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NaOH + H2O2 + CrBr3 = H2O + Br2 + Na2CrO4

Input interpretation

NaOH sodium hydroxide + H_2O_2 hydrogen peroxide + Br_3Cr chromium tribromide ⟶ H_2O water + Br_2 bromine + Na_2CrO_4 sodium chromate
NaOH sodium hydroxide + H_2O_2 hydrogen peroxide + Br_3Cr chromium tribromide ⟶ H_2O water + Br_2 bromine + Na_2CrO_4 sodium chromate

Balanced equation

Balance the chemical equation algebraically: NaOH + H_2O_2 + Br_3Cr ⟶ H_2O + Br_2 + Na_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 H_2O_2 + c_3 Br_3Cr ⟶ c_4 H_2O + c_5 Br_2 + c_6 Na_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, Br and Cr: H: | c_1 + 2 c_2 = 2 c_4 Na: | c_1 = 2 c_6 O: | c_1 + 2 c_2 = c_4 + 4 c_6 Br: | 3 c_3 = 2 c_5 Cr: | c_3 = c_6 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 1 c_4 = 4 c_5 = 3/2 c_6 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 6 c_3 = 2 c_4 = 8 c_5 = 3 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 NaOH + 6 H_2O_2 + 2 Br_3Cr ⟶ 8 H_2O + 3 Br_2 + 2 Na_2CrO_4
Balance the chemical equation algebraically: NaOH + H_2O_2 + Br_3Cr ⟶ H_2O + Br_2 + Na_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 H_2O_2 + c_3 Br_3Cr ⟶ c_4 H_2O + c_5 Br_2 + c_6 Na_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, Br and Cr: H: | c_1 + 2 c_2 = 2 c_4 Na: | c_1 = 2 c_6 O: | c_1 + 2 c_2 = c_4 + 4 c_6 Br: | 3 c_3 = 2 c_5 Cr: | c_3 = c_6 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 1 c_4 = 4 c_5 = 3/2 c_6 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 6 c_3 = 2 c_4 = 8 c_5 = 3 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 NaOH + 6 H_2O_2 + 2 Br_3Cr ⟶ 8 H_2O + 3 Br_2 + 2 Na_2CrO_4

Structures

+ + ⟶ + +
+ + ⟶ + +

Names

sodium hydroxide + hydrogen peroxide + chromium tribromide ⟶ water + bromine + sodium chromate
sodium hydroxide + hydrogen peroxide + chromium tribromide ⟶ water + bromine + sodium chromate

Equilibrium constant

Construct the equilibrium constant, K, expression for: NaOH + H_2O_2 + Br_3Cr ⟶ H_2O + Br_2 + Na_2CrO_4 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: 4 NaOH + 6 H_2O_2 + 2 Br_3Cr ⟶ 8 H_2O + 3 Br_2 + 2 Na_2CrO_4 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 | 4 | -4 H_2O_2 | 6 | -6 Br_3Cr | 2 | -2 H_2O | 8 | 8 Br_2 | 3 | 3 Na_2CrO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 4 | -4 | ([NaOH])^(-4) H_2O_2 | 6 | -6 | ([H2O2])^(-6) Br_3Cr | 2 | -2 | ([Br3Cr])^(-2) H_2O | 8 | 8 | ([H2O])^8 Br_2 | 3 | 3 | ([Br2])^3 Na_2CrO_4 | 2 | 2 | ([Na2CrO4])^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 = ([NaOH])^(-4) ([H2O2])^(-6) ([Br3Cr])^(-2) ([H2O])^8 ([Br2])^3 ([Na2CrO4])^2 = (([H2O])^8 ([Br2])^3 ([Na2CrO4])^2)/(([NaOH])^4 ([H2O2])^6 ([Br3Cr])^2)
Construct the equilibrium constant, K, expression for: NaOH + H_2O_2 + Br_3Cr ⟶ H_2O + Br_2 + Na_2CrO_4 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: 4 NaOH + 6 H_2O_2 + 2 Br_3Cr ⟶ 8 H_2O + 3 Br_2 + 2 Na_2CrO_4 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 | 4 | -4 H_2O_2 | 6 | -6 Br_3Cr | 2 | -2 H_2O | 8 | 8 Br_2 | 3 | 3 Na_2CrO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 4 | -4 | ([NaOH])^(-4) H_2O_2 | 6 | -6 | ([H2O2])^(-6) Br_3Cr | 2 | -2 | ([Br3Cr])^(-2) H_2O | 8 | 8 | ([H2O])^8 Br_2 | 3 | 3 | ([Br2])^3 Na_2CrO_4 | 2 | 2 | ([Na2CrO4])^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 = ([NaOH])^(-4) ([H2O2])^(-6) ([Br3Cr])^(-2) ([H2O])^8 ([Br2])^3 ([Na2CrO4])^2 = (([H2O])^8 ([Br2])^3 ([Na2CrO4])^2)/(([NaOH])^4 ([H2O2])^6 ([Br3Cr])^2)

