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Chapter 15: Problem 42

(a) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

Short Answer

Expert verified
(a) If \(Q_c > K_c\), the reaction must proceed in the reverse direction to create more reactants and reach equilibrium. (b) At the start of the reaction, when only reactants are present and no products have been formed, the value of \(Q_c\) is 0.

Step by step solution

01

(a) Understanding the relationship between \(Q_c\) and \(K_c\)

When a reaction is at equilibrium, the concentrations of the reactants and products will stay constant, and the reaction quotient, \(Q_c\), will be equal to the equilibrium constant, \(K_c\). If the reaction is not at equilibrium, the reaction will proceed in a certain direction to reach equilibrium. There are three possible cases: 1. If \(Q_c < K_c\), the reaction has more reactants than products, and it will proceed in the forward direction to create more products and reach equilibrium. 2. If \(Q_c = K_c\), the reaction is at equilibrium, and the concentrations of the reactants and products will stay constant. 3. If \(Q_c > K_c\), the reaction has more products than reactants, and it will proceed in the reverse direction to create more reactants and reach equilibrium. In the given exercise, we are told that \(Q_c\) is greater than \(K_c\), i.e., \(Q_c > K_c\). Therefore, the reaction must proceed in the reverse direction to create more reactants and reach equilibrium.
02

(b) Calculating the value of \(Q_c\) at the start of the reaction

At the start of a reaction, only reactants are present and no products have been formed. Let's denote reactants as 'A' and 'B', and products as 'C' and 'D'. The balanced chemical equation for the reaction can be written as: \(aA + bB \rightleftharpoons cC + dD\) The reaction quotient, \(Q_c\), is calculated as: \(Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\) At the beginning of the reaction, since no products have been formed, the concentrations of 'C' and 'D' are both 0. Therefore, we have: \(Q_c = \frac{0^c \times 0^d}{[A]^a[B]^b} = 0\) Thus, the value of \(Q_c\) at the start of the reaction, when only reactants are present and no products have been formed, is 0.

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Most popular questions from this chapter

At \(100^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) has the value \(K_{c}=\) \(2.19 \times 10^{-10}\). Are the following mixtures of \(\mathrm{COCl}_{2}, \mathrm{CO}\), and \(\mathrm{Cl}_{2}\) at \(100^{\circ} \mathrm{C}\) at equilibrium? If not, indicate the direction that the reaction must proceed to achieve equilibrium. (a) \(\left[\mathrm{COCl}_{2}\right]=2.00 \times 10^{-3} \mathrm{M},[\mathrm{CO}]=3.3 \times 10^{-6} \mathrm{M}\), \(\left[\mathrm{Cl}_{2}\right]=6.62 \times 10^{-6} \mathrm{M}\) (b) \(\left[\mathrm{COCl}_{2}\right]=4.50 \times 10^{-2} \mathrm{M},[\mathrm{CO}]=1.1 \times 10^{-7} \mathrm{M}\), \(\left[\mathrm{Cl}_{2}\right]=2.25 \times 10^{-6} \mathrm{M}\) (c) \(\left[\mathrm{COCl}_{2}\right]=0.0100 \mathrm{M},[\mathrm{CO}]=\left[\mathrm{Cl}_{2}\right]=1.48 \times 10^{-6} \mathrm{M}\)

Gaseous hydrogen iodide is placed in a closed container at \(425^{\circ} \mathrm{C}\), where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\). At equilibrium it is found that \([\mathrm{HI}]=3.53 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2}\right]=4.79 \times 10^{-4} \mathrm{M}\), and \(\left[\mathrm{I}_{2}\right]=4.79 \times 10^{-4} \mathrm{M}\). What is the value of \(K_{c}\) at this temperature?

Silver chloride, \(\mathrm{AgCl}(\mathrm{s})\), 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 \(C\) and Le Ch芒telier'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 $$ \mathrm{Ag}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}_{2}^{-}(a q) $$ where \(K_{c}=1.8 \times 10^{5}\) at \(25^{\circ} \mathrm{C}\). Although \(\mathrm{AgCl}\) is "not soluble" in water, the complex \(\mathrm{AgCl}_{2}{ }^{\prime}\) is soluble. At \(25^{\circ} \mathrm{C}\), is the solubility of \(\mathrm{AgCl}\) in a \(0.100 \mathrm{M} \mathrm{NaCl}\) solution greater than the solubility of \(\mathrm{AgCl}\) 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 Ch芒telier-type argument? Justify your answer with calculations. (Hint: Any form in which silver is in solution counts as "solubility.")

For the equilibrium $$ \mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{BrCl}(g) $$ at \(400 \mathrm{~K}, K_{c}=7.0\). If \(0.25 \mathrm{~mol}\) of \(\mathrm{Br}_{2}\) and \(0.55 \mathrm{~mol}\) of \(\mathrm{Cl}_{2}\) are introduced into a \(3.0-\mathrm{L}\) container at \(400 \mathrm{~K}\), what will be the equilibrium concentrations of \(\mathrm{Br}_{2}, \mathrm{Cl}_{2}\), and \(\mathrm{BrCl}\) ?

A mixture of \(1.374 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(70.31 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) is heated in a \(2.00\)-L vessel at \(700 \mathrm{~K}\). These substances react according to $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) $$ At equilibrium, the vessel is found to contain \(0.566 \mathrm{~g}\) of \(\mathrm{H}_{2}\). (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{Br}_{2}\), and \(\mathrm{HBr}\). (b) Calculate \(K_{\text {. }}\).
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