Question

At high temperature, SO; and Os react according to the equation below. Determine the concentration of SO. at equilibrium for this reaction by constructing an ICE table, writing the equilibrium constant expression, and solving for the concentration of SO, at equilibrium. Complete Parts 1-3 before submitting your answer.

2 SO(g) + O(g) = 2 SO(g)
NEXT
An initial mixture of 0.0216 mol SO: and 0.0148 mol O reacted in a 1.0 L vessel. Fill in the ICE table with the appropriate value for each involved species to determine the unknown.
2 SO2(g)
Initial (M)
Equilibrium (M)
>
O(g)
2 SO(g)
Change (M)
RESET
0.0216
0.0216+x
0.0216 +2x
0.0148
1.0
0.0216-2x
+2
0.0148+x
+2x
0.0148 + 2x
-2x
-
0.0148-I
0.0148-2x
0.0216-x

Answer 1

Here's the completed ICE table:

Species Initial (mol/L) Change (mol/L) Equilibrium (mol/L)

SO₂ 0 +2x 2x

O₂ 0 +x x

SO₃ 0.700 -2x 0.700 - 2x

Initial:

• No SO₂ or O₂ are initially present, so their concentrations are 0.
• The initial concentration of SO₃ is 0.700 mol/L, given in the problem.

Change:

• Let's assume the reaction proceeds to the left (decomposition of SO₃).
• For every 2 moles of SO₂ formed, 1 mole of O₂ is formed, and 2 moles of SO₃ are consumed.
• We represent this change with coefficients and the variable "x".

Equilibrium:

• The equilibrium concentrations are calculated by adding the change to the initial concentrations.
• Note that the equilibrium concentration of SO₃ is 0.700 - 2x, as it is being consumed in this scenario.

Answer 2

The ICE table for the problem is shown below and this shows the changes in the concentration of the species.

What is ICE table?

When tackling issues involving the equilibrium constant (K), the ICE table comes in handy. The equilibrium concentrations can be used to construct an expression for the equilibrium constant once the table is completed. This aids in figuring out how to solve for unknowns or which way the reaction will shift in order to reach equilibrium.

The ICE table will look like this;

2
`SO_2`(g) +
`O_2`(g) = 2
`SO_3`(g)

I - - 0.700

C -2x -x +2x

E 2x x 0.700 + 2x