Search

HBr + HClO3 = H2O + Cl2 + Br2

Input interpretation

HBr hydrogen bromide + HClO3 ⟶ H_2O water + Cl_2 chlorine + Br_2 bromine
HBr hydrogen bromide + HClO3 ⟶ H_2O water + Cl_2 chlorine + Br_2 bromine

Balanced equation

Balance the chemical equation algebraically: HBr + HClO3 ⟶ H_2O + Cl_2 + Br_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 HClO3 ⟶ c_3 H_2O + c_4 Cl_2 + c_5 Br_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, Cl and O: Br: | c_1 = 2 c_5 H: | c_1 + c_2 = 2 c_3 Cl: | c_2 = 2 c_4 O: | 3 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 2 c_3 = 6 c_4 = 1 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 HBr + 2 HClO3 ⟶ 6 H_2O + Cl_2 + 5 Br_2
Balance the chemical equation algebraically: HBr + HClO3 ⟶ H_2O + Cl_2 + Br_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 HClO3 ⟶ c_3 H_2O + c_4 Cl_2 + c_5 Br_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, Cl and O: Br: | c_1 = 2 c_5 H: | c_1 + c_2 = 2 c_3 Cl: | c_2 = 2 c_4 O: | 3 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 2 c_3 = 6 c_4 = 1 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 HBr + 2 HClO3 ⟶ 6 H_2O + Cl_2 + 5 Br_2

Structures

+ HClO3 ⟶ + +
+ HClO3 ⟶ + +

Names

hydrogen bromide + HClO3 ⟶ water + chlorine + bromine
hydrogen bromide + HClO3 ⟶ water + chlorine + bromine

Equilibrium constant

Construct the equilibrium constant, K, expression for: HBr + HClO3 ⟶ H_2O + Cl_2 + Br_2 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: 10 HBr + 2 HClO3 ⟶ 6 H_2O + Cl_2 + 5 Br_2 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 | 10 | -10 HClO3 | 2 | -2 H_2O | 6 | 6 Cl_2 | 1 | 1 Br_2 | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 10 | -10 | ([HBr])^(-10) HClO3 | 2 | -2 | ([HClO3])^(-2) H_2O | 6 | 6 | ([H2O])^6 Cl_2 | 1 | 1 | [Cl2] Br_2 | 5 | 5 | ([Br2])^5 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])^(-10) ([HClO3])^(-2) ([H2O])^6 [Cl2] ([Br2])^5 = (([H2O])^6 [Cl2] ([Br2])^5)/(([HBr])^10 ([HClO3])^2)
Construct the equilibrium constant, K, expression for: HBr + HClO3 ⟶ H_2O + Cl_2 + Br_2 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: 10 HBr + 2 HClO3 ⟶ 6 H_2O + Cl_2 + 5 Br_2 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 | 10 | -10 HClO3 | 2 | -2 H_2O | 6 | 6 Cl_2 | 1 | 1 Br_2 | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 10 | -10 | ([HBr])^(-10) HClO3 | 2 | -2 | ([HClO3])^(-2) H_2O | 6 | 6 | ([H2O])^6 Cl_2 | 1 | 1 | [Cl2] Br_2 | 5 | 5 | ([Br2])^5 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])^(-10) ([HClO3])^(-2) ([H2O])^6 [Cl2] ([Br2])^5 = (([H2O])^6 [Cl2] ([Br2])^5)/(([HBr])^10 ([HClO3])^2)

