Reversible Reactions and Principles of Chemical Equilibrium

Application of the Law of Mass Action

The law of mass action is demonstrated through the equilibrium of sulphur dioxide and oxygen into sulphur trioxide: 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)

The equilibrium constant expression is given as: K=[SO3]2[SO2]2[O2]K = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}

Experimental data verifies that the equilibrium constant KK remains consistent between different trials:

  • Experiment 1: K1=4.36 M−1K_1 = 4.36\,M^{-1}
  • Experiment 2: K2=4.32 M−1K_2 = 4.32\,M^{-1}

Relationship Between KpK_p and KcK_c

For a reaction at temperature TT, the pressure-based equilibrium constant (KpK_p) and the concentration-based equilibrium constant (KcK_c) are related by: Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}

Variables and constants defined:

  • T=25+273=298 KT = 25 + 273 = 298\,K
  • R=0.08206 L atmmol KR = 0.08206\,\frac{L\,atm}{mol\,K}
  • Δn=moles of gaseous product−moles of gaseous reactant\Delta n = \text{moles of gaseous product} - \text{moles of gaseous reactant}

For the formation of nitrosylchloride (2NO(g)+Cl2(g)⇌2NOCI(g)2NO(g) + Cl_2(g) \rightleftharpoons 2NOCI(g)), Δn=2−(2+1)=−1\Delta n = 2 - (2 + 1) = -1.

The Magnitude and Extent of Reaction

The value of the equilibrium constant KK indicates the tendency of a reaction to occur:

  • K≫1K \gg 1: Indicates that at equilibrium, the system consists mostly of products; the equilibrium lies to the right.
  • K≪1K \ll 1: Indicates that at equilibrium, the system consists mostly of reactants; the equilibrium lies to the left.

The Reaction Quotient (QQ)

The Reaction Quotient (QQ) uses initial concentrations []0[ ]_0 instead of equilibrium concentrations to determine system direction:

  • Q=KQ = K: The system is at equilibrium; no shift occurs.
  • Q>KQ > K: The ratio of products to reactants is too large; the system shifts to the left.
  • Q<KQ < K: The ratio of products to reactants is too small; the system shifts to the right.

Systematic Procedure for Solving Equilibrium Problems

  1. Write the balanced chemical equation.
  2. Write the equilibrium expression using the law of mass action.
  3. List initial concentrations.
  4. Calculate QQ to determine the direction of the shift.
  5. Define the change needed to reach equilibrium (xx) and specify equilibrium concentrations.
  6. Use the quadratic formula to solve for unknowns: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  7. Check calculated concentrations against the known KK value.

Note: Roots resulting in negative concentrations are physically impossible and must be discarded.

Le Chatelier's Principle

Proposed by Henry Lois Le Chatelier, the principle states that if an equilibrium system is subjected to a change, processes occur that tend to counteract partially the imposed change.

Temperature Changes

  • Raising Temperature: Favors the reaction that absorbs heat (endothermic).
  • Lowering Temperature: Favors the reaction that gives off heat (exothermic).

Concentration Changes

  • Increasing Concentration: The system shifts to consume the added component.
  • Decreasing Concentration: The system shifts to produce the missing component.

Pressure Changes

  • Increased Pressure: The system shifts in the direction that results in lower pressure by reducing the volume (fewer moles of gas).
  • Decreased Pressure: The system shifts toward the side with higher volume (more moles of gas).
  • If the change in volume is zero (Δn=0\Delta n = 0), pressure has no effect on the equilibrium position.

Industrial Applications of Equilibrium

  • The Haber Process (N2+3H2⇌2NH3N_2 + 3H_2 \rightleftharpoons 2NH_3 + heat): Highest yields occur at high pressure and low temperature.
  • Sulphur Trioxide Synthesis (2SO2+O2⇌2SO32SO_2 + O_2 \rightleftharpoons 2SO_3): Optimized by high pressure, low temperature, and an excess of air or oxygen.
  • Nitric Oxide Formation (N2+O2⇌2NON_2 + O_2 \rightleftharpoons 2NO): Not affected by pressure changes; highest yields are achieved at high temperatures.
  • Hydrogen Production (CO+H2O⇌CO2+H2CO + H_2O \rightleftharpoons CO_2 + H_2): Highest yield obtained using low temperatures as possible; pressure has no effect.

Questions & Discussion

Question 1: Calculate the value of KpK_p for the formation of nitrosylchloride at 25∘C25^{\circ}C given PNOCI=1.2 atmP_{NOCI} = 1.2\,atm, PNO=5.0×102 atmP_{NO} = 5.0 \times 10^2\,atm, and Pc2=3.0×101 atmP_{c2} = 3.0 \times 10^1\,atm. Response: Kp=(PNOCI)2(PNO)2(PCl2)=(1.2 atm)2(5.0×102 atm)2(3.0×102 atm)2=1.92×103 atmK_p = \frac{(P_{NOCI})^2}{(P_{NO})^2 (P_{Cl2})} = \frac{(1.2\,atm)^2}{(5.0 \times 10^2\,atm)^2 (3.0 \times 10^2\,atm)^2} = 1.92 \times 10^3\,atm.

Question 2: Predict the direction of the reaction H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g) at 448∘C448^{\circ}C (K=50.5K = 50.5) starting with 2.0×10−2 mole2.0 \times 10^{-2}\,mole of HIHI, 1.0×10−2 mole1.0 \times 10^{-2}\,mole of H2H_2, and 3.0×10−2 mole3.0 \times 10^{-2}\,mole of I2I_2 in a 2.0 litre2.0\,litre container. Response: Calculating initial concentrations yields [HI]0=1.0×10−2 M[HI]_0 = 1.0 \times 10^{-2}\,M, [H2]0=5.0×10−3 M[H_2]_0 = 5.0 \times 10^{-3}\,M, and [I2]0=1.5×10−2 M[I_2]_0 = 1.5 \times 10^{-2}\,M. This results in Q=1.3Q = 1.3. Since Q<KQ < K, the reaction proceeds from left to right.

Question 3: Synthesize hydrogen fluoride (H2+F2⇌2HFH_2 + F_2 \rightleftharpoons 2HF) where K=1.15×102K = 1.15 \times 10^2. Starting with 1.00 M1.00\,M H2H_2 and 2.00 M2.00\,M F2F_2, find equilibrium concentrations. Response: Substituting into the expression: 1.15×102=(2x)2(1.00−x)(2.00−x)1.15 \times 10^2 = \frac{(2x)^2}{(1.00-x)(2.00-x)}. Solving the quadratic results in x=0.968 mol/Lx = 0.968\,mol/L. Concentrations: [H2]=3.2×10−2 M[H_2] = 3.2 \times 10^{-2}\,M, [F2]=1.032 M[F_2] = 1.032\,M, and [HF]=1.936 M[HF] = 1.936\,M.