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NaOH + Br2 = H2O + NaBr + NaOBr

Input interpretation

NaOH sodium hydroxide + Br_2 bromine ⟶ H_2O water + NaBr sodium bromide + NaOBr
NaOH sodium hydroxide + Br_2 bromine ⟶ H_2O water + NaBr sodium bromide + NaOBr

Balanced equation

Balance the chemical equation algebraically: NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Br_2 ⟶ c_3 H_2O + c_4 NaBr + c_5 NaOBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O and Br: H: | c_1 = 2 c_3 Na: | c_1 = c_4 + c_5 O: | c_1 = c_3 + c_5 Br: | 2 c_2 = c_4 + c_5 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 = 2 c_2 = 1 c_3 = 1 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr
Balance the chemical equation algebraically: NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Br_2 ⟶ c_3 H_2O + c_4 NaBr + c_5 NaOBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O and Br: H: | c_1 = 2 c_3 Na: | c_1 = c_4 + c_5 O: | c_1 = c_3 + c_5 Br: | 2 c_2 = c_4 + c_5 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 = 2 c_2 = 1 c_3 = 1 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr

Structures

+ ⟶ + + NaOBr
+ ⟶ + + NaOBr

Names

sodium hydroxide + bromine ⟶ water + sodium bromide + NaOBr
sodium hydroxide + bromine ⟶ water + sodium bromide + NaOBr

Equilibrium constant

Construct the equilibrium constant, K, expression for: NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr 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: 2 NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr 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 NaOH | 2 | -2 Br_2 | 1 | -1 H_2O | 1 | 1 NaBr | 1 | 1 NaOBr | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 2 | -2 | ([NaOH])^(-2) Br_2 | 1 | -1 | ([Br2])^(-1) H_2O | 1 | 1 | [H2O] NaBr | 1 | 1 | [NaBr] NaOBr | 1 | 1 | [NaOBr] 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 = ([NaOH])^(-2) ([Br2])^(-1) [H2O] [NaBr] [NaOBr] = ([H2O] [NaBr] [NaOBr])/(([NaOH])^2 [Br2])
Construct the equilibrium constant, K, expression for: NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr 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: 2 NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr 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 NaOH | 2 | -2 Br_2 | 1 | -1 H_2O | 1 | 1 NaBr | 1 | 1 NaOBr | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 2 | -2 | ([NaOH])^(-2) Br_2 | 1 | -1 | ([Br2])^(-1) H_2O | 1 | 1 | [H2O] NaBr | 1 | 1 | [NaBr] NaOBr | 1 | 1 | [NaOBr] 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 = ([NaOH])^(-2) ([Br2])^(-1) [H2O] [NaBr] [NaOBr] = ([H2O] [NaBr] [NaOBr])/(([NaOH])^2 [Br2])

Rate of reaction

Construct the rate of reaction expression for: NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr 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: 2 NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr 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 NaOH | 2 | -2 Br_2 | 1 | -1 H_2O | 1 | 1 NaBr | 1 | 1 NaOBr | 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 NaOH | 2 | -2 | -1/2 (Δ[NaOH])/(Δt) Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NaBr | 1 | 1 | (Δ[NaBr])/(Δt) NaOBr | 1 | 1 | (Δ[NaOBr])/(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 = -1/2 (Δ[NaOH])/(Δt) = -(Δ[Br2])/(Δt) = (Δ[H2O])/(Δt) = (Δ[NaBr])/(Δt) = (Δ[NaOBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr 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: 2 NaOH + Br_2 ⟶ H_2O + NaBr + NaOBr 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 NaOH | 2 | -2 Br_2 | 1 | -1 H_2O | 1 | 1 NaBr | 1 | 1 NaOBr | 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 NaOH | 2 | -2 | -1/2 (Δ[NaOH])/(Δt) Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NaBr | 1 | 1 | (Δ[NaBr])/(Δt) NaOBr | 1 | 1 | (Δ[NaOBr])/(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 = -1/2 (Δ[NaOH])/(Δt) = -(Δ[Br2])/(Δt) = (Δ[H2O])/(Δt) = (Δ[NaBr])/(Δt) = (Δ[NaOBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| sodium hydroxide | bromine | water | sodium bromide | NaOBr formula | NaOH | Br_2 | H_2O | NaBr | NaOBr Hill formula | HNaO | Br_2 | H_2O | BrNa | BrNaO name | sodium hydroxide | bromine | water | sodium bromide | IUPAC name | sodium hydroxide | molecular bromine | water | sodium bromide |
| sodium hydroxide | bromine | water | sodium bromide | NaOBr formula | NaOH | Br_2 | H_2O | NaBr | NaOBr Hill formula | HNaO | Br_2 | H_2O | BrNa | BrNaO name | sodium hydroxide | bromine | water | sodium bromide | IUPAC name | sodium hydroxide | molecular bromine | water | sodium bromide |

Substance properties

| sodium hydroxide | bromine | water | sodium bromide | NaOBr molar mass | 39.997 g/mol | 159.81 g/mol | 18.015 g/mol | 102.89 g/mol | 118.89 g/mol phase | solid (at STP) | liquid (at STP) | liquid (at STP) | solid (at STP) | melting point | 323 °C | -7.2 °C | 0 °C | 755 °C | boiling point | 1390 °C | 58.8 °C | 99.9839 °C | 1396 °C | density | 2.13 g/cm^3 | 3.119 g/cm^3 | 1 g/cm^3 | 3.2 g/cm^3 | solubility in water | soluble | insoluble | | soluble | surface tension | 0.07435 N/m | 0.0409 N/m | 0.0728 N/m | | dynamic viscosity | 0.004 Pa s (at 350 °C) | 9.44×10^-4 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |
| sodium hydroxide | bromine | water | sodium bromide | NaOBr molar mass | 39.997 g/mol | 159.81 g/mol | 18.015 g/mol | 102.89 g/mol | 118.89 g/mol phase | solid (at STP) | liquid (at STP) | liquid (at STP) | solid (at STP) | melting point | 323 °C | -7.2 °C | 0 °C | 755 °C | boiling point | 1390 °C | 58.8 °C | 99.9839 °C | 1396 °C | density | 2.13 g/cm^3 | 3.119 g/cm^3 | 1 g/cm^3 | 3.2 g/cm^3 | solubility in water | soluble | insoluble | | soluble | surface tension | 0.07435 N/m | 0.0409 N/m | 0.0728 N/m | | dynamic viscosity | 0.004 Pa s (at 350 °C) | 9.44×10^-4 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |

Units

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