Consider the following gas-phase reaction: 2 CCl4(g) + H2(g) C2H2(g) + 4 Cl2(g)

Question

Consider the following gas-phase reaction: 2 CCl4(g) + H2(g) C2H2(g) + 4 Cl2(g) Using data from Appendix C of your textbook calculate the temperature, To, at which this reaction will be at equilibrium under standard conditions (Go = 0) and choose whether >Go will increase, decrease, or not change with increasing temperature from the pulldown menu. To = K, and Go will with increasing temperature. For each of the temperatures listed below calculate Go for the reaction above, and select from the pulldown menu whether the reaction under standard conditions will be spontaneous, nonspontaneous, or near equilibrium ("near equilibrium" means that T is within 5 K of To). (a) At T = 641 K Go = kJ/mol, and the reaction is under standard conditions. (b) At T = 1282 K Go = kJ/mol, and the reaction is under standard conditions. (c) At T = 1923 K Go = kJ/mol, and the reaction is under standard conditions.

Explanation / Answer

dH° = (4 * dH Cl2 + dH C2H2) - (2* dH CCl4 + dH H2) = ( 4* 0 + 226.7 kJ/mol) - (2*-102.9 kJ/mol + 0) = 432.5 kJ

dS° = (4 * dS Cl2 + dS C2H2) - (2* dS CCl4 + dS H2) = ( 4*223.1 J/mol K+ 200.9) - (2*309.9 + 130.7) = 342.8 J/mol K   = 0.343 kJ/K

At equilibrium dG° = 0

dG° = dH° - TdS° = 0

T = dH° / dS° = 432.5 kJ / 0.343 kJ/K = 1261 K

Now, for each temperature:

(a) dG° = dH° - TdS° = 435.5 kJ - 641 K*0.343kJ/K = 215.6 kJ > 0 non spontaneous

(b) dG° = dH° - TdS° = 435.5 kJ - 1282 K*0.343kJ/K = - 4.226 < 0 spontaneous

(c) dG° = dH° - TdS° = 435.5 kJ - 1923 K*0.343kJ/K = - 224.1 kJ < 0 spontaneous

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