Question

2. For the reaction, 2SO2(g) + O2(g) 2SO3(g), at 450.0 K the equilibrium con¬stant, Kc, has a value of 4.62. A system was charged to give these initial concentrations, [SO3] = 0.254 M, [O2] = 0.00855 M, [SO2] = 0.500 M. In which direction will it go?

Answer 1

Answer:

the reaction will favor forward, the products.

Explanation:

reaction:

• 2SO2(g) + O2(g) → 2SO3(g)

∴ Kc = 4.62 = [SO3]² / ([O2]*[SO2]²)

initial concentration:

∴ [SO3]i = 0.254 M

∴ [O2]i = 0.00855 M

∴ [SO2]i = 0.500 M

reaction quotient (Q):

⇒ Q = [SO3]² / ([O2]*[SO2]²)

⇒ Q = (0.254)² / ((0.0085)(0.500)²)

⇒ Q = 30.3605

using the reaction quotient to predict the direction of the reaction:

⇒ Q < Kc

In this case, the ratio of products to reagents is less than for the equilibrium system. In other words, the concentration of the reagents is higher than it would be at equilibrium.

Answer 2

Qc > Kc, the reaction favors the reverse reaction.

The equation of the reaction is;

2SO2(g) + O2(g) ⇄ 2SO3(g)

From the information provided;

[SO3] = 0.254 M

[O2] = 0.00855 M

[SO2] = 0.500 M

We can calculate Qc as follows;

Qc = [SO3]^2/ [O2]  [SO2]^2

Qc = [ 0.254]^2/[0.00855] [0.500]^2

Qc = 30

Since Qc > K, the reaction favors the reverse reaction

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