Input interpretation
H_2O water + NaOH sodium hydroxide + FeSO_4 duretter + NaOCl sodium hypochlorite ⟶ NaCl sodium chloride + Na_2SO_4 sodium sulfate + Fe(OH)_3 iron(III) hydroxide
Balanced equation
Balance the chemical equation algebraically: H_2O + NaOH + FeSO_4 + NaOCl ⟶ NaCl + Na_2SO_4 + Fe(OH)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 NaOH + c_3 FeSO_4 + c_4 NaOCl ⟶ c_5 NaCl + c_6 Na_2SO_4 + c_7 Fe(OH)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Na, Fe, S and Cl: H: | 2 c_1 + c_2 = 3 c_7 O: | c_1 + c_2 + 4 c_3 + c_4 = 4 c_6 + 3 c_7 Na: | c_2 + c_4 = c_5 + 2 c_6 Fe: | c_3 = c_7 S: | c_3 = c_6 Cl: | c_4 = 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 = 4 c_3 = 2 c_4 = 1 c_5 = 1 c_6 = 2 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + 4 NaOH + 2 FeSO_4 + NaOCl ⟶ NaCl + 2 Na_2SO_4 + 2 Fe(OH)_3
Structures
+ + + ⟶ + +
Names
water + sodium hydroxide + duretter + sodium hypochlorite ⟶ sodium chloride + sodium sulfate + iron(III) hydroxide
Equilibrium constant
Construct the equilibrium constant, K, expression for: H_2O + NaOH + FeSO_4 + NaOCl ⟶ NaCl + Na_2SO_4 + Fe(OH)_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: H_2O + 4 NaOH + 2 FeSO_4 + NaOCl ⟶ NaCl + 2 Na_2SO_4 + 2 Fe(OH)_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 H_2O | 1 | -1 NaOH | 4 | -4 FeSO_4 | 2 | -2 NaOCl | 1 | -1 NaCl | 1 | 1 Na_2SO_4 | 2 | 2 Fe(OH)_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 1 | -1 | ([H2O])^(-1) NaOH | 4 | -4 | ([NaOH])^(-4) FeSO_4 | 2 | -2 | ([FeSO4])^(-2) NaOCl | 1 | -1 | ([NaOCl])^(-1) NaCl | 1 | 1 | [NaCl] Na_2SO_4 | 2 | 2 | ([Na2SO4])^2 Fe(OH)_3 | 2 | 2 | ([Fe(OH)3])^2 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 = ([H2O])^(-1) ([NaOH])^(-4) ([FeSO4])^(-2) ([NaOCl])^(-1) [NaCl] ([Na2SO4])^2 ([Fe(OH)3])^2 = ([NaCl] ([Na2SO4])^2 ([Fe(OH)3])^2)/([H2O] ([NaOH])^4 ([FeSO4])^2 [NaOCl])
Rate of reaction
Construct the rate of reaction expression for: H_2O + NaOH + FeSO_4 + NaOCl ⟶ NaCl + Na_2SO_4 + Fe(OH)_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: H_2O + 4 NaOH + 2 FeSO_4 + NaOCl ⟶ NaCl + 2 Na_2SO_4 + 2 Fe(OH)_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 H_2O | 1 | -1 NaOH | 4 | -4 FeSO_4 | 2 | -2 NaOCl | 1 | -1 NaCl | 1 | 1 Na_2SO_4 | 2 | 2 Fe(OH)_3 | 2 | 2 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 H_2O | 1 | -1 | -(Δ[H2O])/(Δt) NaOH | 4 | -4 | -1/4 (Δ[NaOH])/(Δt) FeSO_4 | 2 | -2 | -1/2 (Δ[FeSO4])/(Δt) NaOCl | 1 | -1 | -(Δ[NaOCl])/(Δt) NaCl | 1 | 1 | (Δ[NaCl])/(Δt) Na_2SO_4 | 2 | 2 | 1/2 (Δ[Na2SO4])/(Δt) Fe(OH)_3 | 2 | 2 | 1/2 (Δ[Fe(OH)3])/(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[H2O])/(Δt) = -1/4 (Δ[NaOH])/(Δt) = -1/2 (Δ[FeSO4])/(Δt) = -(Δ[NaOCl])/(Δt) = (Δ[NaCl])/(Δt) = 1/2 (Δ[Na2SO4])/(Δt) = 1/2 (Δ[Fe(OH)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
| water | sodium hydroxide | duretter | sodium hypochlorite | sodium chloride | sodium sulfate | iron(III) hydroxide formula | H_2O | NaOH | FeSO_4 | NaOCl | NaCl | Na_2SO_4 | Fe(OH)_3 Hill formula | H_2O | HNaO | FeO_4S | ClNaO | ClNa | Na_2O_4S | FeH_3O_3 name | water | sodium hydroxide | duretter | sodium hypochlorite | sodium chloride | sodium sulfate | iron(III) hydroxide IUPAC name | water | sodium hydroxide | iron(+2) cation sulfate | sodium hypochlorite | sodium chloride | disodium sulfate | ferric trihydroxide