NaOH + NH4Cl = H2O + NaCl + NH3

NaOH (sodium hydroxide) + NH_4Cl (ammonium chloride) ⟶ H_2O (water) + NaCl (sodium chloride) + NH_3 (ammonia)

Balance the chemical equation algebraically: NaOH + NH_4Cl ⟶ H_2O + NaCl + NH_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 NH_4Cl ⟶ c_3 H_2O + c_4 NaCl + c_5 NH_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, Cl and N: H: | c_1 + 4 c_2 = 2 c_3 + 3 c_5 Na: | c_1 = c_4 O: | c_1 = c_3 Cl: | c_2 = c_4 N: | c_2 = c_5 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: | | NaOH + NH_4Cl ⟶ H_2O + NaCl + NH_3

+ ⟶ + +

sodium hydroxide + ammonium chloride ⟶ water + sodium chloride + ammonia

| sodium hydroxide | ammonium chloride | water | sodium chloride | ammonia molecular enthalpy | -425.8 kJ/mol | -314.4 kJ/mol | -285.8 kJ/mol | -411.2 kJ/mol | -45.9 kJ/mol total enthalpy | -425.8 kJ/mol | -314.4 kJ/mol | -285.8 kJ/mol | -411.2 kJ/mol | -45.9 kJ/mol | H_initial = -740.2 kJ/mol | | H_final = -742.9 kJ/mol | | ΔH_rxn^0 | -742.9 kJ/mol - -740.2 kJ/mol = -2.73 kJ/mol (exothermic) | | | |

| sodium hydroxide | ammonium chloride | water | sodium chloride | ammonia molecular free energy | -379.7 kJ/mol | -202.9 kJ/mol | -237.1 kJ/mol | -384.1 kJ/mol | -16.4 kJ/mol total free energy | -379.7 kJ/mol | -202.9 kJ/mol | -237.1 kJ/mol | -384.1 kJ/mol | -16.4 kJ/mol | G_initial = -582.6 kJ/mol | | G_final = -637.6 kJ/mol | | ΔG_rxn^0 | -637.6 kJ/mol - -582.6 kJ/mol = -55 kJ/mol (exergonic) | | | |

| sodium hydroxide | ammonium chloride | water | sodium chloride | ammonia molecular entropy | 64 J/(mol K) | 96 J/(mol K) | 69.91 J/(mol K) | 72 J/(mol K) | 193 J/(mol K) total entropy | 64 J/(mol K) | 96 J/(mol K) | 69.91 J/(mol K) | 72 J/(mol K) | 193 J/(mol K) | S_initial = 160 J/(mol K) | | S_final = 334.9 J/(mol K) | | ΔS_rxn^0 | 334.9 J/(mol K) - 160 J/(mol K) = 174.9 J/(mol K) (endoentropic) | | | |

Construct the equilibrium constant, K, expression for: NaOH + NH_4Cl ⟶ H_2O + NaCl + NH_3 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: NaOH + NH_4Cl ⟶ H_2O + NaCl + NH_3 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 NaOH | 1 | -1 NH_4Cl | 1 | -1 H_2O | 1 | 1 NaCl | 1 | 1 NH_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 1 | -1 | ([NaOH])^(-1) NH_4Cl | 1 | -1 | ([NH4Cl])^(-1) H_2O | 1 | 1 | [H2O] NaCl | 1 | 1 | [NaCl] NH_3 | 1 | 1 | [NH3] 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 = ([NaOH])^(-1) ([NH4Cl])^(-1) [H2O] [NaCl] [NH3] = ([H2O] [NaCl] [NH3])/([NaOH] [NH4Cl])

Construct the rate of reaction expression for: NaOH + NH_4Cl ⟶ H_2O + NaCl + NH_3 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: NaOH + NH_4Cl ⟶ H_2O + NaCl + NH_3 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 NaOH | 1 | -1 NH_4Cl | 1 | -1 H_2O | 1 | 1 NaCl | 1 | 1 NH_3 | 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 NaOH | 1 | -1 | -(Δ[NaOH])/(Δt) NH_4Cl | 1 | -1 | -(Δ[NH4Cl])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NaCl | 1 | 1 | (Δ[NaCl])/(Δt) NH_3 | 1 | 1 | (Δ[NH3])/(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[NaOH])/(Δt) = -(Δ[NH4Cl])/(Δt) = (Δ[H2O])/(Δt) = (Δ[NaCl])/(Δt) = (Δ[NH3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium hydroxide | ammonium chloride | water | sodium chloride | ammonia formula | NaOH | NH_4Cl | H_2O | NaCl | NH_3 Hill formula | HNaO | ClH_4N | H_2O | ClNa | H_3N name | sodium hydroxide | ammonium chloride | water | sodium chloride | ammonia

| sodium hydroxide | ammonium chloride | water | sodium chloride | ammonia molar mass | 39.997 g/mol | 53.49 g/mol | 18.015 g/mol | 58.44 g/mol | 17.031 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | gas (at STP) melting point | 323 °C | 340 °C | 0 °C | 801 °C | -77.73 °C boiling point | 1390 °C | | 99.9839 °C | 1413 °C | -33.33 °C density | 2.13 g/cm^3 | 1.5256 g/cm^3 | 1 g/cm^3 | 2.16 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) solubility in water | soluble | soluble | | soluble | surface tension | 0.07435 N/m | | 0.0728 N/m | | 0.0234 N/m dynamic viscosity | 0.004 Pa s (at 350 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | 1.009×10^-5 Pa s (at 25 °C) odor | | | odorless | odorless |