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KMnO4 + HBr = H2O + Br2 + KBr + MnBr

Input interpretation

KMnO_4 potassium permanganate + HBr hydrogen bromide ⟶ H_2O water + Br_2 bromine + KBr potassium bromide + MnBr
KMnO_4 potassium permanganate + HBr hydrogen bromide ⟶ H_2O water + Br_2 bromine + KBr potassium bromide + MnBr

Balanced equation

Balance the chemical equation algebraically: KMnO_4 + HBr ⟶ H_2O + Br_2 + KBr + MnBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KMnO_4 + c_2 HBr ⟶ c_3 H_2O + c_4 Br_2 + c_5 KBr + c_6 MnBr Set the number of atoms in the reactants equal to the number of atoms in the products for K, Mn, O, Br and H: K: | c_1 = c_5 Mn: | c_1 = c_6 O: | 4 c_1 = c_3 Br: | c_2 = 2 c_4 + c_5 + 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 = 8 c_3 = 4 c_4 = 3 c_5 = 1 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | KMnO_4 + 8 HBr ⟶ 4 H_2O + 3 Br_2 + KBr + MnBr
Balance the chemical equation algebraically: KMnO_4 + HBr ⟶ H_2O + Br_2 + KBr + MnBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KMnO_4 + c_2 HBr ⟶ c_3 H_2O + c_4 Br_2 + c_5 KBr + c_6 MnBr Set the number of atoms in the reactants equal to the number of atoms in the products for K, Mn, O, Br and H: K: | c_1 = c_5 Mn: | c_1 = c_6 O: | 4 c_1 = c_3 Br: | c_2 = 2 c_4 + c_5 + 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 = 8 c_3 = 4 c_4 = 3 c_5 = 1 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | KMnO_4 + 8 HBr ⟶ 4 H_2O + 3 Br_2 + KBr + MnBr

Structures

+ ⟶ + + + MnBr
+ ⟶ + + + MnBr

Names

potassium permanganate + hydrogen bromide ⟶ water + bromine + potassium bromide + MnBr
potassium permanganate + hydrogen bromide ⟶ water + bromine + potassium bromide + MnBr

Equilibrium constant

Construct the equilibrium constant, K, expression for: KMnO_4 + HBr ⟶ H_2O + Br_2 + KBr + MnBr 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: KMnO_4 + 8 HBr ⟶ 4 H_2O + 3 Br_2 + KBr + MnBr 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 KMnO_4 | 1 | -1 HBr | 8 | -8 H_2O | 4 | 4 Br_2 | 3 | 3 KBr | 1 | 1 MnBr | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KMnO_4 | 1 | -1 | ([KMnO4])^(-1) HBr | 8 | -8 | ([HBr])^(-8) H_2O | 4 | 4 | ([H2O])^4 Br_2 | 3 | 3 | ([Br2])^3 KBr | 1 | 1 | [KBr] MnBr | 1 | 1 | [MnBr] 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 = ([KMnO4])^(-1) ([HBr])^(-8) ([H2O])^4 ([Br2])^3 [KBr] [MnBr] = (([H2O])^4 ([Br2])^3 [KBr] [MnBr])/([KMnO4] ([HBr])^8)
Construct the equilibrium constant, K, expression for: KMnO_4 + HBr ⟶ H_2O + Br_2 + KBr + MnBr 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: KMnO_4 + 8 HBr ⟶ 4 H_2O + 3 Br_2 + KBr + MnBr 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 KMnO_4 | 1 | -1 HBr | 8 | -8 H_2O | 4 | 4 Br_2 | 3 | 3 KBr | 1 | 1 MnBr | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KMnO_4 | 1 | -1 | ([KMnO4])^(-1) HBr | 8 | -8 | ([HBr])^(-8) H_2O | 4 | 4 | ([H2O])^4 Br_2 | 3 | 3 | ([Br2])^3 KBr | 1 | 1 | [KBr] MnBr | 1 | 1 | [MnBr] 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 = ([KMnO4])^(-1) ([HBr])^(-8) ([H2O])^4 ([Br2])^3 [KBr] [MnBr] = (([H2O])^4 ([Br2])^3 [KBr] [MnBr])/([KMnO4] ([HBr])^8)

