Input interpretation
![NaOH sodium hydroxide + Br_2 bromine + KCrO2 ⟶ H_2O water + KBr potassium bromide + NaBr sodium bromide + Na_2CrO_4 sodium chromate](../image_source/01b1062b881dacd9b0fb9357fb953ec3.png)
NaOH sodium hydroxide + Br_2 bromine + KCrO2 ⟶ H_2O water + KBr potassium bromide + NaBr sodium bromide + Na_2CrO_4 sodium chromate
Balanced equation
![Balance the chemical equation algebraically: NaOH + Br_2 + KCrO2 ⟶ H_2O + KBr + NaBr + Na_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Br_2 + c_3 KCrO2 ⟶ c_4 H_2O + c_5 KBr + c_6 NaBr + c_7 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, K and Cr: H: | c_1 = 2 c_4 Na: | c_1 = c_6 + 2 c_7 O: | c_1 + 2 c_3 = c_4 + 4 c_7 Br: | 2 c_2 = c_5 + c_6 K: | c_3 = c_5 Cr: | c_3 = c_7 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 3/2 c_3 = 1 c_4 = 2 c_5 = 1 c_6 = 2 c_7 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 8 c_2 = 3 c_3 = 2 c_4 = 4 c_5 = 2 c_6 = 4 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 NaOH + 3 Br_2 + 2 KCrO2 ⟶ 4 H_2O + 2 KBr + 4 NaBr + 2 Na_2CrO_4](../image_source/106ae0d834d8a4c0128c3203e25ebe7f.png)
Balance the chemical equation algebraically: NaOH + Br_2 + KCrO2 ⟶ H_2O + KBr + NaBr + Na_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Br_2 + c_3 KCrO2 ⟶ c_4 H_2O + c_5 KBr + c_6 NaBr + c_7 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, K and Cr: H: | c_1 = 2 c_4 Na: | c_1 = c_6 + 2 c_7 O: | c_1 + 2 c_3 = c_4 + 4 c_7 Br: | 2 c_2 = c_5 + c_6 K: | c_3 = c_5 Cr: | c_3 = c_7 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 3/2 c_3 = 1 c_4 = 2 c_5 = 1 c_6 = 2 c_7 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 8 c_2 = 3 c_3 = 2 c_4 = 4 c_5 = 2 c_6 = 4 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 NaOH + 3 Br_2 + 2 KCrO2 ⟶ 4 H_2O + 2 KBr + 4 NaBr + 2 Na_2CrO_4
Structures
![+ + KCrO2 ⟶ + + +](../image_source/8c832a74032a16268c969a2c34f1b88c.png)
+ + KCrO2 ⟶ + + +
Names
![sodium hydroxide + bromine + KCrO2 ⟶ water + potassium bromide + sodium bromide + sodium chromate](../image_source/4d37df326747951ba9408cc61048b2c5.png)
sodium hydroxide + bromine + KCrO2 ⟶ water + potassium bromide + sodium bromide + sodium chromate
Equilibrium constant
![Construct the equilibrium constant, K, expression for: NaOH + Br_2 + KCrO2 ⟶ H_2O + KBr + 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: 8 NaOH + 3 Br_2 + 2 KCrO2 ⟶ 4 H_2O + 2 KBr + 4 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 | 8 | -8 Br_2 | 3 | -3 KCrO2 | 2 | -2 H_2O | 4 | 4 KBr | 2 | 2 NaBr | 4 | 4 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 | 8 | -8 | ([NaOH])^(-8) Br_2 | 3 | -3 | ([Br2])^(-3) KCrO2 | 2 | -2 | ([KCrO2])^(-2) H_2O | 4 | 4 | ([H2O])^4 KBr | 2 | 2 | ([KBr])^2 NaBr | 4 | 4 | ([NaBr])^4 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])^(-8) ([Br2])^(-3) ([KCrO2])^(-2) ([H2O])^4 ([KBr])^2 ([NaBr])^4 ([Na2CrO4])^2 = (([H2O])^4 ([KBr])^2 ([NaBr])^4 ([Na2CrO4])^2)/(([NaOH])^8 ([Br2])^3 ([KCrO2])^2)](../image_source/901fa442595f762c6ea8a5a13d6fea64.png)
Construct the equilibrium constant, K, expression for: NaOH + Br_2 + KCrO2 ⟶ H_2O + KBr + 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: 8 NaOH + 3 Br_2 + 2 KCrO2 ⟶ 4 H_2O + 2 KBr + 4 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 | 8 | -8 Br_2 | 3 | -3 KCrO2 | 2 | -2 H_2O | 4 | 4 KBr | 2 | 2 NaBr | 4 | 4 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 | 8 | -8 | ([NaOH])^(-8) Br_2 | 3 | -3 | ([Br2])^(-3) KCrO2 | 2 | -2 | ([KCrO2])^(-2) H_2O | 4 | 4 | ([H2O])^4 KBr | 2 | 2 | ([KBr])^2 NaBr | 4 | 4 | ([NaBr])^4 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])^(-8) ([Br2])^(-3) ([KCrO2])^(-2) ([H2O])^4 ([KBr])^2 ([NaBr])^4 ([Na2CrO4])^2 = (([H2O])^4 ([KBr])^2 ([NaBr])^4 ([Na2CrO4])^2)/(([NaOH])^8 ([Br2])^3 ([KCrO2])^2)
Rate of reaction
