Search

I2 + Ba(OH)2 = HI + Ba(IO3)2 + BaI2

Input interpretation

I_2 iodine + Ba(OH)_2 barium hydroxide ⟶ HI hydrogen iodide + BaI_2O_6 barium iodate + BaI_2 barium iodide
I_2 iodine + Ba(OH)_2 barium hydroxide ⟶ HI hydrogen iodide + BaI_2O_6 barium iodate + BaI_2 barium iodide

Balanced equation

Balance the chemical equation algebraically: I_2 + Ba(OH)_2 ⟶ HI + BaI_2O_6 + BaI_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 I_2 + c_2 Ba(OH)_2 ⟶ c_3 HI + c_4 BaI_2O_6 + c_5 BaI_2 Set the number of atoms in the reactants equal to the number of atoms in the products for I, Ba, H and O: I: | 2 c_1 = c_3 + 2 c_4 + 2 c_5 Ba: | c_2 = c_4 + c_5 H: | 2 c_2 = c_3 O: | 2 c_2 = 6 c_4 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 3 c_3 = 6 c_4 = 1 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 I_2 + 3 Ba(OH)_2 ⟶ 6 HI + BaI_2O_6 + 2 BaI_2
Balance the chemical equation algebraically: I_2 + Ba(OH)_2 ⟶ HI + BaI_2O_6 + BaI_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 I_2 + c_2 Ba(OH)_2 ⟶ c_3 HI + c_4 BaI_2O_6 + c_5 BaI_2 Set the number of atoms in the reactants equal to the number of atoms in the products for I, Ba, H and O: I: | 2 c_1 = c_3 + 2 c_4 + 2 c_5 Ba: | c_2 = c_4 + c_5 H: | 2 c_2 = c_3 O: | 2 c_2 = 6 c_4 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 3 c_3 = 6 c_4 = 1 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 I_2 + 3 Ba(OH)_2 ⟶ 6 HI + BaI_2O_6 + 2 BaI_2

Structures

+ ⟶ + +
+ ⟶ + +

Names

iodine + barium hydroxide ⟶ hydrogen iodide + barium iodate + barium iodide
iodine + barium hydroxide ⟶ hydrogen iodide + barium iodate + barium iodide

Equilibrium constant

Construct the equilibrium constant, K, expression for: I_2 + Ba(OH)_2 ⟶ HI + BaI_2O_6 + BaI_2 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: 6 I_2 + 3 Ba(OH)_2 ⟶ 6 HI + BaI_2O_6 + 2 BaI_2 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 I_2 | 6 | -6 Ba(OH)_2 | 3 | -3 HI | 6 | 6 BaI_2O_6 | 1 | 1 BaI_2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression I_2 | 6 | -6 | ([I2])^(-6) Ba(OH)_2 | 3 | -3 | ([Ba(OH)2])^(-3) HI | 6 | 6 | ([HI])^6 BaI_2O_6 | 1 | 1 | [BaI2O6] BaI_2 | 2 | 2 | ([BaI2])^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 = ([I2])^(-6) ([Ba(OH)2])^(-3) ([HI])^6 [BaI2O6] ([BaI2])^2 = (([HI])^6 [BaI2O6] ([BaI2])^2)/(([I2])^6 ([Ba(OH)2])^3)
Construct the equilibrium constant, K, expression for: I_2 + Ba(OH)_2 ⟶ HI + BaI_2O_6 + BaI_2 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: 6 I_2 + 3 Ba(OH)_2 ⟶ 6 HI + BaI_2O_6 + 2 BaI_2 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 I_2 | 6 | -6 Ba(OH)_2 | 3 | -3 HI | 6 | 6 BaI_2O_6 | 1 | 1 BaI_2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression I_2 | 6 | -6 | ([I2])^(-6) Ba(OH)_2 | 3 | -3 | ([Ba(OH)2])^(-3) HI | 6 | 6 | ([HI])^6 BaI_2O_6 | 1 | 1 | [BaI2O6] BaI_2 | 2 | 2 | ([BaI2])^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 = ([I2])^(-6) ([Ba(OH)2])^(-3) ([HI])^6 [BaI2O6] ([BaI2])^2 = (([HI])^6 [BaI2O6] ([BaI2])^2)/(([I2])^6 ([Ba(OH)2])^3)

