Write the equilibrium constant expression for these...

Write the equilibrium constant expression for these reactions: (a) $2 \mathrm{NO}_2(\mathrm{~g})+7 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ (b) $\mathrm{H}_2 \mathrm{CO}_3(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HCO}_3^{-}(a q)$ (c) $2 \mathrm{COF}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{CF}_4(\mathrm{~g})$

Write the equilibrium constant expression for these reactions:
(a) $2 \mathrm{NO}_2(\mathrm{~g})+7 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
(b) $\mathrm{H}_2 \mathrm{CO}_3(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HCO}_3^{-}(a q)$
(c) $2 \mathrm{COF}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{CF}_4(\mathrm{~g})$

Key Concepts

Law of Mass Action

The Law of Mass Action underpins the formulation of the equilibrium constant expression. It asserts that, for a reversible reaction at equilibrium, the rate at which reactants convert into products is proportional to the product of the reactant concentrations, each raised to the power of their coefficient in the balanced equation. This law is fundamental in deriving the equilibrium constant for any given chemical reaction.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between the amounts of reactants and products in a chemical reaction, as indicated by the balanced chemical equation. In equilibrium expressions, these stoichiometric coefficients are used as the exponents for the corresponding concentration or pressure terms, ensuring that the equilibrium constant accurately reflects the reaction’s balanced proportions.

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in constant concentrations of reactants and products over time. At this point, although the reactions continue to occur, they do so at an equal rate, maintaining a steady state without net change in concentrations.

Equilibrium Constant Expression

The equilibrium constant expression is a mathematical representation that relates the concentrations or partial pressures of products and reactants at equilibrium. It is formulated by taking the ratio of the products’ activities (or concentrations/pressures) raised to their stoichiometric coefficients to those of the reactants, which provides a snapshot of the system’s composition at equilibrium.

Transcript

00:01 For each part of this problem, we are going to write out the equilibrium expressions for the equilibrium constants kc and kp.

00:09 We are going to do this whenever it is possible to do so.

00:14 And the way that we know whether or not it is possible to form these expressions is by examining the states of matter for each one of the chemical species involved in the reaction.

00:27 The equilibrium constants are defined in terms of the equilibrium concentrations for, all the species involved in the reaction.

00:36 And we know that we can only have concentrations for species that appear as gases or an aqueous solution.

00:44 And we can only have kp expressions for reactions that involve gases because the kp involves the equilibrium partial pressures of gases, and we can only measure gases in terms of partial pressures.

01:00 So that is how we know whether it's possible to form kc and kp.

01:06 So starting with part a, we see that we have three gases and one liquid.

01:10 So since we have at least one gas, we know that we can form expressions for both kc and kp.

01:15 We know that we do not include liquid species, and in this case liquid water in our expression for kc and kp in part a, because we cannot write liquids in terms of concentrations.

01:28 So starting with the expression for kc, we know that this, these equilibrium concentrations are in terms of molarity, moles per liter.

01:41 So starting with the product side, which goes on the numerator of our expression, the only gas is nh3.

01:47 So that would be a molar concentration on equilibrium of nh3, and we square it due to its stoichiometric coefficient of 2.

01:56 And then we go to the reactant side since the other product is a liquid, and we cannot write it in terms of concentrations.

02:03 For the first reactant, we have the equilibrium concentration of no2, and we square it, and we multiply it by the equilibrium concentration of h2 gas, and we raise it to the power of 7.

02:17 And so now we can write out the expression for kp in terms of the gaseous species...

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