4. Chlorine dioxide gas \( \left(\mathrm{ClO}_{2}\right) \) is a key ingredient for commercial bleaching agents. The bleaching action accomplished by it is an oxidation-reduction process. (a) By drawing the Lewis structure for \( \mathrm{ClO}_{2} \) (only single bond present), explain why \( \mathrm{ClO}_{2} \) is reduced so readily? [3 marks] (b) (i) When a \( \mathrm{ClO}_{2} \) molecule gains an electron, the chlorite ion, \( \mathrm{ClO}_{2}^{-} \), forms. Draw TWO (2) possible Lewis structures for \( \mathrm{ClO}_{2}^{-} \), with one consisting of single bond only and the other with double bond. [2 marks] (ii) On the basis of the formal charges, determine which Lewis structure is expected to be dominant for chlorite ion \( \mathrm{ClO}_{2}^{-} \). (Show your calculation steps.) [6 marks] (c) Predict the geometry (molecular shape) of the most dominant structure of the chlorite ion \( \left(\mathrm{ClO}_{2}^{-}\right)^{-} \). [1 marks] (d) \( 15.0 \mathrm{~g} \) of sodium chlorite \( \left(\mathrm{NaClO}_{2}\right) \) reacts with \( 2.00 \mathrm{~L} \) of chlorine gas \( \left(\mathrm{Cl}_{2}\right) \) at a pressure of 1.50 atm at \( 21^{\circ} \mathrm{C} \) to give the chlorine dioxide gas \( \left(\mathrm{ClO}_{2}(\mathrm{~g})\right) \), referring to the following balanced equation. 2 CHEM 1002SEF/ CHEM S102F - Essential Chemistry (2023) Assignment \[ 2 \mathrm{NaClO}_{2}(a q)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{NaCl}(a q)+2 \mathrm{ClO}_{2}(g) \] (i) Identify the limiting reagent in this reaction. [3 marks] (ii) Determine the mass \( (\mathrm{g}) \) of \( \mathrm{ClO}_{2} \) produced under this condition. [3 marks] 5. Consider the endothermic reaction, \[ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \quad \Delta \mathrm{H}=+ \text { ve } \mathrm{kJ} / \mathrm{mol} \] When 1.00 mole of \( \mathrm{PCl}_{5} \) is introduced into a \( 5.00 \mathrm{~L} \) reaction chamber at \( 500 \mathrm{~K}, 78.50 \% \) of the \( \mathrm{PCl}_{5} \) dissociates to give an equilibrium mixture of \( \mathrm{PCl}_{5}, \mathrm{PCl}_{3} \), and \( \mathrm{Cl}_{2} \). (a) Write the equilibrium constant expression \( \left(K_{c}\right) \) for this reaction. [1 mark] (b) (i) Calculate the initial concentration (M) of \( \mathrm{PCl}_{5} \) in the reaction vessel. [1 mark] (ii) Using an "ICE" table, calculate the equilibrium concentrations (M) of \( \mathrm{PCl}_{5}, \mathrm{PCl}_{3} \), and \( \mathrm{Cl}_{2} \). [6 marks] (c) Calculate the values of \( K_{c} \) and \( K_{p} \) at \( 500 \mathrm{~K} \) for the equilibrium. [4 marks] 6. Consider the following chemical reaction: \[ 2 \mathrm{NaOH}(a q)+\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(a q) \rightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \] A \( 28.64 \mathrm{~mL} \) solution of \( 37500 \mathrm{ppm} \) (mg/L) of \( \mathrm{NaOH} \) (f.w. = \( 39.997 \mathrm{amu} \) ) is mixed with \( 22.02 \mathrm{~mL} \) solution containing \( 1.615 \mathrm{~g} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2} \) (f.w. = \( \left.148.3 \mathrm{amu}\right) \). Assume volumes are additive. (a) (i) Identify the limiting reagent in this reaction. [3 marks] (ii) Calculate the concentrations of the ions ([ \( \left.\mathrm{Na}^{+}\right],\left[\mathrm{Mg}^{2+}\right],[\mathrm{OH}],\left[\mathrm{NO}_{3}^{-}\right] \)) remaining in solution after the reaction is complete. (Hint: (1) consider the change of moles of spectator ions and reacting ions; (2) assume that approximately "no ions" dissociated in aqueous solution of \( \left.\mathrm{Mg}(\mathrm{OH})_{2}\right) \) [9 marks] \begin{tabular}{|l|l|l|l|l|} \hline & \( 2[\mathrm{NaOH}] / \mathrm{mol} \) & {\( \left[\mathrm{Mg}_{(}\left(\mathrm{NO}_{3}^{-}\right)_{2}\right] / \mathrm{mol} \)} & \( 2\left[\mathrm{NaNO}_{3}^{-}\right] / \mathrm{mol} \) & \( \mathrm{Mg}(\mathrm{OH})_{2}(s) / \mathrm{mol} \) \\ \hline Before reaction & & & & \\ \hline Change & & & & \\ \hline After reaction & & & & \\ \hline \end{tabular} (b) Determine the \( \mathrm{pH} \) value of the final solution at \( 25^{\circ} \mathrm{C} \). [2 marks]
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The chlorine atom has one lone pair of electrons. The oxygen atoms each have three lone pairs of electrons. The chlorine atom is in an excited state, with one electron in the 3d orbital. This makes ClO2 a good oxidizing agent because it can easily accept an Show more…
