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The reversible reaction: 2SO2(g) O2(g) darrw-tn.gif 2SO3(g) has come to equilibrium in a vessel of specific volume at a given temperature. Before the reaction began, the concentrations of the reactants were 0.060 mol/L of SO2 and 0.050 mol/L of O2. After equilibrium is reached, the concentration of SO3 is 0.040 mol/L. What is the equilibrium concentration of O2

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dsdrajlin

Answer:

0.030 M

Explanation:

Step 1: Make an ICE chart

        2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

I        0.060      0.050          0

C         -2x            -x            +2x

E     0.060-2x  0.050-x       2x

Step 2: Find the value of x

The concentration of SO₃ at equilibrium is 0.040 mol/L. Then,

2x = 0.040

x = 0.020

Step 3: Calculate the concentration at equilibrium of O₂

[O₂] = 0.050 - x = 0.050 - 0.020 = 0.030 M

pstnonsonjoku

The equilibrium concentration of oxygen is 0.030 M.

A reversible reaction is a reaction that can move either in the forward or in the reverse direction. We have the reaction; 2SO2(g) + O2(g) ⇄ 2SO3(g). We can now set up the ICE table as shown below;

  2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

I        0.060      0.050          0

C         -2x            -x            +2x

E     0.060-2x  0.050-x       2x

At equilibrium;

2x = 0.040 M

x = 0.040 M/2 =  0.020 M

For oxygen;

0.050-x  

0.050 M - 0.020 M = 0.030 M

Learn more about equilibrium: https://brainly.com/question/953809

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