Question

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question 19 (1 point)
use the following information to answer the next question.
the equilibrium constant has been determined in
several reactions as follows:

1. n₂(g) + o₂(g) ↔ 2no(g) ( k_{eq} = 4.8 \times 10^{-31} )
2. h₂(g) + cl₂(g) ↔ 2hcl (g) ( k_{eq} = 4.0 \times 10^{31} )
3. 2co(g) + o₂(g) ↔ 2co₂(g) ( k_{eq} = 2.3 \times 10^{22} )
4. cu(s) + 2ag⁺(aq) ↔ cu²⁺(aq) + 2ag(s) ( k_{eq} = 2.0 \times 10^{15} )

the correct arrangement of the above reactions, in decreasing order of their ability
to form products, is

1, 2, 4, 3
1, 4, 3, 2
2, 1, 3, 4
2, 3, 4, 1
3, 4, 2, 1
question 20 (1 point)
phosphoric acid is found in fertilizers. the equilibrium reaction used to produce
phosphoric acid from water and tetraphosphorus decaoxide is shown below.

Explanation:

Step1: Understand the relationship between \(K_{eq}\) and product - formation ability

The equilibrium constant \(K_{eq}=\frac{[products]}{[reactants]}\). A larger \(K_{eq}\) value indicates a greater tendency of the reaction to form products.

Step2: Compare the \(K_{eq}\) values of the given reactions

• For reaction 1: \(K_{eq1}=4.8\times 10^{-31}\)
• For reaction 2: \(K_{eq2}=4.0\times 10^{31}\)
• For reaction 3: \(K_{eq3}=2.3\times 10^{22}\)
• For reaction 4: \(K_{eq4}=2.0\times 10^{15}\)

Step3: Arrange the reactions in decreasing order of \(K_{eq}\)

We have \(K_{eq2}(4.0\times 10^{31})>K_{eq3}(2.3\times 10^{22})>K_{eq4}(2.0\times 10^{15})>K_{eq1}(4.8\times 10^{-31})\)

Answer:

2, 3, 4, 1