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Chapter 15: Problem 23

The equilibrium constant for the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ is \(K_{c}=1.3 \times 10^{-2}\) at \(1000 \mathrm{~K}\). (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr? (b) Calculate \(K_{c}\) for $2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) .$ (c) Calculate \(K_{c}\) for $\mathrm{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)$.

Short Answer

Expert verified
(a) At this temperature, the equilibrium favors NO and Br2 formation, as the equilibrium constant is less than 1. (b) The equilibrium constant for the reverse reaction, \(K_c^R\), is 76.92. (c) The equilibrium constant for the half reaction, \(K_c^{1/2}\), is 0.114.

Step by step solution

01

Determine favored side of the reaction

Given that equilibrium constant for the main reaction is \(K_c = 1.3 \times 10^{-2}\). Since \(K_c\) is less than 1, the equilibrium favors the reactant side, i.e., NO and Br2. Answer: Equilibrium favors NO and Br2 formation. (b) Calculate the equilibrium constant for the reverse reaction
02

Calculate equilibrium constant for the reverse reaction

We know that the equilibrium constant for the reverse reaction is the reciprocal of the forward reaction's equilibrium constant. So, for the reaction \(2 \text{NOBr} (g) \rightleftharpoons 2 \text{NO} (g) + \text{Br}_2 (g)\): \[ K_c^R = \frac{1}{K_c} = \frac{1}{1.3 \times 10^{-2}} = 76.92 \] Answer: The equilibrium constant for the reverse reaction, \(K_c^R\) is 76.92. (c) Calculate the equilibrium constant for the half reaction of NOBr dissociation
03

Calculate equilibrium constant for half reaction

The equilibrium constant for a half reaction is the square root of the full reaction's equilibrium constant. So, for the half reaction \(\text{NOBr} (g) \rightleftharpoons \text{NO} (g) + \frac{1}{2} \text{Br}_2 (g)\), we will find the square root of the given equilibrium constant: \[ K_c^{1/2} = \sqrt{1.3 \times 10^{-2}} = 0.114 \] Answer: The equilibrium constant for the half reaction, \(K_c^{1/2}\) is 0.114.

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Most popular questions from this chapter

At \(1285^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is $K_{c}=1.04 \times 10^{-3} .\( A \)1.00-\mathrm{L}$ vessel containing an equilibrium mixture of the gases has \(1.50 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel?

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?

(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?

Consider the reaction $$\begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \rightleftharpoons \\ & 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g), \Delta H=-904.4 \mathrm{~kJ} \end{aligned}$$ Does each of the following increase, decrease, or leave unchanged the yield of \(\mathrm{NO}\) at equilibrium? (a) increase \(\left[\mathrm{NH}_{3}\right] ;(\mathbf{b})\) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\mathbf{c})\) decrease \(\left[\mathrm{O}_{2}\right] ;(\mathbf{d})\) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction $4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\( in a fetus, compared to \)K_{c}$ for the same reaction in an adult.
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