Question

In a closed flask, sulfur dioxide and oxygen react to produce sulfur trioxide.

2SO2(g) + O2(g)<--->2SO3(g)
In a graph of reaction rate versus time, what is occurring in this closed system when the curve for the rate of the SO2(g) reaction is horizontal?
The concentration of SO2(g) is decreasing as the reaction continues.
The concentrations of SO2(g) and SO3(g) are equal.
The rate that SO2(g) reacts equals the rate that SO3(g) decomposes.
The SO2(g) has been consumed until there is no more in the flask.

Answer 1

• The rate that SO₂ (g) reacts equals the rate that SO₃ (g) decomposes.

Step-by-step explanation:

The chemical equation for the reaction of sulfur dioxide and oxygen to produce sulfur trioxide is given:

• SO₂ (g) + O ₂ (g) ⇄ SO₃ (g)

The double arrow is indicating that this is an equilibrium reaction, which means that, once the reactants start to react, two reactions occur simultaneously:

• Direct or forward reaction:

SO₂ (g) + O ₂ (g) → SO₃ (g) . . . [the arrow goes from left to right]

• Reverse reaction:

SO₂ (g) + O ₂ (g) ← SO₃ (g) . . . [the arrow goes from right to left]

The chemical equilbrium is a dynamic equilibrium, which means that the species (reactants and products) do not remain static but they continue reacting: the rate of both direct and reverse reactions are equal, so in net terms the concentrations do not change.

A graph of reaction rate versus time shows the concentrations of reactants and products over time. Before reaching the equilibrium, the concentrations of reactants will decrease and the concentrations of products will increase, but when the equilibrium is reached the concentrations will remain constant, because, as explained the rate of both forward and reverse reactions are equal.

For the given equation, that means that SO₂ (g) (reactant in the forward reaction) will be consumed at the same rate that SO₃ (g) (reactant in the reverse reaction) will be decomposing.