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CaO + CH3COOH = H2O + (CH3COO)2Ca

Input interpretation

CaO lime + CH_3CO_2H acetic acid ⟶ H_2O water + (CH3COO)2Ca
CaO lime + CH_3CO_2H acetic acid ⟶ H_2O water + (CH3COO)2Ca

Balanced equation

Balance the chemical equation algebraically: CaO + CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CaO + c_2 CH_3CO_2H ⟶ c_3 H_2O + c_4 (CH3COO)2Ca Set the number of atoms in the reactants equal to the number of atoms in the products for Ca, O, C and H: Ca: | c_1 = c_4 O: | c_1 + 2 c_2 = c_3 + 4 c_4 C: | 2 c_2 = 4 c_4 H: | 4 c_2 = 2 c_3 + 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | CaO + 2 CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca
Balance the chemical equation algebraically: CaO + CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CaO + c_2 CH_3CO_2H ⟶ c_3 H_2O + c_4 (CH3COO)2Ca Set the number of atoms in the reactants equal to the number of atoms in the products for Ca, O, C and H: Ca: | c_1 = c_4 O: | c_1 + 2 c_2 = c_3 + 4 c_4 C: | 2 c_2 = 4 c_4 H: | 4 c_2 = 2 c_3 + 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | CaO + 2 CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca

Structures

+ ⟶ + (CH3COO)2Ca
+ ⟶ + (CH3COO)2Ca

Names

lime + acetic acid ⟶ water + (CH3COO)2Ca
lime + acetic acid ⟶ water + (CH3COO)2Ca

Equilibrium constant

Construct the equilibrium constant, K, expression for: CaO + CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca 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: CaO + 2 CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca 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 CaO | 1 | -1 CH_3CO_2H | 2 | -2 H_2O | 1 | 1 (CH3COO)2Ca | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CaO | 1 | -1 | ([CaO])^(-1) CH_3CO_2H | 2 | -2 | ([CH3CO2H])^(-2) H_2O | 1 | 1 | [H2O] (CH3COO)2Ca | 1 | 1 | [(CH3COO)2Ca] 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 = ([CaO])^(-1) ([CH3CO2H])^(-2) [H2O] [(CH3COO)2Ca] = ([H2O] [(CH3COO)2Ca])/([CaO] ([CH3CO2H])^2)
Construct the equilibrium constant, K, expression for: CaO + CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca 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: CaO + 2 CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca 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 CaO | 1 | -1 CH_3CO_2H | 2 | -2 H_2O | 1 | 1 (CH3COO)2Ca | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CaO | 1 | -1 | ([CaO])^(-1) CH_3CO_2H | 2 | -2 | ([CH3CO2H])^(-2) H_2O | 1 | 1 | [H2O] (CH3COO)2Ca | 1 | 1 | [(CH3COO)2Ca] 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 = ([CaO])^(-1) ([CH3CO2H])^(-2) [H2O] [(CH3COO)2Ca] = ([H2O] [(CH3COO)2Ca])/([CaO] ([CH3CO2H])^2)

Rate of reaction

Construct the rate of reaction expression for: CaO + CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca 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: CaO + 2 CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca 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 CaO | 1 | -1 CH_3CO_2H | 2 | -2 H_2O | 1 | 1 (CH3COO)2Ca | 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 CaO | 1 | -1 | -(Δ[CaO])/(Δt) CH_3CO_2H | 2 | -2 | -1/2 (Δ[CH3CO2H])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) (CH3COO)2Ca | 1 | 1 | (Δ[(CH3COO)2Ca])/(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 = -(Δ[CaO])/(Δt) = -1/2 (Δ[CH3CO2H])/(Δt) = (Δ[H2O])/(Δt) = (Δ[(CH3COO)2Ca])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: CaO + CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca 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: CaO + 2 CH_3CO_2H ⟶ H_2O + (CH3COO)2Ca 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 CaO | 1 | -1 CH_3CO_2H | 2 | -2 H_2O | 1 | 1 (CH3COO)2Ca | 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 CaO | 1 | -1 | -(Δ[CaO])/(Δt) CH_3CO_2H | 2 | -2 | -1/2 (Δ[CH3CO2H])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) (CH3COO)2Ca | 1 | 1 | (Δ[(CH3COO)2Ca])/(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 = -(Δ[CaO])/(Δt) = -1/2 (Δ[CH3CO2H])/(Δt) = (Δ[H2O])/(Δt) = (Δ[(CH3COO)2Ca])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| lime | acetic acid | water | (CH3COO)2Ca formula | CaO | CH_3CO_2H | H_2O | (CH3COO)2Ca Hill formula | CaO | C_2H_4O_2 | H_2O | C4H6CaO4 name | lime | acetic acid | water |
| lime | acetic acid | water | (CH3COO)2Ca formula | CaO | CH_3CO_2H | H_2O | (CH3COO)2Ca Hill formula | CaO | C_2H_4O_2 | H_2O | C4H6CaO4 name | lime | acetic acid | water |

Substance properties

| lime | acetic acid | water | (CH3COO)2Ca molar mass | 56.077 g/mol | 60.052 g/mol | 18.015 g/mol | 158.17 g/mol phase | solid (at STP) | liquid (at STP) | liquid (at STP) | melting point | 2580 °C | 16.2 °C | 0 °C | boiling point | 2850 °C | 117.5 °C | 99.9839 °C | density | 3.3 g/cm^3 | 1.049 g/cm^3 | 1 g/cm^3 | solubility in water | reacts | 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 |
| lime | acetic acid | water | (CH3COO)2Ca molar mass | 56.077 g/mol | 60.052 g/mol | 18.015 g/mol | 158.17 g/mol phase | solid (at STP) | liquid (at STP) | liquid (at STP) | melting point | 2580 °C | 16.2 °C | 0 °C | boiling point | 2850 °C | 117.5 °C | 99.9839 °C | density | 3.3 g/cm^3 | 1.049 g/cm^3 | 1 g/cm^3 | solubility in water | reacts | 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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