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Chapter 15: Problem 20

Can the equilibrium constant ever be zero? Explain.

Short Answer

Expert verified
No, the equilibrium constant (K) can never be zero. This is because K represents the ratio of product concentrations to reactant concentrations at equilibrium, and both must be present for equilibrium to exist. If K were zero, it would imply that either no reactants or no products are present, which contradicts the definition of equilibrium.

Step by step solution

01

Define the Equilibrium Constant

The equilibrium constant, represented by K or K_eq, is a value that describes the ratio of the concentrations of products to the concentrations of reactants at equilibrium in a chemical reaction. It is calculated using the following formula: K = \(\frac{[C]^c[D]^d}{[A]^a[B]^b}\) where [A], [B], [C], and [D] represent the molar concentrations of the reactants and products, and a, b, c, and d are their respective coefficients in the balanced chemical equation. The equilibrium constant is used to determine the extent of a reaction at equilibrium.
02

Consider the Zero Value for Equilibrium Constant

To evaluate if the equilibrium constant K can ever be zero, consider the implications of this value. If K = 0, this would suggest that either the numerator or the denominator of the K equation must also be zero. However, since the concentrations of reactants and products can never be negative, it means that the numerator and denominator should be positive or zero. Therefore, in order to get K = 0, the numerator of the equation must be zero, and the denominator should be a positive number.
03

Analyze the Implications of a Zero Numerator

If the numerator of the K equation (product concentrations) is zero, this would mean that there are no products formed in the reaction at equilibrium. However, this contradicts the definition of equilibrium, which is the state in which the concentrations of reactants and products are constant and the forward and reverse reaction rates are equal.
04

Conclusion

As the equilibrium state requires the existence of both reactants and products, having an equilibrium constant (K) equal to zero would go against this definition. Therefore, the equilibrium constant can never be zero.

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Most popular questions from this chapter

Write the equilibrium-constant expression for the equilibrium $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ The table that follows shows the relative mole percentages of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) at a total pressure of 1 atm for several temperatures. Calculate the value of \(K_{p}\) at each temperature. Is the reaction exothermic or endothermic? Explain. $$\begin{array}{rll}\hline{\text { Temperature }\left({ }^{\circ} \mathrm{C}\right)} & \mathrm{CO}_{2}(\mathrm{~mol} \%) & \mathrm{CO}(\mathrm{mol} \%) \\ \hline 850 & 6.23 & 93.77 \\ 950 & 1.32 & 98.68 \\\1050 & 0.37 & 99.63 \\ 1200 & 0.06 & 99.94 \\\\\hline\end{array}$$

Consider the following equilibrium: \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7}\) at \(700^{\circ} \mathrm{C}\) (a) Calculate \(K_{p}\). (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{~S} ?\) (c) Calculate the values of \(\mathrm{K}_{c}\) and \(K_{p}\) if you rewrote the balanced chemical equation with \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) instead of \(2 \mathrm{~mol}\).

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

Consider \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g),\) \(\Delta H=-904.4 \mathrm{~kJ} .\) How does each of the following changes affect the yield of \(\mathrm{NO}\) at equilibrium? Answer increase, decrease, or no change: (a) increase \(\left[\mathrm{NH}_{3}\right] ;(\mathbf{b})\) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\mathrm{c})\) decrease \(\left[\mathrm{O}_{2}\right]\) (d) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.

Silver chloride, \(\mathrm{AgCl}(s)\), is an "insoluble" strong electrolyte. (a) Write the equation for the dissolution of \(\mathrm{AgCl}(s)\) in \(\mathrm{H}_{2} \mathrm{O}(l)\) (b) Write the expression for \(K_{c}\) for the reaction in part (a). (c) Based on the thermochemical data in Appendix \(\mathrm{C}\) and Le Châtelier's principle, predict whether the solubility of \(\mathrm{AgCl}\) in \(\mathrm{H}_{2} \mathrm{O}\) increases or decreases with increasing temperature. (d) The equilibrium constant for the dissolution of \(\mathrm{AgCl}\) in water is \(1.6 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\). In addition, \(\mathrm{Ag}^{+}(a q)\) can react with \(\mathrm{Cl}^{-}(a q)\) according to the reaction $$\mathrm{Ag}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}_{2}^{-}(a q)$$ where \(K_{c}=1.8 \times 10^{5}\) at \(25^{\circ} \mathrm{C}\). Although \(\mathrm{AgCl}\) is "not soluble" in water, the complex \(\mathrm{AgCl}_{2}^{-}\) is soluble. At \(25^{\circ} \mathrm{C},\) is the solubility of AgCl in a \(0.100 M\) NaCl solution greater than the solubility of AgCl in pure water, due to the formation of soluble \(\mathrm{AgCl}_{2}^{-}\) ions? Or is the \(\mathrm{AgCl}\) solubility in \(0.100 \mathrm{M} \mathrm{NaCl}\) less than in pure water because of a Le Châtelier-type argument? Justify your answer with calculations.
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