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Reactivity 2. How much, how fast and how far? - How far? The extent of chemical change

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Dynamic Equilibrium: This occurs in a closed system when the rate of the forward reaction equals the rate of the reverse reaction, i.e., rateforward=ratereverserate_{forward} = rate_{reverse}. Macroscopic properties remain constant while microscopic processes continue.

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The Equilibrium Law: For a reversible reaction of the form aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, the equilibrium constant expression is Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}.

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Magnitude of KcK_c: If Kc>1010K_c > 10^{10}, the reaction is considered to go to completion. If Kc<10−10K_c < 10^{-10}, the reaction effectively does not take place. If Kc≈1K_c \approx 1, both reactants and products are present in significant amounts.

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Reaction Quotient (QQ): QQ is calculated using the same expression as KcK_c but with concentrations at any point in time. If Q<KcQ < K_c, the reaction proceeds to the right. If Q>KcQ > K_c, the reaction proceeds to the left. If Q=KcQ = K_c, the system is at equilibrium.

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Le Chatelier’s Principle: If a system at equilibrium is subjected to a change in conditions (concentration, pressure, or temperature), the position of equilibrium shifts to counteract the change.

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Effect of Temperature: Temperature is the only factor that changes the value of KcK_c. For an exothermic reaction (ΔH<0\Delta H < 0), increasing temperature decreases KcK_c. For an endothermic reaction (ΔH>0\Delta H > 0), increasing temperature increases KcK_c.

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Effect of Catalysts: Catalysts increase the rate of both forward and reverse reactions equally. They do not change the position of equilibrium or the value of KcK_c; they only allow equilibrium to be reached faster.

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Gibbs Free Energy and Equilibrium: The position of equilibrium corresponds to the minimum value of Gibbs Free Energy (GG) for the system. The relationship is given by ΔG⊖=−RTln⁡K\Delta G^\ominus = -RT \ln K.

📐Formulae

Kc=[P]p[Q]q[A]a[B]bK_c = \frac{[P]^p [Q]^q}{[A]^a [B]^b}

Qc=[P]initialp[Q]initialq[A]initiala[B]initialbQ_c = \frac{[P]_{initial}^p [Q]_{initial}^q}{[A]_{initial}^a [B]_{initial}^b}

ΔG⊖=−RTln⁡K\Delta G^\ominus = -RT \ln K

Kreverse=1KforwardK_{reverse} = \frac{1}{K_{forward}}

Knew=(Koriginal)nK_{new} = (K_{original})^n

💡Examples

Problem 1:

For the reaction N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g), the equilibrium concentrations at 500 K500 \text{ K} are [N2]=0.60 mol dm−3[\text{N}_2] = 0.60 \text{ mol dm}^{-3}, [H2]=0.40 mol dm−3[\text{H}_2] = 0.40 \text{ mol dm}^{-3}, and [NH3]=0.113 mol dm−3[\text{NH}_3] = 0.113 \text{ mol dm}^{-3}. Calculate the value of KcK_c.

Solution:

Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} Kc=(0.113)2(0.60)(0.40)3K_c = \frac{(0.113)^2}{(0.60)(0.40)^3} Kc=0.0127690.0384K_c = \frac{0.012769}{0.0384} Kc=0.332K_c = 0.332

Explanation:

Substitute the given equilibrium concentrations into the equilibrium constant expression. Note that the power for each concentration corresponds to its stoichiometric coefficient in the balanced equation.

Problem 2:

Consider the endothermic reaction PCl5(g)⇌PCl3(g)+Cl2(g)\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g). Predict the effect on the equilibrium position if: (i) The volume of the container is decreased. (ii) The temperature is increased.

Solution:

(i) Decreasing the volume increases the pressure. The system shifts to the side with fewer moles of gas. Since there is 11 mole of gas on the left and 22 moles on the right, the equilibrium shifts to the left. (ii) Since the reaction is endothermic (ΔH>0\Delta H > 0), increasing the temperature shifts the equilibrium in the direction that absorbs heat (the forward reaction). Thus, the equilibrium shifts to the right.

Explanation:

Le Chatelier's Principle states the system will oppose the change. Higher pressure favors the side with lower volume (fewer gas moles). Higher temperature favors the endothermic direction.

Problem 3:

Calculate the standard Gibbs free energy change ΔG⊖\Delta G^\ominus for a reaction at 298 K298 \text{ K} where the equilibrium constant Kc=1.5×105K_c = 1.5 \times 10^5. (Use R=8.31 J K−1 mol−1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1})

Solution:

ΔG⊖=−RTln⁡K\Delta G^\ominus = -RT \ln K ΔG⊖=−(8.31)(298)ln⁡(1.5×105)\Delta G^\ominus = -(8.31)(298) \ln(1.5 \times 10^5) ln⁡(1.5×105)≈11.92\ln(1.5 \times 10^5) \approx 11.92 ΔG⊖=−(8.31)(298)(11.92)\Delta G^\ominus = -(8.31)(298)(11.92) ΔG⊖≈−29518 J mol−1 or −29.5 kJ mol−1\Delta G^\ominus \approx -29518 \text{ J mol}^{-1} \text{ or } -29.5 \text{ kJ mol}^{-1}

Explanation:

Apply the thermodynamic relationship between the equilibrium constant and Gibbs free energy. A large KK (K>1K > 1) results in a negative ΔG⊖\Delta G^\ominus, indicating the reaction is spontaneous under standard conditions.