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?

Short Answer

Expert verified
A water dimer consists of two water molecules connected by hydrogen bonding, exhibiting a bent structure due to VSEPR theory. The main intermolecular force involved is hydrogen bonding, which is an electrostatic attraction between hydrogen and oxygen. The given equilibrium constants indicate that the dimer formation is an exothermic process, as the equilibrium favors reactants at higher temperatures and products at lower temperatures.

Step by step solution

01

Draw the water dimer structure using VSEPR theory

: VSEPR (Valence Shell Electron Pair Repulsion) theory is used to predict the geometric structure of molecules. Water has the molecular formula H2O, and according to VSEPR theory, it exhibits a bent or angular structure because of the 2 lone pairs present on the oxygen atom. In the case of a water dimer, two water molecules are interacting via intermolecular forces. To show these interactions, draw two separate water molecules with their respective bent structures. Then, connect the hydrogen of one water molecule to the oxygen of the other water molecule with a dashed line to represent the hydrogen bond. Similarly, do this for the other hydrogen in the same water molecule as well.
02

Identify the intermolecular forces involved

: The main intermolecular force in water dimer formation is hydrogen bonding. This interaction arises because of the large difference in electronegativity between hydrogen and oxygen, which leads to an electrostatic attraction between the two. Hydrogen bonding is a strong intermolecular force and plays a major role in determining the properties of water, such as its high boiling point and surface tension.
03

Determine if the process is endothermic or exothermic

: We are given the equilibrium constant (\(K_p\)) values for water dimer formation at two different temperatures. At \(300 \ \mathrm{K}\), \(K_p = 0.050\), and at \(350 \ \mathrm{K}\), \(K_p = 0.020\). Notice that the equilibrium constant decreases as the temperature increases, which implies that the reaction favors the reactants (individual water molecules) at higher temperatures and the products (dimers) at lower temperatures. According to Le Châtelier's principle, this is indicative of an exothermic process, as increasing the temperature leads to a shift of the equilibrium towards the reactants, offsetting the temperature increase and releasing heat. Therefore, water dimer formation is an exothermic process.

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Most popular questions from this chapter

\(\mathrm{NiO}\) is to be reduced to nickel metal in an industrial process by use of the reaction $$\mathrm{NiO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{CO}_{2}(g)$$ At \(1600 \mathrm{~K}\), the equilibrium constant for the reaction is $K_{p}=6.0 \times 10^{2} .\( If a CO pressure of \)20 \mathrm{kPa}$ is to be employed in the furnace and total pressure never exceeds \(101.3 \mathrm{kPa}\), will reduction occur?

(a) Is the dissociation of fluorine molecules into atomic fluorine, \(\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{~F}(g)\) an exothermic or endothermic process? (b) If the temperature is raised by \(100 \mathrm{~K}\), does the equilibrium constant for this reaction increase or decrease? (c) If the temperature is raised by \(100 \mathrm{~K},\) does the forward rate constant \(k_{f}\) increase by a larger or smaller amount than the reverse rate constant \(k_{r} ?\)

For the equilibrium $$2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\operatorname{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If \(5.07 \mathrm{kPa}\) of IBr is placed in a 10.0 -L container, what is the partial pressure of all substances after equilibrium is reached?

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}(g)\) (b) $\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SiCl}_{4}(g)$ (c) $\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)$ (d) $\mathrm{O}_{2}(g)+2 \mathrm{CO}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)$ (e) $\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}^{+}(a q)$ (f) $\mathrm{Fe}^{2+}(a q)+\mathrm{Ce}^{4+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)$ (g) $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$

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?

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