Rate of reaction

Construct the rate of reaction expression for: I_2 + Ba(OH)_2 ⟶ HI + BaI_2O_6 + BaI_2 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: 6 I_2 + 3 Ba(OH)_2 ⟶ 6 HI + BaI_2O_6 + 2 BaI_2 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 I_2 | 6 | -6 Ba(OH)_2 | 3 | -3 HI | 6 | 6 BaI_2O_6 | 1 | 1 BaI_2 | 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 I_2 | 6 | -6 | -1/6 (Δ[I2])/(Δt) Ba(OH)_2 | 3 | -3 | -1/3 (Δ[Ba(OH)2])/(Δt) HI | 6 | 6 | 1/6 (Δ[HI])/(Δt) BaI_2O_6 | 1 | 1 | (Δ[BaI2O6])/(Δt) BaI_2 | 2 | 2 | 1/2 (Δ[BaI2])/(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 = -1/6 (Δ[I2])/(Δt) = -1/3 (Δ[Ba(OH)2])/(Δt) = 1/6 (Δ[HI])/(Δt) = (Δ[BaI2O6])/(Δt) = 1/2 (Δ[BaI2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: I_2 + Ba(OH)_2 ⟶ HI + BaI_2O_6 + BaI_2 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: 6 I_2 + 3 Ba(OH)_2 ⟶ 6 HI + BaI_2O_6 + 2 BaI_2 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 I_2 | 6 | -6 Ba(OH)_2 | 3 | -3 HI | 6 | 6 BaI_2O_6 | 1 | 1 BaI_2 | 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 I_2 | 6 | -6 | -1/6 (Δ[I2])/(Δt) Ba(OH)_2 | 3 | -3 | -1/3 (Δ[Ba(OH)2])/(Δt) HI | 6 | 6 | 1/6 (Δ[HI])/(Δt) BaI_2O_6 | 1 | 1 | (Δ[BaI2O6])/(Δt) BaI_2 | 2 | 2 | 1/2 (Δ[BaI2])/(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 = -1/6 (Δ[I2])/(Δt) = -1/3 (Δ[Ba(OH)2])/(Δt) = 1/6 (Δ[HI])/(Δt) = (Δ[BaI2O6])/(Δt) = 1/2 (Δ[BaI2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| iodine | barium hydroxide | hydrogen iodide | barium iodate | barium iodide formula | I_2 | Ba(OH)_2 | HI | BaI_2O_6 | BaI_2 Hill formula | I_2 | BaH_2O_2 | HI | BaI_2O_6 | BaI_2 name | iodine | barium hydroxide | hydrogen iodide | barium iodate | barium iodide IUPAC name | molecular iodine | barium(+2) cation dihydroxide | hydrogen iodide | barium(+2) cation diiodate | barium(+2) cation diiodide
| iodine | barium hydroxide | hydrogen iodide | barium iodate | barium iodide formula | I_2 | Ba(OH)_2 | HI | BaI_2O_6 | BaI_2 Hill formula | I_2 | BaH_2O_2 | HI | BaI_2O_6 | BaI_2 name | iodine | barium hydroxide | hydrogen iodide | barium iodate | barium iodide IUPAC name | molecular iodine | barium(+2) cation dihydroxide | hydrogen iodide | barium(+2) cation diiodate | barium(+2) cation diiodide

Substance properties

| iodine | barium hydroxide | hydrogen iodide | barium iodate | barium iodide molar mass | 253.80894 g/mol | 171.34 g/mol | 127.912 g/mol | 487.13 g/mol | 391.136 g/mol phase | solid (at STP) | solid (at STP) | gas (at STP) | | solid (at STP) melting point | 113 °C | 300 °C | -50.76 °C | | 740 °C boiling point | 184 °C | | -35.55 °C | | density | 4.94 g/cm^3 | 2.2 g/cm^3 | 0.005228 g/cm^3 (at 25 °C) | 4.998 g/cm^3 | 5.15 g/cm^3 solubility in water | | | very soluble | | dynamic viscosity | 0.00227 Pa s (at 116 °C) | | 0.001321 Pa s (at -39 °C) | |
| iodine | barium hydroxide | hydrogen iodide | barium iodate | barium iodide molar mass | 253.80894 g/mol | 171.34 g/mol | 127.912 g/mol | 487.13 g/mol | 391.136 g/mol phase | solid (at STP) | solid (at STP) | gas (at STP) | | solid (at STP) melting point | 113 °C | 300 °C | -50.76 °C | | 740 °C boiling point | 184 °C | | -35.55 °C | | density | 4.94 g/cm^3 | 2.2 g/cm^3 | 0.005228 g/cm^3 (at 25 °C) | 4.998 g/cm^3 | 5.15 g/cm^3 solubility in water | | | very soluble | | dynamic viscosity | 0.00227 Pa s (at 116 °C) | | 0.001321 Pa s (at -39 °C) | |

Units

Random Queries

CAS number of sodium chlorate vs potassium dicyanoaurate(I)
Mohs hardness of melilite vs souzalite
molar mass of lanthanum fluoride
ion class of bromide anion vs tin(II) cation
ammonium sulfate vs xenon pentafluoride hexafluororuthenate
freezing point of dibucaine hydrochloride vs 2-amino-3,5-dichloropyridine
chemical types of iodine trichloride
name of topaz vs trimerite
vertex types of dextromethorphan
molar mass of lead(II) cation vs chlorate anion