a. Write the law of mass action (the equilibrium expression for Keg) [NO3]²_ [0₂] [NO₂]² Keq= b. If gas concentrations are as follows, 2.10 M NO2, 1.75 M O2, and 1.00 M NO3, calculate Keq Keq= = 0.130 M-¹ (NOTE: "M-¹" is another way to write "1/M") [1.00 M₁² [1.75 M] [2.10 M₁² c. Based on the Keq value that you calculated in part b, are the reactants or products predominant (predominant means that there is a greater amount present)? REACTANTS: Keq is much less than 1
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![**Equilibrium and the Law of Mass Action**
**4. For the reaction below:**
\( 2 \text{NO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2 \text{NO}_3 \, (g) \)
**a. Write the law of mass action (the equilibrium expression for \( K_{eq} \))**
\[ K_{eq} = \frac{{[\text{NO}_3]^2}}{{[\text{O}_2][\text{NO}_2]^2}} \]
**b. If gas concentrations are as follows, 2.10 M \(\text{NO}_2\), 1.75 M \(\text{O}_2\), and 1.00 M \(\text{NO}_3\), calculate \( K_{eq} \)**
\[ K_{eq} = \frac{{[1.00 \, \text{M}]^2}}{{[1.75 \, \text{M}][2.10 \, \text{M}]^2}} = 0.130 \, \text{M}^{-1} \]
*(NOTE: "M\(^{-1}\)" is another way to write "1/M")*
**c. Based on the \( K_{eq} \) value that you calculated in part b, are the reactants or products predominant (predominant means that there is a greater amount present)?**
**REACTANTS:** \( K_{eq} \) is much less than 1
This means the equilibrium favors the reactants, indicating that a larger amount of reactants are present compared to products in the equilibrium mixture.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4510828-9e01-409d-b146-fe7482892b4c%2F15dd81b8-a0f3-44ab-90ca-2afbd1fbdc58%2Fk13rum5_processed.png&w=3840&q=75)
2NO2 + O2 <----> 2NO3
1) Law of mass action (equilibrium expression):
Law of mass action states that chemical reaction is directly proportional to the product of reactant concentrations.
Rate of a reaction = k [[A]a[B]b
For above reaction, Rate of a reaction = k [NO3]2
It is applicable for the reversible reactions at equilibrium and at constant temperature, an equilibrium constant remains constant.
Rate of forward reaction = Rate of backward reaction
Kf [A]a[B]b = Kb [C]c [D]d
Equilibrium constant, Keq = Keq = kf/kb = [C]c [D]d/[A]a [B]b
where a,b,c and d are stoichiometric coefficients of A, B, C and D. C and D are the products and A and B are the reactants in the reaction
For above reaction, equilibrium constant expression, Keq = [NO3]2/[O2] [NO2]2
2) [NO2] = 2.10 M
[O2] = 1.75 M
[NO3] = 1.00 M
Keq = [NO3]2/[O2] [NO2]2 = (1.0 M)2/1.75 Mx (2.10 M)2 = 1/7.7175 M= 0.1296 /M 0.13 /M
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