Search

HBr + CrO3 = H2O + Br2 + CrBr3

Input interpretation

HBr hydrogen bromide + CrO_3 chromium trioxide ⟶ H_2O water + Br_2 bromine + Br_3Cr chromium tribromide
HBr hydrogen bromide + CrO_3 chromium trioxide ⟶ H_2O water + Br_2 bromine + Br_3Cr chromium tribromide

Balanced equation

Balance the chemical equation algebraically: HBr + CrO_3 ⟶ H_2O + Br_2 + Br_3Cr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 CrO_3 ⟶ c_3 H_2O + c_4 Br_2 + c_5 Br_3Cr Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, Cr and O: Br: | c_1 = 2 c_4 + 3 c_5 H: | c_1 = 2 c_3 Cr: | c_2 = c_5 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 1 c_3 = 3 c_4 = 3/2 c_5 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 12 c_2 = 2 c_3 = 6 c_4 = 3 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 HBr + 2 CrO_3 ⟶ 6 H_2O + 3 Br_2 + 2 Br_3Cr
Balance the chemical equation algebraically: HBr + CrO_3 ⟶ H_2O + Br_2 + Br_3Cr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 CrO_3 ⟶ c_3 H_2O + c_4 Br_2 + c_5 Br_3Cr Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, Cr and O: Br: | c_1 = 2 c_4 + 3 c_5 H: | c_1 = 2 c_3 Cr: | c_2 = c_5 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 1 c_3 = 3 c_4 = 3/2 c_5 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 12 c_2 = 2 c_3 = 6 c_4 = 3 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 HBr + 2 CrO_3 ⟶ 6 H_2O + 3 Br_2 + 2 Br_3Cr

Structures

+ ⟶ + +
+ ⟶ + +

Names

hydrogen bromide + chromium trioxide ⟶ water + bromine + chromium tribromide
hydrogen bromide + chromium trioxide ⟶ water + bromine + chromium tribromide

Equilibrium constant

Construct the equilibrium constant, K, expression for: HBr + CrO_3 ⟶ H_2O + Br_2 + 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: 12 HBr + 2 CrO_3 ⟶ 6 H_2O + 3 Br_2 + 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 | 12 | -12 CrO_3 | 2 | -2 H_2O | 6 | 6 Br_2 | 3 | 3 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 | 12 | -12 | ([HBr])^(-12) CrO_3 | 2 | -2 | ([CrO3])^(-2) H_2O | 6 | 6 | ([H2O])^6 Br_2 | 3 | 3 | ([Br2])^3 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])^(-12) ([CrO3])^(-2) ([H2O])^6 ([Br2])^3 ([Br3Cr])^2 = (([H2O])^6 ([Br2])^3 ([Br3Cr])^2)/(([HBr])^12 ([CrO3])^2)
Construct the equilibrium constant, K, expression for: HBr + CrO_3 ⟶ H_2O + Br_2 + 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: 12 HBr + 2 CrO_3 ⟶ 6 H_2O + 3 Br_2 + 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 | 12 | -12 CrO_3 | 2 | -2 H_2O | 6 | 6 Br_2 | 3 | 3 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 | 12 | -12 | ([HBr])^(-12) CrO_3 | 2 | -2 | ([CrO3])^(-2) H_2O | 6 | 6 | ([H2O])^6 Br_2 | 3 | 3 | ([Br2])^3 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])^(-12) ([CrO3])^(-2) ([H2O])^6 ([Br2])^3 ([Br3Cr])^2 = (([H2O])^6 ([Br2])^3 ([Br3Cr])^2)/(([HBr])^12 ([CrO3])^2)

Rate of reaction

Construct the rate of reaction expression for: HBr + CrO_3 ⟶ H_2O + Br_2 + 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: 12 HBr + 2 CrO_3 ⟶ 6 H_2O + 3 Br_2 + 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 | 12 | -12 CrO_3 | 2 | -2 H_2O | 6 | 6 Br_2 | 3 | 3 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 | 12 | -12 | -1/12 (Δ[HBr])/(Δt) CrO_3 | 2 | -2 | -1/2 (Δ[CrO3])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(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 = -1/12 (Δ[HBr])/(Δt) = -1/2 (Δ[CrO3])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/3 (Δ[Br2])/(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 + CrO_3 ⟶ H_2O + Br_2 + 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: 12 HBr + 2 CrO_3 ⟶ 6 H_2O + 3 Br_2 + 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 | 12 | -12 CrO_3 | 2 | -2 H_2O | 6 | 6 Br_2 | 3 | 3 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 | 12 | -12 | -1/12 (Δ[HBr])/(Δt) CrO_3 | 2 | -2 | -1/2 (Δ[CrO3])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(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 = -1/12 (Δ[HBr])/(Δt) = -1/2 (Δ[CrO3])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/3 (Δ[Br2])/(Δt) = 1/2 (Δ[Br3Cr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| hydrogen bromide | chromium trioxide | water | bromine | chromium tribromide formula | HBr | CrO_3 | H_2O | Br_2 | Br_3Cr Hill formula | BrH | CrO_3 | H_2O | Br_2 | Br_3Cr name | hydrogen bromide | chromium trioxide | water | bromine | chromium tribromide IUPAC name | hydrogen bromide | trioxochromium | water | molecular bromine | chromium(+3) cation tribromide
| hydrogen bromide | chromium trioxide | water | bromine | chromium tribromide formula | HBr | CrO_3 | H_2O | Br_2 | Br_3Cr Hill formula | BrH | CrO_3 | H_2O | Br_2 | Br_3Cr name | hydrogen bromide | chromium trioxide | water | bromine | chromium tribromide IUPAC name | hydrogen bromide | trioxochromium | water | molecular bromine | chromium(+3) cation tribromide

Substance properties

| hydrogen bromide | chromium trioxide | water | bromine | chromium tribromide molar mass | 80.912 g/mol | 99.993 g/mol | 18.015 g/mol | 159.81 g/mol | 291.71 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | liquid (at STP) | solid (at STP) melting point | -86.8 °C | 196 °C | 0 °C | -7.2 °C | 1130 °C boiling point | -66.38 °C | | 99.9839 °C | 58.8 °C | density | 0.003307 g/cm^3 (at 25 °C) | | 1 g/cm^3 | 3.119 g/cm^3 | 4.68 g/cm^3 solubility in water | miscible | very soluble | | insoluble | slightly soluble 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 | odorless | |
| hydrogen bromide | chromium trioxide | water | bromine | chromium tribromide molar mass | 80.912 g/mol | 99.993 g/mol | 18.015 g/mol | 159.81 g/mol | 291.71 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | liquid (at STP) | solid (at STP) melting point | -86.8 °C | 196 °C | 0 °C | -7.2 °C | 1130 °C boiling point | -66.38 °C | | 99.9839 °C | 58.8 °C | density | 0.003307 g/cm^3 (at 25 °C) | | 1 g/cm^3 | 3.119 g/cm^3 | 4.68 g/cm^3 solubility in water | miscible | very soluble | | insoluble | slightly soluble 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 | odorless | |

Units

Random Queries

color of platinum vs iridium
name of lithium oxide
CAS number of strontium formate dihydrate vs molybdenum carbide
H2SO4 + KI = H2O + K2SO4 + I2 + K2S
alternate names of sylvite vs nantokite
chemical types of 2-thienyllithium
bonding characteristics of indium
alternate names of potassium perchlorate
DOT numbers of antimony pentoxide
PubChem SID number of aluminum sulfate vs lead(II) zirconate