# 8.6 Factors Affecting Equilibrium (The Le Chatelier's Principle) The position of a chemical equilibrium is influenced by various external factors. Understanding these factors is crucial, especially in industrial chemistry, to optimize conditions for maximizing the yield of desired products. Le Chatelier's principle provides a qualitative way to predict these effects. > *Le Chatelier's Principle states that if a change of condition (e.g., concentration, pressure, or temperature) is applied to a system in equilibrium, the system will shift in a direction that counteracts the change.* At equilibrium, the rates of the forward and reverse reactions are equal. Changes in conditions alter the frequency and energy of molecular collisions, which in turn changes the reaction rates and causes the equilibrium position to shift until a new equilibrium is established. ### 8.6.1 The Effect of a Change in Concentration When the concentration of a reactant or product in an equilibrium mixture is changed, the system is no longer at equilibrium. The system will then shift to restore equilibrium. Consider the gas-phase equilibrium: $$ \mathrm{CO}_{2(g)} + \mathrm{H}_{2(g)} \rightleftharpoons \mathrm{CO}_{(g)} + \mathrm{H}_{2}\mathrm{O}_{(g)} $$ * **Adding a Substance:** If more $\mathrm{CO}_2$ is added, the concentration of reactants increases. This increases the rate of the forward reaction. The equilibrium shifts to the *right* (towards the products) to consume the added $\mathrm{CO}_2$. This results in higher concentrations of $\mathrm{CO}$ and $\mathrm{H}_2\mathrm{O}$ at the new equilibrium. * **Removing a Substance:** If a product, like $\mathrm{H}_2\mathrm{O}$, were removed from the system, the rate of the reverse reaction would decrease. The equilibrium would shift to the *right* to produce more $\mathrm{H}_2\mathrm{O}$ and restore the balance. **General Rules for Concentration Changes:** * **Increase reactant concentration:** Equilibrium shifts to the **products** (right). * **Decrease reactant concentration:** Equilibrium shifts to the **reactants** (left). * **Increase product concentration:** Equilibrium shifts to the **reactants** (left). * **Decrease product concentration:** Equilibrium shifts to the **products** (right). **Microscopic Explanation:** The rate of reaction depends on the number of effective collisions. Increasing the concentration of a reactant increases the number of collisions for the forward reaction, causing the equilibrium to shift right until the forward and reverse reaction rates become equal again. ### 8.6.2 The Effect of a Change in Pressure Changes in pressure primarily affect gaseous equilibria where the number of moles of gas on the reactant and product sides are different. * **Increasing Pressure:** When pressure is increased, the system tries to reduce the pressure by shifting to the side with *fewer moles of gas*. This reduces the total number of gas molecules and thus the volume. * **Example:** $$ 2\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2\mathrm{SO}_{3(g)} $$ * On the left, there are $2+1=3$ moles of gas. On the right, there are $2$ moles of gas. * Increasing the pressure will shift the equilibrium to the **right** to produce more $\mathrm{SO}_3$. * **Decreasing Pressure:** When pressure is decreased, the system shifts to the side with *more moles of gas* to counteract the drop in pressure. * **No Effect:** If the number of moles of gas is the same on both sides of the equation, a change in pressure will have **no effect** on the position of the equilibrium. * **Example:** $$ \mathrm{H}_{2(g)} + \mathrm{I}_{2(g)} \rightleftharpoons 2\mathrm{HI}_{(g)} $$ * There are $2$ moles of gas on both the left and right sides. ### 8.6.3 The Effect of a Change in Temperature The effect of temperature depends on whether the reaction is exothermic ($\Delta H$ is negative) or endothermic ($\Delta H$ is positive). *A change in temperature is the only factor that changes the value of the equilibrium constant, $K_c$.* * **Exothermic Reactions (Heat is a product):** * Example: $$ 2\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2\mathrm{SO}_{3(g)} \quad \Delta H^{\circ} = -198 \, \text{kJ} $$ * We can write this as: $2\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2\mathrm{SO}_{3(g)} + \text{heat}$ * **Increasing temperature** (adding heat) shifts the equilibrium to the **left** (reactant side) to absorb the excess heat. The value of $K_c$ **decreases**. * **Decreasing temperature** (removing heat) shifts the equilibrium to the **right** (product side) to produce more heat. The value of $K_c$ **increases**. * **Endothermic Reactions (Heat is a reactant):** * Example: $$ \mathrm{N}_{2}\mathrm{O}_{4(g)} \rightleftharpoons 2\mathrm{NO}_{2(g)} \quad \Delta H^{\circ} = +57.2 \, \text{kJ} $$ * We can write this as: $\mathrm{N}_{2}\mathrm{O}_{4(g)} + \text{heat} \rightleftharpoons 2\mathrm{NO}_{2(g)}$ * **Increasing temperature** shifts the equilibrium to the **right** (product side) to consume the added heat. The value of $K_c$ **increases**. * **Decreasing temperature** shifts the equilibrium