Consider the following reaction: PCl_5(g) PCl_3(g) + Cl_2(g) Using the following

Question

Consider the following reaction: PCl_5(g) PCl_3(g) + Cl_2(g) Using the following data find the Delta H degree, Delta S degree, and Delta G degree at 534 K. For the reaction above state whether the equilibrium shifts to the left or the right (or neither) when the following changes are made in an equilibrium mixture at 25 degree C. T is decreased at constant P V is decreased at constant T Some PCl5 is removed at constant T and V Helium gas is added to the mixture at constant T and V Helium gas is added to the mixture at constant T and P

Explanation / Answer

G0 = -2.303 RT log Kp

Kc = 3.26 x 10-2 at 191 C Calculate Kp

Kp = Kc (RT)n

n = moles product - moles reactant

For any real system NOT at equilibrium:

If Qc > Kc the reaction will go toward more reactants

If Qc < Kc the reaction will go toward more products

enthalpy of reaction

Hf product - Hfreactant

[Hf(PCl3 (g)) + Hf(Cl2 (g))] - [Hf(PCl5 (g))]

[(-287.02) + (0)] - [(-374.9)] = 87.88 kJ

87.88 kJ     (endothermic)

entropy of reaction

Sf product - Sfreactnat

[Sf(PCl3 (g)) + Sf(Cl2 (g))] - [Sf(PCl5 (g))]

[(311.67) + (222.97)] - [(364.6)] = 170.04 J/K

170.04 J/K     (increase in entropy)

From Gf° values:

Gf product - Gfreactant

[Gf(PCl3 (g)) + Gf(Cl2 (g))] - [Gf(PCl5 (g))]

[(-267.78) + (0)] - [(-305)] = 37.22 kJ

37.22 kJ     (nonspontaneous)

From G = H - TS:

G = 87880 KJ - (534 K ) (170.04 J/k)

-2921.36 J = -2.92136 KJ (spontaneous)

1) Qc > Kc backward

2) no change

3) Qc > kc backward

4)no change

5) Qc < kc forward

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