Pharmaceutical Analytical Chemistry Lecture 4 PDF

## Pharmaceutical Analytical Chemistry - I ### Lecture 4 **Associate Prof. Dr Heba Wagdy** ### Outline - Chemical Equilibrium - The equilibrium constant expression - Le Chatelier's principal - Effect of concentration. - Effect of temperature. - Practice Questions. ### Chemical Equilibr...

## Pharmaceutical Analytical Chemistry - I ### Lecture 4 **Associate Prof. Dr Heba Wagdy** ### Outline - Chemical Equilibrium - The equilibrium constant expression - Le Chatelier's principal - Effect of concentration. - Effect of temperature. - Practice Questions. ### Chemical Equilibrium When the rates of the forward and backward reactions are equal and the concentrations of the reactants and products are no longer change with time, chemical equilibrium is reached. - A graph showing forward, reverse, and equal rates plotted over time. - A graph showing the amount of reactants, products, and equilibrium plotted over time. ### The equilibrium constant For a reversible equation at equilibrium (at constant temperature): $aA + bB \rightleftharpoons cC + dD$, where a, b, c, and d are coefficients. According to the law of mass action, the equilibrium expression is represented as follows: $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ where: - K: Equilibrium constant (Temperature dependent). - []: The bracketed terms: Molar concentration, at equilibrium. - Molar(M) = $\frac{moles of solute}{liters of solution}$ ### The equilibrium constant (cont.) - The equilibrium could be homogeneous or heterogeneous. #### Homogeneous Equilibrium: Applies to reactions in which all reacting species are in the same phase. **Example**: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ $K = \frac{[NO_2]^2}{[N_2O_4]}$ #### Heterogeneous Equilibrium: Applies to reactions in which reactants and products are in different phases. **Example**: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ $K =[CO_2]$ - Whenever a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium expression. ### The equilibrium constant (cont.) **Example 1**: Write the equilibrium expression for the following reaction: a) $4NH_3(g) + 7O_2 \rightleftharpoons 4NO_2(g) + 6H_2O(g)$ $K = \frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}$ b) $2H_2O(l) \rightleftharpoons 2H_2(g) + O_2(g)$ $K=[H_2]^2 [O_2]$ c) $2H_2O(g) \rightleftharpoons 2H_2(g) + O_2(g)$ $K = \frac{[O_2][H_2]^2}{[H_2O]^2}$ **Example 2**: $2NO(g)+O_2(g) \rightleftharpoons 2NO_2$ If the equilibrium concentrations at $230^\circ C$ are [NO]= $0.0542 M$, [O2] = $0.127 M$ and [NO₂] = $15.5 M$. Calculate the equilibrium constant at this temperature. $K = \frac{[NO_2]^2}{[NO]^2[O_2]}$ $K = \frac{[15.5]^2}{[0.0542]^2[0.127]}$ $K= 6.44 \times 10^5$ ### Le Châtelier's principle - If an external stress is applied to a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. - **Factors that affect a reacting system at equilibrium:** - Change in Concentration. - Change in Temperature. - Change in Pressure. - Change in Volume. - **Note**: Only a change in temperature changes the value of the equilibrium constant. ### Le Châtelier's principle: Effect of Concentration - If a component (reactant or product) is added to a reaction system at equilibrium (at constant T and P or constant T and V), the equilibrium position will shift in the direction that lowers the concentration of that component. - If a component is removed, to a reaction system at equilibrium (at constant T and P or constant T and V), the equilibrium position will shift in the direction that increases the concentration of that component. - Remember: Change in concentration ONLY change the equilibrium position and not the VALUE. ### Le Châtelier's principle: Effect of Concentration (cont.) **Example 1**: $AS_4O_6(s) + 6C(s) \rightleftharpoons AS_4(g) + 6CO(g)$ Predict the direction of the shift of the equilibrium position in response to each of the following changes in conditions and justify your answer. a) Addition of carbon monoxide. The equilibrium position will shift to the left/backward. Since carbon monoxide is one of the products, so if carbon monoxide is added to a reaction system at equilibrium, the equilibrium position will shift in the direction backward to lower its concentration. b) Addition or removal of carbon or tetraarsenic hexoxide ($As_4O_6$). No effect on the equilibrium position. Since their physical state is solid and their concentration is not included in the equilibrium expression. c) Removal of gaseous arsenic ($As_4$). The equilibrium position will shift to the right/forward. Since gaseous arsenic is one of the products, so if gaseous arsenic is removed from a reaction system at equilibrium, the equilibrium