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Chapter 15: Problem 19

Can the equilibrium constant ever be a negative number? Explain.

Short Answer

Expert verified
The equilibrium constant (K) can never be a negative number because it represents the ratio of positive concentrations of products to positive concentrations of reactants, with positive stoichiometric coefficients as powers. This implies that K will always be a positive number or equal to 1 when the concentrations of products and reactants are equal at equilibrium.

Step by step solution

01

Definition of the Equilibrium Constant

The equilibrium constant (K) is a measure of the extent of a reversible chemical reaction when it reaches a state of equilibrium. It is the ratio of the concentrations of products to the concentrations of reactants, each raised to the power of their respective stoichiometric coefficients. The equilibrium constant only depends on the temperature of the system.
02

Formula of the Equilibrium Constant

The equilibrium constant is expressed using the following formula: \[ K = \frac{[\text{Products}]}{[\text{Reactants}]} \] Here, the brackets represent the concentrations of products and reactants at equilibrium, and the stoichiometric coefficients are used as powers. Remember that gases and aqueous solutions are included in the equilibrium constant expression, while pure solids and pure liquids are not.
03

Sign of the Equilibrium Constant

The concentrations of products and reactants in the equilibrium constant expression are always expressed in positive values, usually as molar concentrations or partial pressures. Since we are taking the ratio of positive values with positive stoichiometric coefficients as powers, the equilibrium constant will always be a positive number. It can also be equal to 1, which indicates that the concentrations of products and reactants are equal at equilibrium.
04

Conclusion

The equilibrium constant can never be a negative number because it is the ratio of positive concentrations of products to positive concentrations of reactants, with positive stoichiometric coefficients as powers.

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Most popular questions from this chapter

Methane, \(\mathrm{CH}_{4}\), reacts with \(\mathrm{I}_{2}\) according to the reaction \(\mathrm{CH}_{4}(g)+\mathrm{l}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{l}(g)+\mathrm{HI}(g) .\) At \(630 \mathrm{~K}, K_{p}\) for this reaction is \(2.26 \times 10^{-4}\). A reaction was set up at \(630 \mathrm{~K}\) with initial partial pressures of methane of 105.1 torr and of 7.96 torr for \(\mathrm{I}_{2}\). Calculate the pressures, in torr, of all reactants and products at equilibrium.

At \(373 \mathrm{~K}, K_{p}=0.416\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

An equilibrium mixture of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and \(\mathrm{HI}\) at \(458{ }^{\circ} \mathrm{C}\) contains \(0.112 \mathrm{~mol} \mathrm{H}_{2}, 0.112 \mathrm{~mol} \mathrm{I}_{2},\) and \(0.775 \mathrm{~mol} \mathrm{HI}\) in a \(5.00-\mathrm{L}\) vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of 0.200 mol of HI?

(a) How does a reaction quotient differ from an equilibrium constant? (b) If \(Q_{c}

At \(2000^{\circ} \mathrm{C}\) the equilibrium constant for the reaction $$2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ is \(K_{c}=2.4 \times 10^{3}\). If the initial concentration of \(\mathrm{NO}\) is \(0.175 \mathrm{M}\) what are the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2},\) and \(\mathrm{O}_{2} ?\)
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