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1. Chemical Calculations 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) MM = 17.0 g/mol MM = 32.0 g/mol MM = 30.0 g/mol MM = 18.0 g/mol a) How many moles of H atoms are present in 45.1 g of NH3? b) How many grams of NO correspond to 8.25 x 10^24 molecules of NO? c) How many moles of O2 correspond to 25.0 g O2? 2. Balance the following chemical equation. Classify the reaction as combination, decomposition, displacement, exchange, or combustion. FeCl3(aq) + Na2CO3(aq) → Fe2(CO3)3(aq) + NaCl(aq) 3. Assign oxidation numbers to every element in the redox reaction below. What gets oxidized and what gets reduced in the following chemical equation? Cl2(g) + 2 NaBr(aq) → 2 NaCl(aq) + Br2(g) Oxidized: Cl2(g) Reduced: NaBr(aq) 4. Using the kinetic molecular theory, explain why the solubility of a gas increases as pressure increases. 5. Write a balanced chemical equation (using equilibrium arrows when necessary) for the ionization of each of the following substances when they dissolve in water. Mg(NO3)2(s), magnesium nitrate – a strong electrolyte: 2 Mg(NO3)2(s) → 2 Mg^2+(aq) + 4 NO3^-(aq) HF(l), hydrofluoric acid – a weak electrolyte: HF(l) ⇌ H+(aq) + F^-(aq) 6. Using the solubility rules in the lecture notes or textbook, predict whether the following are soluble or insoluble. a) Ca3(PO4)2 - Insoluble b) BaSO4 - Insoluble c) PbCl2 - Insoluble d) Na2S - Soluble 7. Predict the products and write the balanced chemical equation for the precipitation reaction of ammonium sulfate with lead(II) nitrate. Use solubility rules to determine the phase labels. (NH4)2SO4(aq) + Pb(NO3)2(aq) → 2 NH4NO3(aq) + PbSO4(s) 8. Solution Concentrations a) How many grams of glucose must be added to 275 g of water in order to prepare a 25.0% (m/m) glucose solution. b) Calculate the volume (in mL) of a 2.5 M AgNO3 solution needed to obtain 0.065 moles of AgNO3. c) You have 355 mL of a 1.75 M glucose solution. What is the diluted concentration after adding 875 mL of water to the glucose solution? 9. Equilibrium & Le Châtelier’s Principle. 2 SO2(g) + 2 H2O(g) → 2 H2S(g) + 3 O2(g) ΔH = –125 kJ a) At equilibrium, what are the two conditions that must be met? b) Write the equilibrium constant expression for the equilibrium reaction shown above. c) Calculate Keq at 125°C for the equilibrium reaction above if a reaction mixture at equilibrium contains [SO2(g)] = 0.025 M, [H2O(g)] = 0.025 M, [H2S(g)] = 2.5 M, and [O2(g)] = 2.0 M Using the equilibrium reaction above, predict the direction that equilibrium shifts under the following conditions. d) Increasing the concentration of H2O(g): e) Decreasing the concentration of SO2(g): f) Increasing pressure: g) Decreasing temperature: 10. Acid-Base Ionization Reactions (Don’t forget to show charges on any acid or base ions) a) Write a chemical equation showing HPO42– acting as a Brønsted-Lowry acid in H2O. HPO42–(aq) + H2O(l) → H2PO4–(aq) + OH–(aq) b) Write a chemical equation showing HPO42– acting as a Brønsted-Lowry base in H2O. HPO42–(aq) + H2O(l) ⇌ PO43–(aq) + H3O+(aq) c) Write a chemical equation showing NH4+ acting as a Brønsted-Lowry acid in H2O. NH4+(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq) d) Write a chemical equation showing CH3COOH acting as a Brønsted-Lowry acid in H2O. CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO–(aq) 11. What is the conjugate acid of HCO3–? H2CO3 What is the conjugate base of CH3NH3+? CH3NH2 12. Conjugate Acids & Bases and Relative Strengths of Acids a) Identify the Brønsted acid & its conjugate base and the Brønsted base & its conjugate acid below. H2PO4–(aq) + HS–(aq) → H3PO4(aq) + S2–(aq) Brønsted acid: H2PO4– Conjugate base: HPO42– Brønsted base: HS– Conjugate acid: H2S b) Using the table on p.7 of the Chapter 10 Lecture Notes, what is the stronger acid? CIRCLE: H3PO4 OR HS– c) If the Ka value for HNO2 is 4.5 x 10–4 and the Ka value for HCN is 4.9 x 10–10, which one is the stronger acid? CIRCLE: HNO2 OR HCN 13. Acid-Base Chemistry Calculations a) If the solution pH is 5.58, calculate [OH–]. Is this solution acidic or basic? b) What is the pH of a solution if [OH–] = 1.9 x 10–3? Is this solution acidic or basic? c) Calculate the pH of this phosphate buffer if it is composed of 0.125 M H2PO4– and 0.675 M HPO42–. For H2PO4–, Ka = 6.31 x 10–8
Dinesh S.
If you irradiate a solid magnesium surface with electromagnetic radiation having a wavelength of 206 nm, determine the kinetic energy of the electrons that are emitted, in kilojoules per mole. The work function of magnesium is 354.9 kJ/mol. The ionization energy of magnesium is 738.2 kJ/mol.
THE SNAr MECHANISM: NUCLEOPHILIC AROMATIC SUBSTITUTION BY ADDITION-ELIMINATION Aromatic compounds bearing one or more strong electron-withdrawing substituents as well as a leaving group can sometimes undergo nucleophilic substitution instead of electrophilic substitution. The standard mechanisms of $\mathrm{S}_{\mathrm{N}} 2$ and $\mathrm{S}_{\mathrm{N}} 1$ reactions are not possible because of the $s p^{2}$ -hybridization at the carbons of a benzene ring, but a mechanism that involves nucleophilic addition followed by elimination, called $S_{\mathrm{N}} \mathrm{Ar}$ (nucleophilic aromatic substitution), is possible. For example chlorobenzene is not reactive toward $\mathrm{S}_{\mathrm{N}} 2$ or $\mathrm{S}_{\mathrm{N}} 1$ substitution by hydroxide. (FIGURE CANNOT COPY) However, if one or more strong electron-withdrawing groups is bonded ortho or para to the leaving group, substitution by a nucleophile is possible, as the following reactions show. (FIGURE CANNOT COPY) As the number of ortho and para electron-withdrawing groups increases, the temperature required for the reaction decreases, signifying an easier reaction. Meta groups do not produce a similar effect. $m$ -Chloronitrobenzene, for example, is unreactive. The mechanism that operates in these reactions is an addition-elimination mechanism involving the formation of a carbanion with delocalized clectrons, called a Meisenheimer intermediate. The process is called nucleophilic aromatic substitution $\left(\mathrm{S}_{\mathrm{N}} \mathrm{Ar}\right) .$ In the first step of the following example, addition of a hydroxide ion to $p$ -nitrochlorobenzene produces the carbanion; then elimination of a chloride ion yields the substitution product as the aromaticity of the ring is recovered. (FIGURE CANNOT COPY) Carbanion (Meisenheimer intermediate) Structures of the contributing resonance forms are shown further below. The carbanion is stabilized by electron-withdrawing groups in the positions ortho and para to the halogen atom. If we examine the following resonance structures for a Meisenheimer intermediate, we can see how: (FIGURE CANNOT COPY) Especially stable (Negative charges are both on oxygen atoms.) What is the product of the following reaction? (FIGURE CANNOT COPY) STRATEGY AND ANSWER: NaH is a strong base that will convert 4 -methylphenol to its phenoxide salt. $1-(p-$ Toluenesulfonyl)-2,6-dinitrobenzene contains both a good leaving group and two strong electron-withdrawing groups. Thus the likely reaction is a nucleophilic aromatic substitution $\left(S_{N} A r\right),$ leading to the following diaryl ether. 1-Fluoro-2,4-dinitrobenzene is highly reactive toward nucleophilic substitution through an S_yAr mechanism. (In Section 24.5 B we shall see how this reagent is used in the Sanger method for determining the structures of proteins.) What product would be formed when 1-fluoro-2,4-dinitrobenzene reacts with each of the following reagents? (a) EtONa (b) $\mathrm{NH}_{3}$ (c) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$ (d) EtSNa THE CHEMISTRY OF... Bacterial Dehalogenation of a PCB Derivative Polychlorinated biphenyls (PCBs) are compounds that were once used in a variety of electrical devices, industrial applications, and polymers. Their use and production were banned in $1979,$ however, owing to the toxicity of PCBs and their tendency to accumulate in the food chain. 4-Chlorobenzoic acidis adegradation product of some PCBs. It is now known that certain bacteria are able to dehalogenate 4-chlorobenzoic acid by an enzymatic nucleophilic aromatic substitution reaction. The product is 4-hydroxybenzoic acid, and a mechanism for this enzyme-catalyzed process is shown here. The sequence begins with the thioester of 4-chlorobenzoic acid derived from coenzyme A (CoA): (FIGURE CANNOT COPY) Some key features of this enzymatic SyAr mechanism are the following. The nucleophile that attacks the chlorinated benzene ring is a carboxylate anion of the enzyme. When the carboxylate attacks, positively charged groups within the enzyme stabilize the additional electron density that develops in the thioester carbonyl group of the Meisenheimer intermediate. Collapse of the Meisenheimer intermediate, with rearomatization of the ring and loss of the chloride ion, results in an intermediate where the substrate is covalently bonded to the enzyme as an ester. Hydrolysis of this ester linkage involves a water molecule whose nucleophilicity has been enhanced by a basic site within the enzyme. Hydrolysis of the ester releases 4-hydroxybenzoic acid and leaves the enzyme ready to catalyze another reaction cycle.
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