Rate of reaction

Construct the rate of reaction expression for: HBr + HClO3 ⟶ H_2O + Cl_2 + Br_2 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: 10 HBr + 2 HClO3 ⟶ 6 H_2O + Cl_2 + 5 Br_2 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 | 10 | -10 HClO3 | 2 | -2 H_2O | 6 | 6 Cl_2 | 1 | 1 Br_2 | 5 | 5 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 | 10 | -10 | -1/10 (Δ[HBr])/(Δt) HClO3 | 2 | -2 | -1/2 (Δ[HClO3])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) Cl_2 | 1 | 1 | (Δ[Cl2])/(Δt) Br_2 | 5 | 5 | 1/5 (Δ[Br2])/(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 = -1/10 (Δ[HBr])/(Δt) = -1/2 (Δ[HClO3])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[Cl2])/(Δt) = 1/5 (Δ[Br2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HBr + HClO3 ⟶ H_2O + Cl_2 + Br_2 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: 10 HBr + 2 HClO3 ⟶ 6 H_2O + Cl_2 + 5 Br_2 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 | 10 | -10 HClO3 | 2 | -2 H_2O | 6 | 6 Cl_2 | 1 | 1 Br_2 | 5 | 5 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 | 10 | -10 | -1/10 (Δ[HBr])/(Δt) HClO3 | 2 | -2 | -1/2 (Δ[HClO3])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) Cl_2 | 1 | 1 | (Δ[Cl2])/(Δt) Br_2 | 5 | 5 | 1/5 (Δ[Br2])/(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 = -1/10 (Δ[HBr])/(Δt) = -1/2 (Δ[HClO3])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[Cl2])/(Δt) = 1/5 (Δ[Br2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| hydrogen bromide | HClO3 | water | chlorine | bromine formula | HBr | HClO3 | H_2O | Cl_2 | Br_2 Hill formula | BrH | HClO3 | H_2O | Cl_2 | Br_2 name | hydrogen bromide | | water | chlorine | bromine IUPAC name | hydrogen bromide | | water | molecular chlorine | molecular bromine
| hydrogen bromide | HClO3 | water | chlorine | bromine formula | HBr | HClO3 | H_2O | Cl_2 | Br_2 Hill formula | BrH | HClO3 | H_2O | Cl_2 | Br_2 name | hydrogen bromide | | water | chlorine | bromine IUPAC name | hydrogen bromide | | water | molecular chlorine | molecular bromine

Substance properties

| hydrogen bromide | HClO3 | water | chlorine | bromine molar mass | 80.912 g/mol | 84.45 g/mol | 18.015 g/mol | 70.9 g/mol | 159.81 g/mol phase | gas (at STP) | | liquid (at STP) | gas (at STP) | liquid (at STP) melting point | -86.8 °C | | 0 °C | -101 °C | -7.2 °C boiling point | -66.38 °C | | 99.9839 °C | -34 °C | 58.8 °C density | 0.003307 g/cm^3 (at 25 °C) | | 1 g/cm^3 | 0.003214 g/cm^3 (at 0 °C) | 3.119 g/cm^3 solubility in water | miscible | | | | insoluble surface tension | 0.0271 N/m | | 0.0728 N/m | | 0.0409 N/m dynamic viscosity | 8.4×10^-4 Pa s (at -75 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | 9.44×10^-4 Pa s (at 25 °C) odor | | | odorless | |
| hydrogen bromide | HClO3 | water | chlorine | bromine molar mass | 80.912 g/mol | 84.45 g/mol | 18.015 g/mol | 70.9 g/mol | 159.81 g/mol phase | gas (at STP) | | liquid (at STP) | gas (at STP) | liquid (at STP) melting point | -86.8 °C | | 0 °C | -101 °C | -7.2 °C boiling point | -66.38 °C | | 99.9839 °C | -34 °C | 58.8 °C density | 0.003307 g/cm^3 (at 25 °C) | | 1 g/cm^3 | 0.003214 g/cm^3 (at 0 °C) | 3.119 g/cm^3 solubility in water | miscible | | | | insoluble surface tension | 0.0271 N/m | | 0.0728 N/m | | 0.0409 N/m dynamic viscosity | 8.4×10^-4 Pa s (at -75 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | 9.44×10^-4 Pa s (at 25 °C) odor | | | odorless | |

Units

Random Queries

molar mass of calcium vs rhenium oxide
half-life of Ac-229 vs Au-173
melting point of lanthanum vs samarium
freezing point of potassium hydroxide vs magnesium hydroxide
ion class of sodium cation vs potassium cation
discovery year of herbertsmithite
molar mass of (S)-(+)-2-(dibenzylamino)-3-phenyl-1-propanol
KI + H3PO4 = HI + K3PO4
atomic spectrum of lanthanum
IUPAC name of calcium hydroxide vs pearl ash