position will shift in the direction forward to increase its concentration. ### Le Châtelier's principle: Effect of Temperature - The change in temperature on equilibrium led to change in equilibrium position and in the value of K. - Treat energy as a reactant in an endothermic process ($ΔH° > 0$) or as a product in an exothermic process ($ΔH° < 0$) and predict the direction of the shift in the same way as when an actual reactant or product is added or removed. **Note**: Although Le Châtelier's principle cannot predict the size of the change in K, it does correctly predict the direction of the change. ### Le Châtelier's principle: Effect of Temperature (cont.) **Example 1**: For each of the following reactions, predict the direction of the shift of the equilibrium position in response to as the temperature is increased. a) $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ $ΔH°= 181kJ$ This is an endothermic reaction ($ΔH°+ve$). Energy can be viewed as a reactant and as the temperature increases, the equilibrium position will shift to the right. b) $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ $ΔH°= -198kJ$ This is an exothermic reaction ($ΔH°-ve$) (energy can be regarded as a product). As the temperature is increased, the value of K decreases (the equilibrium shifts to the left). ### Le Châtelier's principle: Effect of Temperature (cont.) Consider the following equilibrium process between dinitrogen tetrafluoride ($N_2F_4$ ) and nitrogen difluoride ($NF_2$). $N_2F_4(g) \rightleftharpoons 2NF2(g)$ $ΔH° = 38.5 kJ/mol$ Answer the following: 1) Write the equilibrium constant. $N_2F_4(g) \rightleftharpoons 2 NF2(g)$ $K= \frac{[NF_2]^2}{[N_2F_4]}$ 2) Predict the changes in the equilibrium if: a) The reacting mixture _**is**_ heated. The reaction is an endothermic process; therefore the system will shift to the right / to the product. b) Some $N_2F_4$ gas is removed from the reacting mixture. $N_2F_4$ gas is a reactant, and the system will shift to left/to until equilibrium is reestablished. c) A catalyst is added to the reacting mixture. No change If a system is already at equilibrium, as in this case, the addition of a catalyst will not affect either the concentrations of $NF_2$ and $N_2F_4$ or the equilibrium constant. ### Equilibria and Equilibrium Constants of Importance to Analytical Chemistry | Type of Equilibrium | Name and Symbol of Equilibrium Constant | Typical Example| Equilibrium-Constant Expression | | --- | --- | --- | --- | | Dissociation of water | Ion-product constant, $K_w$ | $2H_2O \rightleftharpoons H_3O^+ + OH^-$ | **K** = $[H_3O^+][OH^-]$ | | Heterogeneous equilibrium between a slightly soluble substance and its ions in a saturated solution | Solubility product, $K_{sp}$ | $BaSO_4(s) \rightleftharpoons Ba^{2+} + SO_4^{2-}$ | $K_{sp} = [Ba^{2+}][SO^{2-}]$ | | Dissociation of a weak acid or base | Dissociation constant, $K_a$ or $K_b$ |$CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH_3COO^- $ <br> $CH_3COO^- + H_2O \rightleftharpoons OH^- + CH_3COOH$| $K_a = \frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}$ <br> $K_b = \frac{[OH^-][CH_3COOH]}{[CH_3COO^-]}$ | | Formation of a complex ion | Formation constant, $β_n$ | $Ni^{2+} + 4CN^- \rightleftharpoons [Ni(CN)_4]^{2-}$ | $β_n = \frac{[Ni(CN)_4]^{2-}}{[Ni^{2+}][CN^-]^4}$ | | Oxidation/reduction equilibrium | | $MnO_4^- + 5Fe^{2+} + 8H^+ \rightleftharpoons Mn^{2+} + 5Fe^{3+} + 4H_2O$ | $K_{redox} = \frac{[Mn^{2+}][Fe^{3+}]^5}{[MnO_4^-][Fe^{2+}][H^+]^8}$ | | Distribution equilibrium for a solute between immiscible solvents | | $I_2(aq) \rightleftharpoons I_2(org)$ | $K_d = \frac{[I_2]_{org}}{[I_2]_{aq}}$ | ### Practice Questions **Question 1**: $N_2 (g) + Cl_2 (g) \rightleftharpoons NCI_3(g)$ An analysis of an equilibrium mixture is performed at a certain temperature. It is found that $[NCI_3] = 1.9 \times 10^{-1} M$, $[N_2] = 1.4 \times 10^{-3} M$ and $[Cl_2] = 4.3 \times 10^{-4} M$. Calculate K for the reaction at this temperature. **Question 2**: $H_2 + F_2 \rightleftharpoons 2HF$ The equilibrium constant, K, for the reaction has the value $ 2.1 \times 10^3$ at a particular Temperature. When the system is at equilibrium at this temperature, the concentrations of $H_2$ and $F_2$ are both found to be $0.0021 M$. What is the Concentration of HF in the equilibrium system under these conditions? **Question 3**: $2SO_3 \rightleftharpoons 2SO_2 (g) + O_2(g)$ $H= 197KJ$ What will happen to the number of moles of $SO_3$ in equilibrium with $SO_2$ and $O_2$ in each of the following cases? a) Oxygen gas is added. b) The temperature is decreased. c) Gaseous sulfur dioxide is removed. **Question 4**: Online Assignment - Please log in to Pearson myLab.