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CaCO3 + HBr = H2CO3 + CaBr2

Input interpretation

CaCO_3 calcium carbonate + HBr hydrogen bromide ⟶ H_2CO_3 carbonic acid + CaBr_2 calcium bromide
CaCO_3 calcium carbonate + HBr hydrogen bromide ⟶ H_2CO_3 carbonic acid + CaBr_2 calcium bromide

Balanced equation

Balance the chemical equation algebraically: CaCO_3 + HBr ⟶ H_2CO_3 + CaBr_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CaCO_3 + c_2 HBr ⟶ c_3 H_2CO_3 + c_4 CaBr_2 Set the number of atoms in the reactants equal to the number of atoms in the products for C, Ca, O, Br and H: C: | c_1 = c_3 Ca: | c_1 = c_4 O: | 3 c_1 = 3 c_3 Br: | c_2 = 2 c_4 H: | c_2 = 2 c_3 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: | | CaCO_3 + 2 HBr ⟶ H_2CO_3 + CaBr_2
Balance the chemical equation algebraically: CaCO_3 + HBr ⟶ H_2CO_3 + CaBr_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CaCO_3 + c_2 HBr ⟶ c_3 H_2CO_3 + c_4 CaBr_2 Set the number of atoms in the reactants equal to the number of atoms in the products for C, Ca, O, Br and H: C: | c_1 = c_3 Ca: | c_1 = c_4 O: | 3 c_1 = 3 c_3 Br: | c_2 = 2 c_4 H: | c_2 = 2 c_3 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: | | CaCO_3 + 2 HBr ⟶ H_2CO_3 + CaBr_2

Structures

+ ⟶ +
+ ⟶ +

Names

calcium carbonate + hydrogen bromide ⟶ carbonic acid + calcium bromide
calcium carbonate + hydrogen bromide ⟶ carbonic acid + calcium bromide

Equilibrium constant

Construct the equilibrium constant, K, expression for: CaCO_3 + HBr ⟶ H_2CO_3 + CaBr_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: CaCO_3 + 2 HBr ⟶ H_2CO_3 + CaBr_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 CaCO_3 | 1 | -1 HBr | 2 | -2 H_2CO_3 | 1 | 1 CaBr_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CaCO_3 | 1 | -1 | ([CaCO3])^(-1) HBr | 2 | -2 | ([HBr])^(-2) H_2CO_3 | 1 | 1 | [H2CO3] CaBr_2 | 1 | 1 | [CaBr2] 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 = ([CaCO3])^(-1) ([HBr])^(-2) [H2CO3] [CaBr2] = ([H2CO3] [CaBr2])/([CaCO3] ([HBr])^2)
Construct the equilibrium constant, K, expression for: CaCO_3 + HBr ⟶ H_2CO_3 + CaBr_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: CaCO_3 + 2 HBr ⟶ H_2CO_3 + CaBr_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 CaCO_3 | 1 | -1 HBr | 2 | -2 H_2CO_3 | 1 | 1 CaBr_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CaCO_3 | 1 | -1 | ([CaCO3])^(-1) HBr | 2 | -2 | ([HBr])^(-2) H_2CO_3 | 1 | 1 | [H2CO3] CaBr_2 | 1 | 1 | [CaBr2] 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 = ([CaCO3])^(-1) ([HBr])^(-2) [H2CO3] [CaBr2] = ([H2CO3] [CaBr2])/([CaCO3] ([HBr])^2)

Rate of reaction

Construct the rate of reaction expression for: CaCO_3 + HBr ⟶ H_2CO_3 + CaBr_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: CaCO_3 + 2 HBr ⟶ H_2CO_3 + CaBr_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 CaCO_3 | 1 | -1 HBr | 2 | -2 H_2CO_3 | 1 | 1 CaBr_2 | 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 CaCO_3 | 1 | -1 | -(Δ[CaCO3])/(Δt) HBr | 2 | -2 | -1/2 (Δ[HBr])/(Δt) H_2CO_3 | 1 | 1 | (Δ[H2CO3])/(Δt) CaBr_2 | 1 | 1 | (Δ[CaBr2])/(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[CaCO3])/(Δt) = -1/2 (Δ[HBr])/(Δt) = (Δ[H2CO3])/(Δt) = (Δ[CaBr2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: CaCO_3 + HBr ⟶ H_2CO_3 + CaBr_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: CaCO_3 + 2 HBr ⟶ H_2CO_3 + CaBr_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 CaCO_3 | 1 | -1 HBr | 2 | -2 H_2CO_3 | 1 | 1 CaBr_2 | 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 CaCO_3 | 1 | -1 | -(Δ[CaCO3])/(Δt) HBr | 2 | -2 | -1/2 (Δ[HBr])/(Δt) H_2CO_3 | 1 | 1 | (Δ[H2CO3])/(Δt) CaBr_2 | 1 | 1 | (Δ[CaBr2])/(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[CaCO3])/(Δt) = -1/2 (Δ[HBr])/(Δt) = (Δ[H2CO3])/(Δt) = (Δ[CaBr2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| calcium carbonate | hydrogen bromide | carbonic acid | calcium bromide formula | CaCO_3 | HBr | H_2CO_3 | CaBr_2 Hill formula | CCaO_3 | BrH | CH_2O_3 | Br_2Ca name | calcium carbonate | hydrogen bromide | carbonic acid | calcium bromide IUPAC name | calcium carbonate | hydrogen bromide | carbonic acid | calcium dibromide
| calcium carbonate | hydrogen bromide | carbonic acid | calcium bromide formula | CaCO_3 | HBr | H_2CO_3 | CaBr_2 Hill formula | CCaO_3 | BrH | CH_2O_3 | Br_2Ca name | calcium carbonate | hydrogen bromide | carbonic acid | calcium bromide IUPAC name | calcium carbonate | hydrogen bromide | carbonic acid | calcium dibromide

Substance properties

| calcium carbonate | hydrogen bromide | carbonic acid | calcium bromide molar mass | 100.09 g/mol | 80.912 g/mol | 62.024 g/mol | 199.89 g/mol phase | solid (at STP) | gas (at STP) | | solid (at STP) melting point | 1340 °C | -86.8 °C | | 730 °C boiling point | | -66.38 °C | | 810 °C density | 2.71 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | | 3.353 g/cm^3 solubility in water | insoluble | miscible | | soluble surface tension | | 0.0271 N/m | | dynamic viscosity | | 8.4×10^-4 Pa s (at -75 °C) | |
| calcium carbonate | hydrogen bromide | carbonic acid | calcium bromide molar mass | 100.09 g/mol | 80.912 g/mol | 62.024 g/mol | 199.89 g/mol phase | solid (at STP) | gas (at STP) | | solid (at STP) melting point | 1340 °C | -86.8 °C | | 730 °C boiling point | | -66.38 °C | | 810 °C density | 2.71 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) | | 3.353 g/cm^3 solubility in water | insoluble | miscible | | soluble surface tension | | 0.0271 N/m | | dynamic viscosity | | 8.4×10^-4 Pa s (at -75 °C) | |

Units

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