Rate of reaction

Construct the rate of reaction expression for: KMnO_4 + HBr ⟶ H_2O + Br_2 + KBr + MnBr 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: KMnO_4 + 8 HBr ⟶ 4 H_2O + 3 Br_2 + KBr + MnBr 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 KMnO_4 | 1 | -1 HBr | 8 | -8 H_2O | 4 | 4 Br_2 | 3 | 3 KBr | 1 | 1 MnBr | 1 | 1 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 KMnO_4 | 1 | -1 | -(Δ[KMnO4])/(Δt) HBr | 8 | -8 | -1/8 (Δ[HBr])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(Δt) KBr | 1 | 1 | (Δ[KBr])/(Δt) MnBr | 1 | 1 | (Δ[MnBr])/(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[KMnO4])/(Δt) = -1/8 (Δ[HBr])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/3 (Δ[Br2])/(Δt) = (Δ[KBr])/(Δt) = (Δ[MnBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KMnO_4 + HBr ⟶ H_2O + Br_2 + KBr + MnBr 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: KMnO_4 + 8 HBr ⟶ 4 H_2O + 3 Br_2 + KBr + MnBr 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 KMnO_4 | 1 | -1 HBr | 8 | -8 H_2O | 4 | 4 Br_2 | 3 | 3 KBr | 1 | 1 MnBr | 1 | 1 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 KMnO_4 | 1 | -1 | -(Δ[KMnO4])/(Δt) HBr | 8 | -8 | -1/8 (Δ[HBr])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) Br_2 | 3 | 3 | 1/3 (Δ[Br2])/(Δt) KBr | 1 | 1 | (Δ[KBr])/(Δt) MnBr | 1 | 1 | (Δ[MnBr])/(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[KMnO4])/(Δt) = -1/8 (Δ[HBr])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/3 (Δ[Br2])/(Δt) = (Δ[KBr])/(Δt) = (Δ[MnBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium permanganate | hydrogen bromide | water | bromine | potassium bromide | MnBr formula | KMnO_4 | HBr | H_2O | Br_2 | KBr | MnBr Hill formula | KMnO_4 | BrH | H_2O | Br_2 | BrK | BrMn name | potassium permanganate | hydrogen bromide | water | bromine | potassium bromide | IUPAC name | potassium permanganate | hydrogen bromide | water | molecular bromine | potassium bromide |
| potassium permanganate | hydrogen bromide | water | bromine | potassium bromide | MnBr formula | KMnO_4 | HBr | H_2O | Br_2 | KBr | MnBr Hill formula | KMnO_4 | BrH | H_2O | Br_2 | BrK | BrMn name | potassium permanganate | hydrogen bromide | water | bromine | potassium bromide | IUPAC name | potassium permanganate | hydrogen bromide | water | molecular bromine | potassium bromide |

Substance properties

| potassium permanganate | hydrogen bromide | water | bromine | potassium bromide | MnBr molar mass | 158.03 g/mol | 80.912 g/mol | 18.015 g/mol | 159.81 g/mol | 119 g/mol | 134.84 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) | liquid (at STP) | solid (at STP) | melting point | 240 °C | -86.8 °C | 0 °C | -7.2 °C | 734 °C | boiling point | | -66.38 °C | 99.9839 °C | 58.8 °C | 1435 °C | density | 1 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | 1 g/cm^3 | 3.119 g/cm^3 | 2.75 g/cm^3 | solubility in water | | miscible | | insoluble | 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 | | |
| potassium permanganate | hydrogen bromide | water | bromine | potassium bromide | MnBr molar mass | 158.03 g/mol | 80.912 g/mol | 18.015 g/mol | 159.81 g/mol | 119 g/mol | 134.84 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) | liquid (at STP) | solid (at STP) | melting point | 240 °C | -86.8 °C | 0 °C | -7.2 °C | 734 °C | boiling point | | -66.38 °C | 99.9839 °C | 58.8 °C | 1435 °C | density | 1 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | 1 g/cm^3 | 3.119 g/cm^3 | 2.75 g/cm^3 | solubility in water | | miscible | | insoluble | 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

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