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HBr + Na2Cr2O7 = H2O + Br2 + NaBr + CrBr3

Input interpretation

HBr hydrogen bromide + Na_2Cr_2O_7 sodium bichromate ⟶ H_2O water + Br_2 bromine + NaBr sodium bromide + Br_3Cr chromium tribromide
HBr hydrogen bromide + Na_2Cr_2O_7 sodium bichromate ⟶ H_2O water + Br_2 bromine + NaBr sodium bromide + Br_3Cr chromium tribromide

Balanced equation

Balance the chemical equation algebraically: HBr + Na_2Cr_2O_7 ⟶ H_2O + Br_2 + NaBr + Br_3Cr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 Na_2Cr_2O_7 ⟶ c_3 H_2O + c_4 Br_2 + c_5 NaBr + c_6 Br_3Cr Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, Cr, Na and O: Br: | c_1 = 2 c_4 + c_5 + 3 c_6 H: | c_1 = 2 c_3 Cr: | 2 c_2 = c_6 Na: | 2 c_2 = c_5 O: | 7 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 14 c_2 = 1 c_3 = 7 c_4 = 3 c_5 = 2 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HBr + Na_2Cr_2O_7 ⟶ 7 H_2O + 3 Br_2 + 2 NaBr + 2 Br_3Cr
Balance the chemical equation algebraically: HBr + Na_2Cr_2O_7 ⟶ H_2O + Br_2 + NaBr + Br_3Cr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 Na_2Cr_2O_7 ⟶ c_3 H_2O + c_4 Br_2 + c_5 NaBr + c_6 Br_3Cr Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, Cr, Na and O: Br: | c_1 = 2 c_4 + c_5 + 3 c_6 H: | c_1 = 2 c_3 Cr: | 2 c_2 = c_6 Na: | 2 c_2 = c_5 O: | 7 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 14 c_2 = 1 c_3 = 7 c_4 = 3 c_5 = 2 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HBr + Na_2Cr_2O_7 ⟶ 7 H_2O + 3 Br_2 + 2 NaBr + 2 Br_3Cr

Structures

+ ⟶ + + +
+ ⟶ + + +

Names

hydrogen bromide + sodium bichromate ⟶ water + bromine + sodium bromide + chromium tribromide
hydrogen bromide + sodium bichromate ⟶ water + bromine + sodium bromide + chromium tribromide

Equilibrium constant

Construct the equilibrium constant, K, expression for: HBr + Na_2Cr_2O_7 ⟶ H_2O + Br_2 + NaBr + Br_3Cr 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: 14 HBr + Na_2Cr_2O_7 ⟶ 7 H_2O + 3 Br_2 + 2 NaBr + 2 Br_3Cr 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 HBr | 14 | -14 Na_2Cr_2O_7 | 1 | -1 H_2O | 7 | 7 Br_2 | 3 | 3 NaBr | 2 | 2 Br_3Cr | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 14 | -14 | ([HBr])^(-14) Na_2Cr_2O_7 | 1 | -1 | ([Na2Cr2O7])^(-1) H_2O | 7 | 7 | ([H2O])^7 Br_2 | 3 | 3 | ([Br2])^3 NaBr | 2 | 2 | ([NaBr])^2 Br_3Cr | 2 | 2 | ([Br3Cr])^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 = ([HBr])^(-14) ([Na2Cr2O7])^(-1) ([H2O])^7 ([Br2])^3 ([NaBr])^2 ([Br3Cr])^2 = (([H2O])^7 ([Br2])^3 ([NaBr])^2 ([Br3Cr])^2)/(([HBr])^14 [Na2Cr2O7])
Construct the equilibrium constant, K, expression for: HBr + Na_2Cr_2O_7 ⟶ H_2O + Br_2 + NaBr + Br_3Cr 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: 14 HBr + Na_2Cr_2O_7 ⟶ 7 H_2O + 3 Br_2 + 2 NaBr + 2 Br_3Cr 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 HBr | 14 | -14 Na_2Cr_2O_7 | 1 | -1 H_2O | 7 | 7 Br_2 | 3 | 3 NaBr | 2 | 2 Br_3Cr | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 14 | -14 | ([HBr])^(-14) Na_2Cr_2O_7 | 1 | -1 | ([Na2Cr2O7])^(-1) H_2O | 7 | 7 | ([H2O])^7 Br_2 | 3 | 3 | ([Br2])^3 NaBr | 2 | 2 | ([NaBr])^2 Br_3Cr | 2 | 2 | ([Br3Cr])^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 = ([HBr])^(-14) ([Na2Cr2O7])^(-1) ([H2O])^7 ([Br2])^3 ([NaBr])^2 ([Br3Cr])^2 = (([H2O])^7 ([Br2])^3 ([NaBr])^2 ([Br3Cr])^2)/(([HBr])^14 [Na2Cr2O7])

