Answer:
- <em><u>The rate that SO₂ (g) reacts equals the rate that SO₃ (g) decomposes</u></em>.
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]
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.
Given:
257J of heat
5500g of mercury
increase by 5.5
degrees Celsius
Required:
Specific heat of
mercury
Solution:
H
= mCpT
257J = (5500g of
mercury) Cp (5.5 degrees Celsius)
Cp = 8.5 x 10^-3
Joules per gram per degree Celsius