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K2Cr2O7 + HBr = H2O + Br2 + KBr + CrBr3

Input interpretation

K_2Cr_2O_7 (potassium dichromate) + HBr (hydrogen bromide) ⟶ H_2O (water) + Br_2 (bromine) + KBr (potassium bromide) + Br_3Cr (chromium tribromide)
K_2Cr_2O_7 (potassium dichromate) + HBr (hydrogen bromide) ⟶ H_2O (water) + Br_2 (bromine) + KBr (potassium bromide) + Br_3Cr (chromium tribromide)

Balanced equation

Balance the chemical equation algebraically: K_2Cr_2O_7 + HBr ⟶ H_2O + Br_2 + KBr + Br_3Cr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 K_2Cr_2O_7 + c_2 HBr ⟶ c_3 H_2O + c_4 Br_2 + c_5 KBr + c_6 Br_3Cr Set the number of atoms in the reactants equal to the number of atoms in the products for Cr, K, O, Br and H: Cr: | 2 c_1 = c_6 K: | 2 c_1 = c_5 O: | 7 c_1 = c_3 Br: | c_2 = 2 c_4 + c_5 + 3 c_6 H: | c_2 = 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 = 14 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: | | K_2Cr_2O_7 + 14 HBr ⟶ 7 H_2O + 3 Br_2 + 2 KBr + 2 Br_3Cr
Balance the chemical equation algebraically: K_2Cr_2O_7 + HBr ⟶ H_2O + Br_2 + KBr + Br_3Cr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 K_2Cr_2O_7 + c_2 HBr ⟶ c_3 H_2O + c_4 Br_2 + c_5 KBr + c_6 Br_3Cr Set the number of atoms in the reactants equal to the number of atoms in the products for Cr, K, O, Br and H: Cr: | 2 c_1 = c_6 K: | 2 c_1 = c_5 O: | 7 c_1 = c_3 Br: | c_2 = 2 c_4 + c_5 + 3 c_6 H: | c_2 = 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 = 14 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: | | K_2Cr_2O_7 + 14 HBr ⟶ 7 H_2O + 3 Br_2 + 2 KBr + 2 Br_3Cr

Structures

+ ⟶ + + +
+ ⟶ + + +

Names

potassium dichromate + hydrogen bromide ⟶ water + bromine + potassium bromide + chromium tribromide
potassium dichromate + hydrogen bromide ⟶ water + bromine + potassium bromide + chromium tribromide

Equilibrium constant

Construct the equilibrium constant, K, expression for: K_2Cr_2O_7 + HBr ⟶ H_2O + Br_2 + KBr + 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: K_2Cr_2O_7 + 14 HBr ⟶ 7 H_2O + 3 Br_2 + 2 KBr + 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 K_2Cr_2O_7 | 1 | -1 HBr | 14 | -14 H_2O | 7 | 7 Br_2 | 3 | 3 KBr | 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 K_2Cr_2O_7 | 1 | -1 | ([K2Cr2O7])^(-1) HBr | 14 | -14 | ([HBr])^(-14) H_2O | 7 | 7 | ([H2O])^7 Br_2 | 3 | 3 | ([Br2])^3 KBr | 2 | 2 | ([KBr])^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 = ([K2Cr2O7])^(-1) ([HBr])^(-14) ([H2O])^7 ([Br2])^3 ([KBr])^2 ([Br3Cr])^2 = (([H2O])^7 ([Br2])^3 ([KBr])^2 ([Br3Cr])^2)/([K2Cr2O7] ([HBr])^14)
Construct the equilibrium constant, K, expression for: K_2Cr_2O_7 + HBr ⟶ H_2O + Br_2 + KBr + 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: K_2Cr_2O_7 + 14 HBr ⟶ 7 H_2O + 3 Br_2 + 2 KBr + 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 K_2Cr_2O_7 | 1 | -1 HBr | 14 | -14 H_2O | 7 | 7 Br_2 | 3 | 3 KBr | 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 K_2Cr_2O_7 | 1 | -1 | ([K2Cr2O7])^(-1) HBr | 14 | -14 | ([HBr])^(-14) H_2O | 7 | 7 | ([H2O])^7 Br_2 | 3 | 3 | ([Br2])^3 KBr | 2 | 2 | ([KBr])^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 = ([K2Cr2O7])^(-1) ([HBr])^(-14) ([H2O])^7 ([Br2])^3 ([KBr])^2 ([Br3Cr])^2 = (([H2O])^7 ([Br2])^3 ([KBr])^2 ([Br3Cr])^2)/([K2Cr2O7] ([HBr])^14)