Rate of reaction

Construct the rate of reaction expression for: HBr + Na_2Cr_2O_7 ⟶ H_2O + Br_2 + NaBr + Br_3Cr 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: 14 HBr + Na_2Cr_2O_7 ⟶ 7 H_2O + 3 Br_2 + 2 NaBr + 2 Br_3Cr 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 HBr | 14 | -14 Na_2Cr_2O_7 | 1 | -1 H_2O | 7 | 7 Br_2 | 3 | 3 NaBr | 2 | 2 Br_3Cr | 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 HBr | 14 | -14 | -1/14 (Δ[HBr])/(Δt) Na_2Cr_2O_7 | 1 | -1 | -(Δ[Na2Cr2O7])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(Δt) NaBr | 2 | 2 | 1/2 (Δ[NaBr])/(Δt) Br_3Cr | 2 | 2 | 1/2 (Δ[Br3Cr])/(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 = -1/14 (Δ[HBr])/(Δt) = -(Δ[Na2Cr2O7])/(Δt) = 1/7 (Δ[H2O])/(Δt) = 1/3 (Δ[Br2])/(Δt) = 1/2 (Δ[NaBr])/(Δt) = 1/2 (Δ[Br3Cr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HBr + Na_2Cr_2O_7 ⟶ H_2O + Br_2 + NaBr + Br_3Cr 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: 14 HBr + Na_2Cr_2O_7 ⟶ 7 H_2O + 3 Br_2 + 2 NaBr + 2 Br_3Cr 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 HBr | 14 | -14 Na_2Cr_2O_7 | 1 | -1 H_2O | 7 | 7 Br_2 | 3 | 3 NaBr | 2 | 2 Br_3Cr | 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 HBr | 14 | -14 | -1/14 (Δ[HBr])/(Δt) Na_2Cr_2O_7 | 1 | -1 | -(Δ[Na2Cr2O7])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(Δt) NaBr | 2 | 2 | 1/2 (Δ[NaBr])/(Δt) Br_3Cr | 2 | 2 | 1/2 (Δ[Br3Cr])/(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 = -1/14 (Δ[HBr])/(Δt) = -(Δ[Na2Cr2O7])/(Δt) = 1/7 (Δ[H2O])/(Δt) = 1/3 (Δ[Br2])/(Δt) = 1/2 (Δ[NaBr])/(Δt) = 1/2 (Δ[Br3Cr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| hydrogen bromide | sodium bichromate | water | bromine | sodium bromide | chromium tribromide formula | HBr | Na_2Cr_2O_7 | H_2O | Br_2 | NaBr | Br_3Cr Hill formula | BrH | Cr_2Na_2O_7 | H_2O | Br_2 | BrNa | Br_3Cr name | hydrogen bromide | sodium bichromate | water | bromine | sodium bromide | chromium tribromide IUPAC name | hydrogen bromide | disodium oxido-(oxido-dioxo-chromio)oxy-dioxo-chromium | water | molecular bromine | sodium bromide | chromium(+3) cation tribromide
| hydrogen bromide | sodium bichromate | water | bromine | sodium bromide | chromium tribromide formula | HBr | Na_2Cr_2O_7 | H_2O | Br_2 | NaBr | Br_3Cr Hill formula | BrH | Cr_2Na_2O_7 | H_2O | Br_2 | BrNa | Br_3Cr name | hydrogen bromide | sodium bichromate | water | bromine | sodium bromide | chromium tribromide IUPAC name | hydrogen bromide | disodium oxido-(oxido-dioxo-chromio)oxy-dioxo-chromium | water | molecular bromine | sodium bromide | chromium(+3) cation tribromide

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