(CH3COO)2Ca = CaCO3 + CH3COCH3

(CH3COO)2Ca ⟶ CaCO_3 calcium carbonate + CH_3COCH_3 acetone

Balance the chemical equation algebraically: (CH3COO)2Ca ⟶ CaCO_3 + CH_3COCH_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 (CH3COO)2Ca ⟶ c_2 CaCO_3 + c_3 CH_3COCH_3 Set the number of atoms in the reactants equal to the number of atoms in the products for C, H, O and Ca: C: | 4 c_1 = c_2 + 3 c_3 H: | 6 c_1 = 6 c_3 O: | 4 c_1 = 3 c_2 + c_3 Ca: | c_1 = c_2 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 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | (CH3COO)2Ca ⟶ CaCO_3 + CH_3COCH_3

(CH3COO)2Ca ⟶ +

(CH3COO)2Ca ⟶ calcium carbonate + acetone

Construct the equilibrium constant, K, expression for: (CH3COO)2Ca ⟶ CaCO_3 + CH_3COCH_3 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: (CH3COO)2Ca ⟶ CaCO_3 + CH_3COCH_3 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 (CH3COO)2Ca | 1 | -1 CaCO_3 | 1 | 1 CH_3COCH_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression (CH3COO)2Ca | 1 | -1 | ([(CH3COO)2Ca])^(-1) CaCO_3 | 1 | 1 | [CaCO3] CH_3COCH_3 | 1 | 1 | [CH3COCH3] 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 = ([(CH3COO)2Ca])^(-1) [CaCO3] [CH3COCH3] = ([CaCO3] [CH3COCH3])/([(CH3COO)2Ca])

Construct the rate of reaction expression for: (CH3COO)2Ca ⟶ CaCO_3 + CH_3COCH_3 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: (CH3COO)2Ca ⟶ CaCO_3 + CH_3COCH_3 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 (CH3COO)2Ca | 1 | -1 CaCO_3 | 1 | 1 CH_3COCH_3 | 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 (CH3COO)2Ca | 1 | -1 | -(Δ[(CH3COO)2Ca])/(Δt) CaCO_3 | 1 | 1 | (Δ[CaCO3])/(Δt) CH_3COCH_3 | 1 | 1 | (Δ[CH3COCH3])/(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[(CH3COO)2Ca])/(Δt) = (Δ[CaCO3])/(Δt) = (Δ[CH3COCH3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| (CH3COO)2Ca | calcium carbonate | acetone formula | (CH3COO)2Ca | CaCO_3 | CH_3COCH_3 Hill formula | C4H6CaO4 | CCaO_3 | C_3H_6O name | | calcium carbonate | acetone

| (CH3COO)2Ca | calcium carbonate | acetone molar mass | 158.17 g/mol | 100.09 g/mol | 58.08 g/mol phase | | solid (at STP) | liquid (at STP) melting point | | 1340 °C | -94 °C boiling point | | | 56 °C (measured at 101308 Pa) density | | 2.71 g/cm^3 | 0.791 g/cm^3 solubility in water | | insoluble | miscible surface tension | | | 0.0228 N/m dynamic viscosity | | | 3.06×10^-4 Pa s (at 25 °C) odor | | | mint-like