to the **left** (reactant side) to release heat. The value of $K_c$ **decreases**. ### 8.6.4 The Effect of Addition of a Catalyst A catalyst increases the rate of both the forward and reverse reactions equally. * A catalyst helps the system reach equilibrium **faster**. * A catalyst has **no effect** on the position of the equilibrium or the value of the equilibrium constant, $K_c$. ## Worked Example ### Example 8.5 $K_c$ for the reaction below is $1.0 \times 10^{-3}$ at $230^{\circ} \mathrm{C}$. $$ 2 \mathrm{ICl}_{(g)} \rightleftharpoons \mathrm{Cl}_{2(g)} + \mathrm{I}_{2(g)} $$ An equilibrium mixture in a $2 \, \text{dm}^3$ container contains $1.6$ moles of $\mathrm{ICl}$, $0.05$ moles of $\mathrm{I}_2$, and $0.05$ moles of $\mathrm{Cl}_2$. Determine the new equilibrium concentrations after an additional $1.0$ mole of $\mathrm{ICl}$ is added to the container. ### Solution 1. **Determine Initial Moles After Addition:** * The system is disturbed by adding $1.0$ mole of $\mathrm{ICl}$. * Initial moles of $\mathrm{ICl}$ = $1.6 + 1.0 = 2.6$ mol * Initial moles of $\mathrm{I}_2 = 0.05$ mol * Initial moles of $\mathrm{Cl}_2 = 0.05$ mol * The equilibrium will shift to the right to consume the added $\mathrm{ICl}$. 2. **Set Up an ICE Table (in moles):** Let 'x' be the change in moles of $\mathrm{I}_2$ and $\mathrm{Cl}_2$. | | $2 \mathrm{ICl}_{(g)}$ | $\rightleftharpoons$ | $\mathrm{Cl}_{2(g)}$ | + | $\mathrm{I}_{2(g)}$ | | ----------- | :--------------------: | :------------------: | :------------------: | - | :-----------------: | | **Initial** | 2.6 | | 0.05 | | 0.05 | | **Change** | -2x | | +x | | +x | | **Equil.** | 2.6 - 2x | | 0.05 + x | | 0.05 + x | 3. **Write the $K_c$ Expression and Substitute:** * Volume $V = 2 \, \text{dm}^3$. * Concentrations at equilibrium: * $[\mathrm{ICl}] = \frac{2.6 - 2x}{2}$ * $[\mathrm{Cl}_2] = \frac{0.05 + x}{2}$ * $[\mathrm{I}_2] = \frac{0.05 + x}{2}$ $$ K_c = \frac{[\mathrm{Cl}_2][\mathrm{I}_2]}{[\mathrm{ICl}]^2} $$ $$ 1.0 \times 10^{-3} = \frac{\left(\frac{0.05+x}{2}\right)\left(\frac{0.05+x}{2}\right)}{\left(\frac{2.6-2x}{2}\right)^2} = \frac{(0.05+x)^2}{(2.6-2x)^2} $$ 4. **Solve for x:** * Take the square root of both sides. $$ \sqrt{1.0 \times 10^{-3}} = \frac{0.05+x}{2.6-2x} $$ $$ 0.0316 = \frac{0.05+x}{2.6-2x} $$ $$ 0.0316 \times (2.6 - 2x) = 0.05 + x $$ $$ 0.0822 - 0.0632x = 0.05 + x $$ $$ 0.0322 = 1.0632x $$ $$ x \approx 0.030 \, \text{mol} $$ *(Note: Using the text's rounded value of x = 0.029 for consistency in the final step)* 5. **Calculate New Equilibrium Concentrations:** * Using $x = 0.029 \, \text{mol}$ * $[\mathrm{ICl}] = \frac{2.6 - 2(0.029)}{2} = \frac{2.542}{2} = \mathbf{1.271 \, \text{mol} \, \text{dm}^{-3}}$ * $[\mathrm{Cl}_2] = \frac{0.05 + 0.029}{2} = \frac{0.079}{2} = \mathbf{0.0395 \, \text{mol} \, \text{dm}^{-3}}$ * $[\mathrm{I}_2] = \frac{0.05 + 0.029}{2} = \frac{0.079}{2} = \mathbf{0.0395 \, \text{mol} \, \text{dm}^{-3}}$ ## Possible Questions/Answers * **Q:** For the reaction $2 \mathrm{ICl}_{(g)} \rightleftharpoons \mathrm{Cl}_{2(g)} + \mathrm{I}_{2(g)}$, what are the new equilibrium concentrations if one mole of $\mathrm{ICl}$ is *removed* from the initial mixture? **A:** The equilibrium will shift to the left. The provided answer is: $[\mathrm{I}_2] = 0.0208 \, M$, $[\mathrm{Cl}_2] = 0.0208 \, M$, $[\mathrm{ICl}] = 0.658 \, M$. * **Q:** For the formation of methanol, $\mathrm{CO}_{(g)} + 2\mathrm{H}_{2(g)} \rightleftharpoons \mathrm{CH}_3\mathrm{OH}_{(g)}$, is decreasing the pressure a reasonable approach to increase the yield? **A:** No. There are $1+2=3$ moles of gas on the reactant side and $1$ mole on the product side. Decreasing the pressure will cause the equilibrium to shift to the side with more moles of gas, which is the left (reactant) side. This would *decrease* the yield of methanol. To increase the yield, pressure should be increased. * **Q:** The reaction $2\mathrm{I}_{(g)} \rightleftharpoons \mathrm{I}_{2(g)}$ is exothermic (bond formation releases energy). What would be the effect on the position of equilibrium when the temperature is decreased? **A:** Since the reaction is exothermic, decreasing the temperature will favor the forward reaction to produce more heat. The equilibrium will shift to the **right**, increasing the amount of $\mathrm{I}_2$. | Change | Effect on Equilibrium Position | Effect on $K_c$ | | :-------------------------- | :---------------------------------- | :------------------------------------------------ | | **Increase Reactant Conc.** | Shifts towards products (Right) | No change | | **Increase Product Conc.** | Shifts towards reactants (Left) | No change | | **Increase Pressure** | Shifts to side with fewer gas moles | No change | | **Increase Temperature** | Shifts in endothermic direction | **Changes** (Increases for endo, decreases for exo) | | **Add a Catalyst** | No change | No change | **Significance:** These principles are vital in industrial processes like the Haber-Bosch process for ammonia synthesis, where high pressure and moderate temperature are used to maximize yield, balanced with reaction rate and cost.