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CH3COOH + Ba(OH)2 = H2O + C(H3COO)2Ba

Input interpretation

CH_3CO_2H acetic acid + Ba(OH)_2 barium hydroxide ⟶ H_2O water + C(H3COO)2Ba
CH_3CO_2H acetic acid + Ba(OH)_2 barium hydroxide ⟶ H_2O water + C(H3COO)2Ba

Balanced equation

Balance the chemical equation algebraically: CH_3CO_2H + Ba(OH)_2 ⟶ H_2O + C(H3COO)2Ba Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CH_3CO_2H + c_2 Ba(OH)_2 ⟶ c_3 H_2O + c_4 C(H3COO)2Ba Set the number of atoms in the reactants equal to the number of atoms in the products for C, H, O and Ba: C: | 2 c_1 = 3 c_4 H: | 4 c_1 + 2 c_2 = 2 c_3 + 6 c_4 O: | 2 c_1 + 2 c_2 = c_3 + 4 c_4 Ba: | c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 1 c_3 = 1 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 2 c_3 = 2 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 CH_3CO_2H + 2 Ba(OH)_2 ⟶ 2 H_2O + 2 C(H3COO)2Ba
Balance the chemical equation algebraically: CH_3CO_2H + Ba(OH)_2 ⟶ H_2O + C(H3COO)2Ba Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CH_3CO_2H + c_2 Ba(OH)_2 ⟶ c_3 H_2O + c_4 C(H3COO)2Ba Set the number of atoms in the reactants equal to the number of atoms in the products for C, H, O and Ba: C: | 2 c_1 = 3 c_4 H: | 4 c_1 + 2 c_2 = 2 c_3 + 6 c_4 O: | 2 c_1 + 2 c_2 = c_3 + 4 c_4 Ba: | c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 1 c_3 = 1 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 2 c_3 = 2 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 CH_3CO_2H + 2 Ba(OH)_2 ⟶ 2 H_2O + 2 C(H3COO)2Ba

Structures

+ ⟶ + C(H3COO)2Ba
+ ⟶ + C(H3COO)2Ba

Names

acetic acid + barium hydroxide ⟶ water + C(H3COO)2Ba
acetic acid + barium hydroxide ⟶ water + C(H3COO)2Ba

Equilibrium constant

Construct the equilibrium constant, K, expression for: CH_3CO_2H + Ba(OH)_2 ⟶ H_2O + C(H3COO)2Ba 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: 3 CH_3CO_2H + 2 Ba(OH)_2 ⟶ 2 H_2O + 2 C(H3COO)2Ba 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 CH_3CO_2H | 3 | -3 Ba(OH)_2 | 2 | -2 H_2O | 2 | 2 C(H3COO)2Ba | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CH_3CO_2H | 3 | -3 | ([CH3CO2H])^(-3) Ba(OH)_2 | 2 | -2 | ([Ba(OH)2])^(-2) H_2O | 2 | 2 | ([H2O])^2 C(H3COO)2Ba | 2 | 2 | ([C(H3COO)2Ba])^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 = ([CH3CO2H])^(-3) ([Ba(OH)2])^(-2) ([H2O])^2 ([C(H3COO)2Ba])^2 = (([H2O])^2 ([C(H3COO)2Ba])^2)/(([CH3CO2H])^3 ([Ba(OH)2])^2)
Construct the equilibrium constant, K, expression for: CH_3CO_2H + Ba(OH)_2 ⟶ H_2O + C(H3COO)2Ba 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: 3 CH_3CO_2H + 2 Ba(OH)_2 ⟶ 2 H_2O + 2 C(H3COO)2Ba 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 CH_3CO_2H | 3 | -3 Ba(OH)_2 | 2 | -2 H_2O | 2 | 2 C(H3COO)2Ba | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CH_3CO_2H | 3 | -3 | ([CH3CO2H])^(-3) Ba(OH)_2 | 2 | -2 | ([Ba(OH)2])^(-2) H_2O | 2 | 2 | ([H2O])^2 C(H3COO)2Ba | 2 | 2 | ([C(H3COO)2Ba])^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 = ([CH3CO2H])^(-3) ([Ba(OH)2])^(-2) ([H2O])^2 ([C(H3COO)2Ba])^2 = (([H2O])^2 ([C(H3COO)2Ba])^2)/(([CH3CO2H])^3 ([Ba(OH)2])^2)