Rate of reaction

Construct the rate of reaction expression for: NaOH + H_2O_2 + Br_3Cr ⟶ H_2O + Br_2 + Na_2CrO_4 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: 4 NaOH + 6 H_2O_2 + 2 Br_3Cr ⟶ 8 H_2O + 3 Br_2 + 2 Na_2CrO_4 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 | 4 | -4 H_2O_2 | 6 | -6 Br_3Cr | 2 | -2 H_2O | 8 | 8 Br_2 | 3 | 3 Na_2CrO_4 | 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 NaOH | 4 | -4 | -1/4 (Δ[NaOH])/(Δt) H_2O_2 | 6 | -6 | -1/6 (Δ[H2O2])/(Δt) Br_3Cr | 2 | -2 | -1/2 (Δ[Br3Cr])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(Δt) Na_2CrO_4 | 2 | 2 | 1/2 (Δ[Na2CrO4])/(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 = -1/4 (Δ[NaOH])/(Δt) = -1/6 (Δ[H2O2])/(Δt) = -1/2 (Δ[Br3Cr])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/3 (Δ[Br2])/(Δt) = 1/2 (Δ[Na2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: NaOH + H_2O_2 + Br_3Cr ⟶ H_2O + Br_2 + Na_2CrO_4 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: 4 NaOH + 6 H_2O_2 + 2 Br_3Cr ⟶ 8 H_2O + 3 Br_2 + 2 Na_2CrO_4 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 | 4 | -4 H_2O_2 | 6 | -6 Br_3Cr | 2 | -2 H_2O | 8 | 8 Br_2 | 3 | 3 Na_2CrO_4 | 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 NaOH | 4 | -4 | -1/4 (Δ[NaOH])/(Δt) H_2O_2 | 6 | -6 | -1/6 (Δ[H2O2])/(Δt) Br_3Cr | 2 | -2 | -1/2 (Δ[Br3Cr])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(Δt) Na_2CrO_4 | 2 | 2 | 1/2 (Δ[Na2CrO4])/(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 = -1/4 (Δ[NaOH])/(Δt) = -1/6 (Δ[H2O2])/(Δt) = -1/2 (Δ[Br3Cr])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/3 (Δ[Br2])/(Δt) = 1/2 (Δ[Na2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| sodium hydroxide | hydrogen peroxide | chromium tribromide | water | bromine | sodium chromate formula | NaOH | H_2O_2 | Br_3Cr | H_2O | Br_2 | Na_2CrO_4 Hill formula | HNaO | H_2O_2 | Br_3Cr | H_2O | Br_2 | CrNa_2O_4 name | sodium hydroxide | hydrogen peroxide | chromium tribromide | water | bromine | sodium chromate IUPAC name | sodium hydroxide | hydrogen peroxide | chromium(+3) cation tribromide | water | molecular bromine | disodium dioxido(dioxo)chromium
| sodium hydroxide | hydrogen peroxide | chromium tribromide | water | bromine | sodium chromate formula | NaOH | H_2O_2 | Br_3Cr | H_2O | Br_2 | Na_2CrO_4 Hill formula | HNaO | H_2O_2 | Br_3Cr | H_2O | Br_2 | CrNa_2O_4 name | sodium hydroxide | hydrogen peroxide | chromium tribromide | water | bromine | sodium chromate IUPAC name | sodium hydroxide | hydrogen peroxide | chromium(+3) cation tribromide | water | molecular bromine | disodium dioxido(dioxo)chromium

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