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NaOH + Br2 + NaCrO3 = H2O + NaBr + Na2CrO4

Input interpretation

NaOH sodium hydroxide + Br_2 bromine + NaCrO3 ⟶ H_2O water + NaBr sodium bromide + Na_2CrO_4 sodium chromate
NaOH sodium hydroxide + Br_2 bromine + NaCrO3 ⟶ H_2O water + NaBr sodium bromide + Na_2CrO_4 sodium chromate

Balanced equation

Balance the chemical equation algebraically: NaOH + Br_2 + NaCrO3 ⟶ H_2O + NaBr + Na_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Br_2 + c_3 NaCrO3 ⟶ c_4 H_2O + c_5 NaBr + c_6 Na_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, Br and Cr: H: | c_1 = 2 c_4 Na: | c_1 + c_3 = c_5 + 2 c_6 O: | c_1 + 3 c_3 = c_4 + 4 c_6 Br: | 2 c_2 = c_5 Cr: | c_3 = c_6 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 = 4 c_2 = 1 c_3 = 2 c_4 = 2 c_5 = 2 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 NaOH + Br_2 + 2 NaCrO3 ⟶ 2 H_2O + 2 NaBr + 2 Na_2CrO_4
Balance the chemical equation algebraically: NaOH + Br_2 + NaCrO3 ⟶ H_2O + NaBr + Na_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Br_2 + c_3 NaCrO3 ⟶ c_4 H_2O + c_5 NaBr + c_6 Na_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, Br and Cr: H: | c_1 = 2 c_4 Na: | c_1 + c_3 = c_5 + 2 c_6 O: | c_1 + 3 c_3 = c_4 + 4 c_6 Br: | 2 c_2 = c_5 Cr: | c_3 = c_6 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 = 4 c_2 = 1 c_3 = 2 c_4 = 2 c_5 = 2 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 NaOH + Br_2 + 2 NaCrO3 ⟶ 2 H_2O + 2 NaBr + 2 Na_2CrO_4

Structures

+ + NaCrO3 ⟶ + +
+ + NaCrO3 ⟶ + +

Names

sodium hydroxide + bromine + NaCrO3 ⟶ water + sodium bromide + sodium chromate
sodium hydroxide + bromine + NaCrO3 ⟶ water + sodium bromide + sodium chromate

Equilibrium constant

Construct the equilibrium constant, K, expression for: NaOH + Br_2 + NaCrO3 ⟶ H_2O + NaBr + Na_2CrO_4 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: 4 NaOH + Br_2 + 2 NaCrO3 ⟶ 2 H_2O + 2 NaBr + 2 Na_2CrO_4 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 | 4 | -4 Br_2 | 1 | -1 NaCrO3 | 2 | -2 H_2O | 2 | 2 NaBr | 2 | 2 Na_2CrO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 4 | -4 | ([NaOH])^(-4) Br_2 | 1 | -1 | ([Br2])^(-1) NaCrO3 | 2 | -2 | ([NaCrO3])^(-2) H_2O | 2 | 2 | ([H2O])^2 NaBr | 2 | 2 | ([NaBr])^2 Na_2CrO_4 | 2 | 2 | ([Na2CrO4])^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 = ([NaOH])^(-4) ([Br2])^(-1) ([NaCrO3])^(-2) ([H2O])^2 ([NaBr])^2 ([Na2CrO4])^2 = (([H2O])^2 ([NaBr])^2 ([Na2CrO4])^2)/(([NaOH])^4 [Br2] ([NaCrO3])^2)
Construct the equilibrium constant, K, expression for: NaOH + Br_2 + NaCrO3 ⟶ H_2O + NaBr + Na_2CrO_4 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: 4 NaOH + Br_2 + 2 NaCrO3 ⟶ 2 H_2O + 2 NaBr + 2 Na_2CrO_4 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 | 4 | -4 Br_2 | 1 | -1 NaCrO3 | 2 | -2 H_2O | 2 | 2 NaBr | 2 | 2 Na_2CrO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 4 | -4 | ([NaOH])^(-4) Br_2 | 1 | -1 | ([Br2])^(-1) NaCrO3 | 2 | -2 | ([NaCrO3])^(-2) H_2O | 2 | 2 | ([H2O])^2 NaBr | 2 | 2 | ([NaBr])^2 Na_2CrO_4 | 2 | 2 | ([Na2CrO4])^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 = ([NaOH])^(-4) ([Br2])^(-1) ([NaCrO3])^(-2) ([H2O])^2 ([NaBr])^2 ([Na2CrO4])^2 = (([H2O])^2 ([NaBr])^2 ([Na2CrO4])^2)/(([NaOH])^4 [Br2] ([NaCrO3])^2)

