NH4Cl + LiOH = H2O + NH3 + LiCl

NH_4Cl ammonium chloride + LiOH lithium hydroxide ⟶ H_2O water + NH_3 ammonia + LiCl lithium chloride

Balance the chemical equation algebraically: NH_4Cl + LiOH ⟶ H_2O + NH_3 + LiCl Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_4Cl + c_2 LiOH ⟶ c_3 H_2O + c_4 NH_3 + c_5 LiCl Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, N, Li and O: Cl: | c_1 = c_5 H: | 4 c_1 + c_2 = 2 c_3 + 3 c_4 N: | c_1 = c_4 Li: | c_2 = c_5 O: | c_2 = c_3 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | NH_4Cl + LiOH ⟶ H_2O + NH_3 + LiCl

+ ⟶ + +

ammonium chloride + lithium hydroxide ⟶ water + ammonia + lithium chloride

| ammonium chloride | lithium hydroxide | water | ammonia | lithium chloride molecular enthalpy | -314.4 kJ/mol | -487.5 kJ/mol | -285.8 kJ/mol | -45.9 kJ/mol | -408.6 kJ/mol total enthalpy | -314.4 kJ/mol | -487.5 kJ/mol | -285.8 kJ/mol | -45.9 kJ/mol | -408.6 kJ/mol | H_initial = -801.9 kJ/mol | | H_final = -740.3 kJ/mol | | ΔH_rxn^0 | -740.3 kJ/mol - -801.9 kJ/mol = 61.57 kJ/mol (endothermic) | | | |

| ammonium chloride | lithium hydroxide | water | ammonia | lithium chloride molecular free energy | -202.9 kJ/mol | -441.5 kJ/mol | -237.1 kJ/mol | -16.4 kJ/mol | -384.4 kJ/mol total free energy | -202.9 kJ/mol | -441.5 kJ/mol | -237.1 kJ/mol | -16.4 kJ/mol | -384.4 kJ/mol | G_initial = -644.4 kJ/mol | | G_final = -637.9 kJ/mol | | ΔG_rxn^0 | -637.9 kJ/mol - -644.4 kJ/mol = 6.5 kJ/mol (endergonic) | | | |

Construct the equilibrium constant, K, expression for: NH_4Cl + LiOH ⟶ H_2O + NH_3 + LiCl Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: NH_4Cl + LiOH ⟶ H_2O + NH_3 + LiCl Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i NH_4Cl | 1 | -1 LiOH | 1 | -1 H_2O | 1 | 1 NH_3 | 1 | 1 LiCl | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NH_4Cl | 1 | -1 | ([NH4Cl])^(-1) LiOH | 1 | -1 | ([LiOH])^(-1) H_2O | 1 | 1 | [H2O] NH_3 | 1 | 1 | [NH3] LiCl | 1 | 1 | [LiCl] The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([NH4Cl])^(-1) ([LiOH])^(-1) [H2O] [NH3] [LiCl] = ([H2O] [NH3] [LiCl])/([NH4Cl] [LiOH])

Construct the rate of reaction expression for: NH_4Cl + LiOH ⟶ H_2O + NH_3 + LiCl Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: NH_4Cl + LiOH ⟶ H_2O + NH_3 + LiCl Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i NH_4Cl | 1 | -1 LiOH | 1 | -1 H_2O | 1 | 1 NH_3 | 1 | 1 LiCl | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term NH_4Cl | 1 | -1 | -(Δ[NH4Cl])/(Δt) LiOH | 1 | -1 | -(Δ[LiOH])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NH_3 | 1 | 1 | (Δ[NH3])/(Δt) LiCl | 1 | 1 | (Δ[LiCl])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -(Δ[NH4Cl])/(Δt) = -(Δ[LiOH])/(Δt) = (Δ[H2O])/(Δt) = (Δ[NH3])/(Δt) = (Δ[LiCl])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| ammonium chloride | lithium hydroxide | water | ammonia | lithium chloride formula | NH_4Cl | LiOH | H_2O | NH_3 | LiCl Hill formula | ClH_4N | HLiO | H_2O | H_3N | ClLi name | ammonium chloride | lithium hydroxide | water | ammonia | lithium chloride

| ammonium chloride | lithium hydroxide | water | ammonia | lithium chloride molar mass | 53.49 g/mol | 23.95 g/mol | 18.015 g/mol | 17.031 g/mol | 42.4 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | solid (at STP) melting point | 340 °C | 462 °C | 0 °C | -77.73 °C | 605 °C boiling point | | | 99.9839 °C | -33.33 °C | 1382 °C density | 1.5256 g/cm^3 | 1.46 g/cm^3 | 1 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | 2.07 g/cm^3 solubility in water | soluble | | | | surface tension | | | 0.0728 N/m | 0.0234 N/m | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | 0.00525 Pa s (at 20 °C) odor | | odorless | odorless | |