Rate of reaction

Construct the rate of reaction expression for: CH_3CO_2H + Ba(OH)_2 ⟶ H_2O + C(H3COO)2Ba 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: 3 CH_3CO_2H + 2 Ba(OH)_2 ⟶ 2 H_2O + 2 C(H3COO)2Ba 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 CH_3CO_2H | 3 | -3 Ba(OH)_2 | 2 | -2 H_2O | 2 | 2 C(H3COO)2Ba | 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 CH_3CO_2H | 3 | -3 | -1/3 (Δ[CH3CO2H])/(Δt) Ba(OH)_2 | 2 | -2 | -1/2 (Δ[Ba(OH)2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) C(H3COO)2Ba | 2 | 2 | 1/2 (Δ[C(H3COO)2Ba])/(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 = -1/3 (Δ[CH3CO2H])/(Δt) = -1/2 (Δ[Ba(OH)2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/2 (Δ[C(H3COO)2Ba])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: CH_3CO_2H + Ba(OH)_2 ⟶ H_2O + C(H3COO)2Ba 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: 3 CH_3CO_2H + 2 Ba(OH)_2 ⟶ 2 H_2O + 2 C(H3COO)2Ba 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 CH_3CO_2H | 3 | -3 Ba(OH)_2 | 2 | -2 H_2O | 2 | 2 C(H3COO)2Ba | 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 CH_3CO_2H | 3 | -3 | -1/3 (Δ[CH3CO2H])/(Δt) Ba(OH)_2 | 2 | -2 | -1/2 (Δ[Ba(OH)2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) C(H3COO)2Ba | 2 | 2 | 1/2 (Δ[C(H3COO)2Ba])/(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 = -1/3 (Δ[CH3CO2H])/(Δt) = -1/2 (Δ[Ba(OH)2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/2 (Δ[C(H3COO)2Ba])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| acetic acid | barium hydroxide | water | C(H3COO)2Ba formula | CH_3CO_2H | Ba(OH)_2 | H_2O | C(H3COO)2Ba Hill formula | C_2H_4O_2 | BaH_2O_2 | H_2O | C3H6BaO4 name | acetic acid | barium hydroxide | water | IUPAC name | acetic acid | barium(+2) cation dihydroxide | water |
| acetic acid | barium hydroxide | water | C(H3COO)2Ba formula | CH_3CO_2H | Ba(OH)_2 | H_2O | C(H3COO)2Ba Hill formula | C_2H_4O_2 | BaH_2O_2 | H_2O | C3H6BaO4 name | acetic acid | barium hydroxide | water | IUPAC name | acetic acid | barium(+2) cation dihydroxide | water |

Substance properties

| acetic acid | barium hydroxide | water | C(H3COO)2Ba molar mass | 60.052 g/mol | 171.34 g/mol | 18.015 g/mol | 243.4 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | melting point | 16.2 °C | 300 °C | 0 °C | boiling point | 117.5 °C | | 99.9839 °C | density | 1.049 g/cm^3 | 2.2 g/cm^3 | 1 g/cm^3 | solubility in water | miscible | | | surface tension | 0.0288 N/m | | 0.0728 N/m | dynamic viscosity | 0.001056 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | vinegar-like | | odorless |
| acetic acid | barium hydroxide | water | C(H3COO)2Ba molar mass | 60.052 g/mol | 171.34 g/mol | 18.015 g/mol | 243.4 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | melting point | 16.2 °C | 300 °C | 0 °C | boiling point | 117.5 °C | | 99.9839 °C | density | 1.049 g/cm^3 | 2.2 g/cm^3 | 1 g/cm^3 | solubility in water | miscible | | | surface tension | 0.0288 N/m | | 0.0728 N/m | dynamic viscosity | 0.001056 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | vinegar-like | | odorless |

Units

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