Rate of reaction

Construct the rate of reaction expression for: NaOH + Br_2 + NaCrO3 ⟶ H_2O + NaBr + Na_2CrO_4 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: 4 NaOH + Br_2 + 2 NaCrO3 ⟶ 2 H_2O + 2 NaBr + 2 Na_2CrO_4 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 | 4 | -4 Br_2 | 1 | -1 NaCrO3 | 2 | -2 H_2O | 2 | 2 NaBr | 2 | 2 Na_2CrO_4 | 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 NaOH | 4 | -4 | -1/4 (Δ[NaOH])/(Δt) Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) NaCrO3 | 2 | -2 | -1/2 (Δ[NaCrO3])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) NaBr | 2 | 2 | 1/2 (Δ[NaBr])/(Δt) Na_2CrO_4 | 2 | 2 | 1/2 (Δ[Na2CrO4])/(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 = -1/4 (Δ[NaOH])/(Δt) = -(Δ[Br2])/(Δt) = -1/2 (Δ[NaCrO3])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/2 (Δ[NaBr])/(Δt) = 1/2 (Δ[Na2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: NaOH + Br_2 + NaCrO3 ⟶ H_2O + NaBr + Na_2CrO_4 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: 4 NaOH + Br_2 + 2 NaCrO3 ⟶ 2 H_2O + 2 NaBr + 2 Na_2CrO_4 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 | 4 | -4 Br_2 | 1 | -1 NaCrO3 | 2 | -2 H_2O | 2 | 2 NaBr | 2 | 2 Na_2CrO_4 | 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 NaOH | 4 | -4 | -1/4 (Δ[NaOH])/(Δt) Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) NaCrO3 | 2 | -2 | -1/2 (Δ[NaCrO3])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) NaBr | 2 | 2 | 1/2 (Δ[NaBr])/(Δt) Na_2CrO_4 | 2 | 2 | 1/2 (Δ[Na2CrO4])/(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 = -1/4 (Δ[NaOH])/(Δt) = -(Δ[Br2])/(Δt) = -1/2 (Δ[NaCrO3])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/2 (Δ[NaBr])/(Δt) = 1/2 (Δ[Na2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| sodium hydroxide | bromine | NaCrO3 | water | sodium bromide | sodium chromate formula | NaOH | Br_2 | NaCrO3 | H_2O | NaBr | Na_2CrO_4 Hill formula | HNaO | Br_2 | CrNaO3 | H_2O | BrNa | CrNa_2O_4 name | sodium hydroxide | bromine | | water | sodium bromide | sodium chromate IUPAC name | sodium hydroxide | molecular bromine | | water | sodium bromide | disodium dioxido(dioxo)chromium
| sodium hydroxide | bromine | NaCrO3 | water | sodium bromide | sodium chromate formula | NaOH | Br_2 | NaCrO3 | H_2O | NaBr | Na_2CrO_4 Hill formula | HNaO | Br_2 | CrNaO3 | H_2O | BrNa | CrNa_2O_4 name | sodium hydroxide | bromine | | water | sodium bromide | sodium chromate IUPAC name | sodium hydroxide | molecular bromine | | water | sodium bromide | disodium dioxido(dioxo)chromium

Substance properties

| sodium hydroxide | bromine | NaCrO3 | water | sodium bromide | sodium chromate molar mass | 39.997 g/mol | 159.81 g/mol | 122.98 g/mol | 18.015 g/mol | 102.89 g/mol | 161.97 g/mol phase | solid (at STP) | liquid (at STP) | | liquid (at STP) | solid (at STP) | solid (at STP) melting point | 323 °C | -7.2 °C | | 0 °C | 755 °C | 780 °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 | 2.698 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 | NaCrO3 | water | sodium bromide | sodium chromate molar mass | 39.997 g/mol | 159.81 g/mol | 122.98 g/mol | 18.015 g/mol | 102.89 g/mol | 161.97 g/mol phase | solid (at STP) | liquid (at STP) | | liquid (at STP) | solid (at STP) | solid (at STP) melting point | 323 °C | -7.2 °C | | 0 °C | 755 °C | 780 °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 | 2.698 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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