Question

Consider the following gas-phase reaction:

C₂H₂(g) + 4 Cl₂(g) 2 CCl₄(g) + H₂(g)

Using data from Appendix C of your textbook calculate the temperature, T_o, at which this reaction will be at equilibrium under standard conditions (G^o = 0) and choose whether >G^o will increase, decrease, or not change with increasing temperature from the pulldown menu.

T_o = K, and G^o will

---Select---

increase

decrease

not change with increasing temperature.

For each of the temperatures listed below calculate G^o 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 T_o).

(a) At T = 1282 K G^o = kJ/mol, and the reaction is

---Select---

spontaneous

nonspontaneous

near equilibrium under standard conditions.

(b) At T = 1923 K G^o = kJ/mol, and the reaction is

---Select---

spontaneous

nonspontaneous

near equilibrium under standard conditions.

(c) At T = 641 K G^o = kJ/mol, and the reaction is

---Select---

spontaneous

nonspontaneous

near equilibrium under standard conditions.

Answer 1

C₂H₂(g) + 4 Cl₂(g)-----------> 2 CCl₄(g) + H₂(g)

H   = Hf products -Hf reactants

        = (0 +2*-75 ) -(226.7+0)   = -376.7Kj/mole

S   = 2*310 +130.6 -(200..8+4*223)   = -342.2J/mole-K

G    = H-TS

            = -376.7-298*-0.3422 = -274.72KJ   decrease

a. At T = 1282 K

G    = H-TS

         = -376.7-1282*-0.3422

           = 62Kj

G>o non spantaneous reaction

(b) At T = 1923 K

G    = H-TS

            =-376.7-1923*-0.3422    = 281.35Kj/mole  

G>o non spantaneous reaction

(c) At T = 641 K

    G    = H-TS

              = -376.7-641*-0.3422   = -157.35Kj/mole

G <0 spantaneous reaction