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Chapter 15: Problem 71

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g)+\mathrm{Cl}(g)$$ At \(25^{\circ} \mathrm{C}\), the rate constants for the forward and reverse reactions are $1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\( and \)9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}$, respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?(\mathbf{b})\) Are reactants or products more plentiful at equilibrium?

Short Answer

Expert verified
The equilibrium constant (K) at \(25^{\circ}C\) can be calculated as the ratio of the forward and reverse rate constants (k_f and k_r, respectively): \(K = \frac{k_f}{k_r} = \frac{1.4 \times 10^{-28}}{9.3 \times 10^{10}} \approx 1.5 \times 10^{-39}\). Since K is a very small value (K << 1), this indicates that the reaction favors the reactants, and thus, reactants are more plentiful at equilibrium than products.

Step by step solution

01

Write down the equilibrium constant expression

According to the equilibrium constant expression, the equilibrium constant (K) can be calculated as the ratio of the forward and reverse rate constants (k_f and k_r, respectively): \[K = \frac{k_f}{k_r}\]
02

Calculate the equilibrium constant

Now we can plug in the given values of forward and reverse rate constants: \(k_f = 1.4 \times 10^{-28} M^{-1} s^{-1}\) \(k_r = 9.3 \times 10^{10} M^{-1} s^{-1}\) \[K = \frac{1.4 \times 10^{-28}}{9.3 \times 10^{10}}\] Calculate the value of K: \[K \approx 1.5 \times 10^{-39}\]
03

Determine whether reactants or products are more plentiful at equilibrium

Since K is a very small value (K << 1), this indicates that the reaction favors the reactants and not the products. Therefore, reactants are more plentiful at equilibrium than products. In conclusion, the value of the equilibrium constant at \(25^{\circ}C\) is approximately \(1.5 \times 10^{-39}\), and reactants are more plentiful at equilibrium.

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Most popular questions from this chapter

Water molecules in the atmosphere can form hydrogenbonded dimers, \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) The presence of these dimers is thought to be important in the nucleation of ice crystals in the atmosphere and in the formation of acid rain. (a) Using VSF.PR theory, draw the structure of a water dimer, using dashed lines to indicate intermolecular interactions. (b) What kind of intermolecular forces are involved in water dimer formation? (c) The \(K_{p}\) for water dimer formation in the gas phase is 0.050 at $300 \mathrm{~K}\( and 0.020 at \)350 \mathrm{~K}$. Is water dimer formation endothermic or exothermic?

In Section \(11.5,\) we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the corresponding expression for $K_{p \cdot}(\mathbf{b})\( By using data in Appendix \)\mathrm{B}$, give the value of \(K_{p}\) for this reaction at \(30^{\circ} \mathrm{C}\). (c) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

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 \(K_{c}\)

At \(120^{\circ} \mathrm{C}, K_{c}=0.090\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2}\) are $0.100 \mathrm{M}\( and \)0.075 \mathrm{M}$, respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

A chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate molecules bind to a protein involved in cancer. The drug molecules bind the protein in a 1: 1 ratio to form a drug- protein complex. The protein concentration in aqueous solution at $25^{\circ} \mathrm{C}\( is \)1.50 \times 10^{-6} \mathrm{M}$. Drug A is introduced into the protein solution at an initial concentration of $2.00 \times 10^{-6} \mathrm{M}$. Drug B is introduced into a separate, identical protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\). At equilibrium, the drug A-protein solution has an A-protein complex concentration of \(1.00 \times 10^{-6} \mathrm{M}\), and the drug \(\mathrm{B}\) solution has a B-protein complex concentration of $1.40 \times 10^{-6} \mathrm{M}\(. Calculate the \)K_{c}$ value for the A-protein binding reaction and for the B-protein binding reaction. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?
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