Rate of reaction

Construct the rate of reaction expression for: K_2Cr_2O_7 + HBr ⟶ H_2O + Br_2 + KBr + 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: K_2Cr_2O_7 + 14 HBr ⟶ 7 H_2O + 3 Br_2 + 2 KBr + 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 K_2Cr_2O_7 | 1 | -1 HBr | 14 | -14 H_2O | 7 | 7 Br_2 | 3 | 3 KBr | 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 K_2Cr_2O_7 | 1 | -1 | -(Δ[K2Cr2O7])/(Δt) HBr | 14 | -14 | -1/14 (Δ[HBr])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(Δt) KBr | 2 | 2 | 1/2 (Δ[KBr])/(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 = -(Δ[K2Cr2O7])/(Δt) = -1/14 (Δ[HBr])/(Δt) = 1/7 (Δ[H2O])/(Δt) = 1/3 (Δ[Br2])/(Δt) = 1/2 (Δ[KBr])/(Δt) = 1/2 (Δ[Br3Cr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: K_2Cr_2O_7 + HBr ⟶ H_2O + Br_2 + KBr + 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: K_2Cr_2O_7 + 14 HBr ⟶ 7 H_2O + 3 Br_2 + 2 KBr + 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 K_2Cr_2O_7 | 1 | -1 HBr | 14 | -14 H_2O | 7 | 7 Br_2 | 3 | 3 KBr | 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 K_2Cr_2O_7 | 1 | -1 | -(Δ[K2Cr2O7])/(Δt) HBr | 14 | -14 | -1/14 (Δ[HBr])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(Δt) KBr | 2 | 2 | 1/2 (Δ[KBr])/(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 = -(Δ[K2Cr2O7])/(Δt) = -1/14 (Δ[HBr])/(Δt) = 1/7 (Δ[H2O])/(Δt) = 1/3 (Δ[Br2])/(Δt) = 1/2 (Δ[KBr])/(Δt) = 1/2 (Δ[Br3Cr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium dichromate | hydrogen bromide | water | bromine | potassium bromide | chromium tribromide formula | K_2Cr_2O_7 | HBr | H_2O | Br_2 | KBr | Br_3Cr Hill formula | Cr_2K_2O_7 | BrH | H_2O | Br_2 | BrK | Br_3Cr name | potassium dichromate | hydrogen bromide | water | bromine | potassium bromide | chromium tribromide IUPAC name | dipotassium oxido-(oxido-dioxochromio)oxy-dioxochromium | hydrogen bromide | water | molecular bromine | potassium bromide | chromium(+3) cation tribromide
| potassium dichromate | hydrogen bromide | water | bromine | potassium bromide | chromium tribromide formula | K_2Cr_2O_7 | HBr | H_2O | Br_2 | KBr | Br_3Cr Hill formula | Cr_2K_2O_7 | BrH | H_2O | Br_2 | BrK | Br_3Cr name | potassium dichromate | hydrogen bromide | water | bromine | potassium bromide | chromium tribromide IUPAC name | dipotassium oxido-(oxido-dioxochromio)oxy-dioxochromium | hydrogen bromide | water | molecular bromine | potassium bromide | chromium(+3) cation tribromide

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