Pharmaceutical Analytical Chemistry I Lecture 4 PDF

# 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**). -  ### 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) $m = \dfrac{12.5}{180} x \dfrac{1000}{87.5} = 0.794 \ m$