![Construct the rate of reaction expression for: NaOH + Br_2 + KCrO2 ⟶ H_2O + KBr + 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: 8 NaOH + 3 Br_2 + 2 KCrO2 ⟶ 4 H_2O + 2 KBr + 4 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 | 8 | -8 Br_2 | 3 | -3 KCrO2 | 2 | -2 H_2O | 4 | 4 KBr | 2 | 2 NaBr | 4 | 4 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 | 8 | -8 | -1/8 (Δ[NaOH])/(Δt) Br_2 | 3 | -3 | -1/3 (Δ[Br2])/(Δt) KCrO2 | 2 | -2 | -1/2 (Δ[KCrO2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) KBr | 2 | 2 | 1/2 (Δ[KBr])/(Δt) NaBr | 4 | 4 | 1/4 (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 = -1/8 (Δ[NaOH])/(Δt) = -1/3 (Δ[Br2])/(Δt) = -1/2 (Δ[KCrO2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/2 (Δ[KBr])/(Δt) = 1/4 (Δ[NaBr])/(Δt) = 1/2 (Δ[Na2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/2dd4e2f0f8357f5deba31b2484b84abb.png)
Construct the rate of reaction expression for: NaOH + Br_2 + KCrO2 ⟶ H_2O + KBr + 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: 8 NaOH + 3 Br_2 + 2 KCrO2 ⟶ 4 H_2O + 2 KBr + 4 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 | 8 | -8 Br_2 | 3 | -3 KCrO2 | 2 | -2 H_2O | 4 | 4 KBr | 2 | 2 NaBr | 4 | 4 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 | 8 | -8 | -1/8 (Δ[NaOH])/(Δt) Br_2 | 3 | -3 | -1/3 (Δ[Br2])/(Δt) KCrO2 | 2 | -2 | -1/2 (Δ[KCrO2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) KBr | 2 | 2 | 1/2 (Δ[KBr])/(Δt) NaBr | 4 | 4 | 1/4 (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 = -1/8 (Δ[NaOH])/(Δt) = -1/3 (Δ[Br2])/(Δt) = -1/2 (Δ[KCrO2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/2 (Δ[KBr])/(Δt) = 1/4 (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 | KCrO2 | water | potassium bromide | sodium bromide | sodium chromate formula | NaOH | Br_2 | KCrO2 | H_2O | KBr | NaBr | Na_2CrO_4 Hill formula | HNaO | Br_2 | CrKO2 | H_2O | BrK | BrNa | CrNa_2O_4 name | sodium hydroxide | bromine | | water | potassium bromide | sodium bromide | sodium chromate IUPAC name | sodium hydroxide | molecular bromine | | water | potassium bromide | sodium bromide | disodium dioxido(dioxo)chromium](../image_source/96980bb78f31627bb87cd6f116a41373.png)
| sodium hydroxide | bromine | KCrO2 | water | potassium bromide | sodium bromide | sodium chromate formula | NaOH | Br_2 | KCrO2 | H_2O | KBr | NaBr | Na_2CrO_4 Hill formula | HNaO | Br_2 | CrKO2 | H_2O | BrK | BrNa | CrNa_2O_4 name | sodium hydroxide | bromine | | water | potassium bromide | sodium bromide | sodium chromate IUPAC name | sodium hydroxide | molecular bromine | | water | potassium bromide | sodium bromide | disodium dioxido(dioxo)chromium
Substance properties
![| sodium hydroxide | bromine | KCrO2 | water | potassium bromide | sodium bromide | sodium chromate molar mass | 39.997 g/mol | 159.81 g/mol | 123.09 g/mol | 18.015 g/mol | 119 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) | solid (at STP) melting point | 323 °C | -7.2 °C | | 0 °C | 734 °C | 755 °C | 780 °C boiling point | 1390 °C | 58.8 °C | | 99.9839 °C | 1435 °C | 1396 °C | density | 2.13 g/cm^3 | 3.119 g/cm^3 | | 1 g/cm^3 | 2.75 g/cm^3 | 3.2 g/cm^3 | 2.698 g/cm^3 solubility in water | soluble | insoluble | | | soluble | 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 | | |](../image_source/ec3d6abc8eb9dc934ef6047a8b5e21eb.png)
| sodium hydroxide | bromine | KCrO2 | water | potassium bromide | sodium bromide | sodium chromate molar mass | 39.997 g/mol | 159.81 g/mol | 123.09 g/mol | 18.015 g/mol | 119 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) | solid (at STP) melting point | 323 °C | -7.2 °C | | 0 °C | 734 °C | 755 °C | 780 °C boiling point | 1390 °C | 58.8 °C | | 99.9839 °C | 1435 °C | 1396 °C | density | 2.13 g/cm^3 | 3.119 g/cm^3 | | 1 g/cm^3 | 2.75 g/cm^3 | 3.2 g/cm^3 | 2.698 g/cm^3 solubility in water | soluble | insoluble | | | soluble | 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