Question

hydrogen chloride and oxygen react to form water and chlorine, like this: 4hcl(g)+o₂(g)→2h₂o(g)+2cl₂(g) use this chemical equation to answer the questions in the table below. suppose 100. mmol of hcl and 25.0 mmol of o₂ are added to an empty flask. how much h₂o will be in the flask at equilibrium? none. some, but less than 50.0 mmol. 50.0 mmol. more than 50.0 mmol. suppose 120. mmol of h₂o and 120. mmol of cl₂ are added to an empty flask. how much hcl will be in the flask at equilibrium? none. some, but less than 240. mmol. 240. mmol. more than 240. mmol.

Explanation:

First Question:

Step1: Analyze the stoichiometry

From the equation \(4HCl(g)+O_{2}(g)
ightarrow2H_{2}O(g) + 2Cl_{2}(g)\), the mole ratio of \(HCl\) to \(O_{2}\) is \(4:1\). Given \(n(HCl)=100\space mmol\) and \(n(O_{2}) = 25.0\space mmol\), they are in the stoichiometric ratio. But in a reversible reaction (implied by equilibrium), the reaction does not go to completion.
If the reaction went to completion, \(n(H_{2}O)\) would be \(n(H_{2}O)=\frac{2}{4}n(HCl)=\frac{2}{4}\times100\space mmol = 50\space mmol\) (using the ratio of coefficients of \(H_{2}O\) and \(HCl\) in the balanced equation). Since it is an equilibrium reaction, the amount of \(H_{2}O\) formed is less than the amount if the reaction went to completion.

Second Question:

Step1: Consider the reverse reaction

The reverse reaction is \(2H_{2}O(g)+2Cl_{2}(g)
ightarrow4HCl(g)+O_{2}(g)\). If the reverse reaction went to completion, \(n(HCl)\) would be \(n(HCl)=\frac{4}{2}n(H_{2}O)=\frac{4}{2}\times120\space mmol=240\space mmol\) (using the ratio of coefficients of \(HCl\) and \(H_{2}O\) in the reverse - direction of the balanced equation). But since it is an equilibrium reaction, the reverse reaction does not go to completion.

Answer:

1. Some, but less than \(50.0\space mmol\)
2. Some, but less than \(240\space mmol\)