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tdsb.elearningontario.ca question 17 (1 point) use the following inform…

Question

tdsb.elearningontario.ca
question 17 (1 point)
use the following information to answer the next question.
nitric oxide (no(g)) is a colourless gas. the main
sources of nitric oxide are power plants and automobile
engines. the gas is produced by many methods. one of
these methods is given below:
so₂(g) + no₂(g) ↔ no(g) + so₃(g)
suppose that at 700 k, the initial concentration of so₂(g) is 0.25 mol/l and that of
no₂(g) is also 0.25 mol/l. if the equilibrium concentration of so₂(g) is 0.080 mol/l,
the value of equilibrium constant at that temperature will be
○ 0.037
○ 0.22
○ 2.7
○ 4.0
○ 4.5
question 18 (1 point)
a very large value of kₑₚ indicates that __________
○ the reaction reaches equilibrium quickly
○ the reaction does not reach equilibrium quickly
○ the reaction makes a lot of product
○ the reaction does not make a lot of product
○ the reaction is irreversible
question 19 (1 point)

Explanation:

Step1: Calculate the change in concentration

The initial concentration of \(SO_2(g)\) is \(0.25\ mol/L\) and the equilibrium concentration is \(0.080\ mol/L\). The change in concentration of \(SO_2(g)\) (\(\Delta[SO_2]\)) is \(0.25 - 0.080= 0.17\ mol/L\).
Since the stoichiometry of the reaction \(SO_2(g)+NO_2(g)
ightleftharpoons NO(g) + SO_3(g)\) is \(1:1:1:1\), the change in concentration of \(NO_2(g)\) is also \(0.17\ mol/L\). The equilibrium concentration of \(NO_2(g)\) is \(0.25 - 0.17 = 0.080\ mol/L\). The equilibrium concentration of \(NO(g)\) and \(SO_3(g)\) is \(0.17\ mol/L\) each.

Step2: Write the equilibrium constant expression

The equilibrium constant expression \(K=\frac{[NO][SO_3]}{[SO_2][NO_2]}\)

Step3: Substitute the equilibrium concentrations

Substitute \([SO_2]=0.080\ mol/L\), \([NO_2]=0.080\ mol/L\), \([NO]=0.17\ mol/L\) and \([SO_3]=0.17\ mol/L\) into the equilibrium constant expression.
\(K=\frac{0.17\times0.17}{0.080\times0.080}=\frac{0.0289}{0.0064}\approx4.5\)

Brief Explanations

The equilibrium constant \(K_{eq}\) is given by \(K_{eq}=\frac{[products]}{[reactants]}\). A very large value of \(K_{eq}\) means that \(\frac{[products]}{[reactants]}\) is large, which implies that at equilibrium, there is a high concentration of products compared to reactants.

Answer:

4.5

For question 18: