Pharmaceutical Analytical Chemistry I Lecture 4 PDF

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El Saleheya El Gadida University

Dr. Ahmed Abdel-Monem

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chemical equilibrium pharmaceutical chemistry analytical chemistry acid-base chemistry

Summary

This lecture covers chemical equilibrium, including homogeneous and heterogeneous reactions, reversible and irreversible reactions, and the law of mass action. It also details the Le Chatelier's principle concerning reaction shifts based on changes in pressure and temperature, and introduces fundamental theories of acids, bases, and ionic equilibrium.

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# Pharmaceutical Analytical Chemistry I ## Lecture (4) **Dr. Ahmed Abdel-Monem** ## Chemical Equilibrium ### Classification of chemical reactions 1. **Homogeneous reaction**: Only one phase is present - $H_2(g) + I_2(g) \rightarrow 2HI(g)$ (all phases are gases) - $CH_3COCI + CH_3OH \rig...

# Pharmaceutical Analytical Chemistry I ## Lecture (4) **Dr. Ahmed Abdel-Monem** ## Chemical Equilibrium ### Classification of chemical reactions 1. **Homogeneous reaction**: Only one phase is present - $H_2(g) + I_2(g) \rightarrow 2HI(g)$ (all phases are gases) - $CH_3COCI + CH_3OH \rightarrow CH_3COOCH_3 + HCl$ (Liquids) Acetyl chloride Methyl alcohol methyl acetate 2. **Heterogeneous reactions**: The mixture is not uniform - $C_2H_5OH(L) \rightleftharpoons C_2H_4(g) + H_2O$ ethanol ethane ### Another Classification of chemical reactions 1. **Irreversible reaction**: When reaction proceeds in one direction - $A + B \rightarrow AB$ 2. **Reversible reactions**: Reactions that proceed either in the forward or in the backward reactions - $N_2(g) + 3H_2 (g) \rightleftharpoons 2NH_3(g)$ - **All Reversible processes tend to attain a state of equilibrium**: (rate at which the forward reaction is proceeding = the rate at which the reverse reaction is proceeding). ### Law of mass action - $A_2(g) + B_2(g) \rightleftharpoons 2AB (g)$ - The rate of the forward reaction is: $Rate_f = K_f[A_2][B_2]$ - The rate of the reverse reaction is: $Rate_r = K_r[AB]^2$ - **At equilibrium**: $Rate_f = Rate_r$ - $K_f[A_2][B_2] = K_r[AB]^2$ - Therefore: $K = \dfrac{[AB]^2}{[A_2][B_2]}$ - **Where**: $K_f/K_r = K$(equilibrium constant) - If $K$ is a larger number, $[AB]^2$ will be larger, the **forward reaction** is fairly complete. - If $K$ is a smaller number, the **reverse reaction** is fairly complete. ### Example 1: - **For the reaction**: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ - The conc. of the substances present in an equilibrium mixture at 25°C are [N₂O₄]=4.27x 10⁻² mol/L & [NO₂]=1.41x10⁻² mol/L - **What is the value of $K_c$ for this temp?** - **Answer**: $K_c = \dfrac{[NO_2]^2}{[N_2O_4]}$ - $= (1.41x10^{-2})^2 / (4.27x10^{-2})$ - $= 4.66x10^{-3}$ mol/L ## Le Chatelier's Principle - It predicts how a system in **equilibrium** will respond to changes in experimental conditions (e.g. temp. or pressure) - It states that: "A system in equilibrium reacts to a stress in a way that **counteracts** the stress and establishes a new **equilibrium state**." - It deals with change in: 1. Concentration change 2. Pressure changes 3. Temperature changes 4. Effect of Catalyst 5. Addition of inert Gas #### 1. Concentration change: - If conc. Is increased, the equilibrium will shift to in a way that will decrease the conc. of the substance that was added. - **e.g**. $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ - Increase $H_2$ Or $I_2$ will shift the reaction toward the formation of HI. - Removal of $H_2$ or $I_2$ ← Reaction shift to decomposition of HI. #### 2. Pressure change: - Increase the pressure causes a shift in the direction that will decrease the number of moles of gas. - **e.g**. $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ - When the pressure is increased (the volume of the system decreased), the position of the equilibrium shift to the right. [ i.e. increase $2SO_3$ ] - For the reaction in which $Δn = 0$, pressure change have no effect on the position of the equilibrium - **e.g**. $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ #### 3. Temperature change: - **For the reaction**: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ - $ΔH = -92.4$ KJ - Since $ΔH$ is negative, the reaction to the right evolves heat and reaction is written as - $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) + 92.4$ KJ - Then the highest yields of $NH_3$ will be obtained at the lowest temp. - **Also consider the reaction**: $CO_2(g) + H_2(g) \rightleftharpoons CO(g) + H_2O(g)$ - $ΔH = +41.2$ KJ - Since $ΔH$ is positive, the reaction written as - $CO_2(g) + H_2(g) + 41.2$ KJ $\rightleftharpoons CO(g) + H_2O(g)$ - Increasing the temp. shift the reaction to the right, and Vice versa. #### 4. Effect of catalyst: - Addition of catalyst caused the reaction to achieve equilibrium faster, but **doesn't alter** the position of equilibrium. #### 5. Addition of inert gas: - If an inert gas is introduced into a reaction vessel containing other gases, it will **not affect** the position of equilibrium because it will not alter the partial pressures or the concentrations of any of the substances already present. ## Ionic Equilibrium ### Theories of Acids and bases 1. **Arrhenius theory** - According to this theory: - An acid produce $H^+$ in aqueous solution - A base produce $OH^-$ in aqueous solution - The reaction between an acid and a base in which water is produced is called **neutralization** - $HCl (aq) \rightarrow H^+(aq) + Cl^-(aq)$ **acid** - $NaOH(aq) \rightarrow Na^+(aq) + OH^-(aq)$ **base** - $H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) \rightleftharpoons Na^+(aq) + Cl^-(aq) + H_2O$ **neutralization** - One of the most serious limitations of this theory is its treatment of the weak base**ammonia**. 2. **Bronsted - Lawry Concept** - In this concept he define acid and base (which may be molecules or ions) in terms of the exchange of a proton. - an acid donates a proton to a base which accept it. - In losing a proton, acid 1 becomes base 1(the conjugated base of acid 1). - In gaining a proton, base 2 become acid2 (the conjugated acid of base 2) - $Acid_1 + base_2 \rightleftharpoons acid_2 + base_1$ - $CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH_3COO^-$ - $H_2O + NH_3 \rightleftharpoons NH_4^+ + OH^-$ - $H_3O^+ + OH^- \rightleftharpoons H_2O + H_2O$ - The acid or base is the stronger, while its conjugated base or acid is the weaker. 3. **The Lewis Concept** - The formation of the coordinate bond is the basis for defining acid-base reaction. - **a base** (a **nucleophilic** substance)supplies **an unshared electron pair** for the formation of a coordinate bond with **an acid**(**an electrophilic substance**). - ![chemical formula](https://www.google.com/search?q=chemical+formula+for+lewis+acid+and+base&tbm=isch&ved=2ahUKEwi64J-y2v79AhW-H7cAHb1oD6oQ2-cCegQIABAA&oq=chemical+formula+for+lewis+acid+and+base&gs_lcp=CgNpbWcQAzIECAAQQzICCAAyAggAMgIIADICCAAyAggAMgIIADICCAAyAggAMgIIADICCAAyAggAMgIIADoECAAQQzoFCAAQgAQ6BAgAEEM6BAgAEENQBRCwA1CpC2D6AmgAcAB4AIABkgGIAZ4BkgEDMgILjAUMgNKBQg3NDI0NjE5NzI2LjGYAQCgAQGqAQdnd3Mtd2l6&sclient=img&ei=Qm-RYoeLL6G7kwW-0Z2wCA&bih=667&biw=1366&hl=en#imgrc=U75q74vF2K58M) ### The dissociation of water - The dissociation of water is reversible and to a very limited extent as illustrated by its very weak conductivity to an electric current. - $H_2O \rightleftharpoons H^+ + OH^-$ - According to law of mass action - $\dfrac{[H^+][OH^-]}{[H_2O]} = K$ - Since the fraction of water ionized is very minute or negligible, The value of [$H_2O$] is equal to 1 - $[H^+][OH^-] = K_w$ - $K_w$ is known as "The ionic product of water" - under ordinary experimental conditions, at 25°C, - $K_w = 1x10^{-14}$ - $[H^+][OH^-] = 1x10^{-14}$ - Since the dissociation of water gives rise to equal number of hydrogen and hydroxyl ions, So equation 2 can be written as: - $[H^+]^2 = K_w = 1 x 10^{-14}$ - $[H^+]=10^{-7}$ - It follow that - If $[H^+]=[OH^-]=10^{-7}$-------the solution is neutral - If $[H^+]$ is more than $10^{-7}$-------the solution is acidic - If $[H^+]$ is less than $10^{-7}$-------the solution is alkaline ### The pH concept - The conc. of $H^+$ in a solution may be expressed in terms of the pH scale. - The pH scale of a solution is defined as : - $pH = log \dfrac{1}{[H^+]} = -log [H^+]$ - **e.g**. In solution of $[H^+]= 10^{-3}M$ - $pH = - log [H^+]= -(-3) = 3$ - **For a neutral solution**: - $[H+] = 1 x 10^{-7}$ - $pH = -log 10^{-7} = 7$ - $pH + POH = 14$ ## Solutions - **Solutions** are homogeneous mixtures of two or more substances. - Dissolving medium is called the **solvent**. - Dissolved species are called the **solute**. - There are three states of matter (solid, liquid, and gas) which when mixed two at a time gives nine different kinds of mixtures. - Seven of the possibilities can be homogeneous. - Two of the possibilities must be heterogeneous. ### Seven Homogeneous Possibilities | Solute | Solvent | Example | |---|---|---| | Solid | Liquid | salt water | | Liquid | Liquid | mixed drinks | | Gas | Liquid | carbonated beverages | | Liquid | Solid | dental amalgams | | Solid | Solid | alloys | | Gas | Solid | metal pipes | | Gas | Gas | air | ### Two Heterogeneous Possibilities | Solute | Solvent | Example | |---|---|---| | Solid | Gas | dust in air | | Liquid | Gas | clouds, fog | ### Ways of expressing concentration - Concentration may be expressed: - Qualitatively - Quantitatively #### Qualitative terms - **Dilute Solution** has a relatively **small concentration** of solute. - **Concentrated solution** has a relatively **high concentration** of solute. - N.B. A solution that contains as much solute as can be dissolved is called a **saturated solution** #### Quantitative terms - **Weight to volume**: - **Molarity (M)**: “The no. of moles of solute dissolved in 1 liter of solution.” - $Molarity \ M = \dfrac{moles \ of \ solute}{liters \ of \ solution}$ - $No \ of \ moles \ n = \dfrac{mass}{M.Wt (solute)}$ - $M = \dfrac{W}{M.Wt} \times \dfrac{1000}{V}$ - **Weight to weight**: - **Weight percent (wt%)**: "Number of grams of solute which present in 100 gram of solution." - $mass \ \% = \dfrac{mass \ of \ solute}{mass \ of \ solution} \times 100\%$ - **e.g**. 10% by weight glucose means: 10 gm glucose + 90 gm H2O = 100 gm solution. - % Solute = (10/100) x 100 = 10 %. - % Solvent = (90/100) x 100 = 90 % - **Mole fraction (X)**: "The ratio of the number of moles of one component to the total number of moles of all components in the mixture solution." - In a two component solution A & B, the mole fraction of one component, A, has the symbol XA - $X_A = \dfrac{number \ of \ moles \ of \ A}{number \ of \ moles \ of \ A + number \ of \ moles \ of \ B}$ - The mole fraction of component B - $X_B$ - $X_B = \dfrac{number \ of \ moles \ of \ B}{number \ of \ moles \ of \ A + number \ of \ moles \ of \ B}$ - Note that $X_A + X_B = 1$ - **Molality (m)**: "is a concentration unit based on the number of moles of solute per **kilogram** of **solvent**. It is similar to molarity but not the same" - $Molality \ m = \dfrac{moles \ of \ solute}{kilograms \ of \ solvent}$ - $m = \dfrac{mas (Solute) \times 1000}{M.Wt \ W(Solvent)}$ ### Example: - **What is the molality of** 12.5% solution of glucose C6H12O6, in water? (Mwt. of glucose is 180.0) 1) In 12.5 % solution 12.5 gm $C_6H_{12}O_6$ is dissolved in 100 gm solution. - W solvent = 100 - 12.5 = 87.5 g $H_2O$ 2) no. of moles glucose = 12.5/180 3) $m = \dfrac{12.5}{180} x \dfrac{1000}{87.5} = 0.794 \ m$

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