if ∆H^o_rxn is positive.

Now calculate the value of DH^ousing the values of DH_f^o given here:

DH_f^o(C₂H₄Cl₂, g) = -132.2 J/Kmol ; DH_f^o (C₂H₄,g) = 52.4 J/Kmol

∆H^o_{rxn =} 1 mol (0kj/ mol) + 1 mol (52.4kj/ mol) – 1 mol (-132.2kj/ mol)

= 184.6 KJ

DH^o = 184.6 KJ (give units)

e.)What is the value of K_p at 298 K?

K_P = Comment on the magnitude (size) of this number. What does it mean?

f.)Would the spontaneity of this reaction change as temperature changes? If so, please describe what specifically happens to the spontaneity as T changes. If it does not change, say why not. If there is a “crossover temperature” calculate it.

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