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:

Q >> Kc

We have more products than reactans. To reach the equilibrium, the balance will shift to the left.

Explanation:

Step 1: Data given

Temperature = 450.0 K

Kc = 4.62

When Kc > Q, we have more reactants than products. To reach the equilibrium, the balance will shift to the right

When Kc < Q, we have more products than reactans. To reach the equilibrium, the balance will shift to the left.

When Kc = Q,the equiation isatequilibrium

[SO3] = 0.254 M

[O2] = 0.00855 M

[SO2] = 0.500 M

Step 2: The balanced equation

2SO2(g) + O2(g) ⇄ 2SO3(g)

Step 3: Calculate the Q

Q = [SO3]² / [O2][SO2]²

Q = 0.254²/ (0.500 * 0.00855²)

Q = 1765

Q >> Kc

We have more products than reactans. To reach the equilibrium, the balance will shift to the left.

Answer 2

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.