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S + Ba(OH)2 = H2O + BaSO4 + BaS

Input interpretation

S mixed sulfur + Ba(OH)_2 barium hydroxide ⟶ H_2O water + BaSO_4 barium sulfate + BaS barium sulfide
S mixed sulfur + Ba(OH)_2 barium hydroxide ⟶ H_2O water + BaSO_4 barium sulfate + BaS barium sulfide

Balanced equation

Balance the chemical equation algebraically: S + Ba(OH)_2 ⟶ H_2O + BaSO_4 + BaS Add stoichiometric coefficients, c_i, to the reactants and products: c_1 S + c_2 Ba(OH)_2 ⟶ c_3 H_2O + c_4 BaSO_4 + c_5 BaS Set the number of atoms in the reactants equal to the number of atoms in the products for S, Ba, H and O: S: | c_1 = c_4 + c_5 Ba: | c_2 = c_4 + c_5 H: | 2 c_2 = 2 c_3 O: | 2 c_2 = c_3 + 4 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 = 4 c_2 = 4 c_3 = 4 c_4 = 1 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 S + 4 Ba(OH)_2 ⟶ 4 H_2O + BaSO_4 + 3 BaS
Balance the chemical equation algebraically: S + Ba(OH)_2 ⟶ H_2O + BaSO_4 + BaS Add stoichiometric coefficients, c_i, to the reactants and products: c_1 S + c_2 Ba(OH)_2 ⟶ c_3 H_2O + c_4 BaSO_4 + c_5 BaS Set the number of atoms in the reactants equal to the number of atoms in the products for S, Ba, H and O: S: | c_1 = c_4 + c_5 Ba: | c_2 = c_4 + c_5 H: | 2 c_2 = 2 c_3 O: | 2 c_2 = c_3 + 4 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 = 4 c_2 = 4 c_3 = 4 c_4 = 1 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 S + 4 Ba(OH)_2 ⟶ 4 H_2O + BaSO_4 + 3 BaS

Structures

+ ⟶ + +
+ ⟶ + +

Names

mixed sulfur + barium hydroxide ⟶ water + barium sulfate + barium sulfide
mixed sulfur + barium hydroxide ⟶ water + barium sulfate + barium sulfide

Equilibrium constant

Construct the equilibrium constant, K, expression for: S + Ba(OH)_2 ⟶ H_2O + BaSO_4 + BaS 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: 4 S + 4 Ba(OH)_2 ⟶ 4 H_2O + BaSO_4 + 3 BaS 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 S | 4 | -4 Ba(OH)_2 | 4 | -4 H_2O | 4 | 4 BaSO_4 | 1 | 1 BaS | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression S | 4 | -4 | ([S])^(-4) Ba(OH)_2 | 4 | -4 | ([Ba(OH)2])^(-4) H_2O | 4 | 4 | ([H2O])^4 BaSO_4 | 1 | 1 | [BaSO4] BaS | 3 | 3 | ([BaS])^3 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 = ([S])^(-4) ([Ba(OH)2])^(-4) ([H2O])^4 [BaSO4] ([BaS])^3 = (([H2O])^4 [BaSO4] ([BaS])^3)/(([S])^4 ([Ba(OH)2])^4)
Construct the equilibrium constant, K, expression for: S + Ba(OH)_2 ⟶ H_2O + BaSO_4 + BaS 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: 4 S + 4 Ba(OH)_2 ⟶ 4 H_2O + BaSO_4 + 3 BaS 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 S | 4 | -4 Ba(OH)_2 | 4 | -4 H_2O | 4 | 4 BaSO_4 | 1 | 1 BaS | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression S | 4 | -4 | ([S])^(-4) Ba(OH)_2 | 4 | -4 | ([Ba(OH)2])^(-4) H_2O | 4 | 4 | ([H2O])^4 BaSO_4 | 1 | 1 | [BaSO4] BaS | 3 | 3 | ([BaS])^3 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 = ([S])^(-4) ([Ba(OH)2])^(-4) ([H2O])^4 [BaSO4] ([BaS])^3 = (([H2O])^4 [BaSO4] ([BaS])^3)/(([S])^4 ([Ba(OH)2])^4)

Rate of reaction

Construct the rate of reaction expression for: S + Ba(OH)_2 ⟶ H_2O + BaSO_4 + BaS 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: 4 S + 4 Ba(OH)_2 ⟶ 4 H_2O + BaSO_4 + 3 BaS 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 S | 4 | -4 Ba(OH)_2 | 4 | -4 H_2O | 4 | 4 BaSO_4 | 1 | 1 BaS | 3 | 3 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 S | 4 | -4 | -1/4 (Δ[S])/(Δt) Ba(OH)_2 | 4 | -4 | -1/4 (Δ[Ba(OH)2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) BaSO_4 | 1 | 1 | (Δ[BaSO4])/(Δt) BaS | 3 | 3 | 1/3 (Δ[BaS])/(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 = -1/4 (Δ[S])/(Δt) = -1/4 (Δ[Ba(OH)2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = (Δ[BaSO4])/(Δt) = 1/3 (Δ[BaS])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: S + Ba(OH)_2 ⟶ H_2O + BaSO_4 + BaS 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: 4 S + 4 Ba(OH)_2 ⟶ 4 H_2O + BaSO_4 + 3 BaS 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 S | 4 | -4 Ba(OH)_2 | 4 | -4 H_2O | 4 | 4 BaSO_4 | 1 | 1 BaS | 3 | 3 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 S | 4 | -4 | -1/4 (Δ[S])/(Δt) Ba(OH)_2 | 4 | -4 | -1/4 (Δ[Ba(OH)2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) BaSO_4 | 1 | 1 | (Δ[BaSO4])/(Δt) BaS | 3 | 3 | 1/3 (Δ[BaS])/(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 = -1/4 (Δ[S])/(Δt) = -1/4 (Δ[Ba(OH)2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = (Δ[BaSO4])/(Δt) = 1/3 (Δ[BaS])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| mixed sulfur | barium hydroxide | water | barium sulfate | barium sulfide formula | S | Ba(OH)_2 | H_2O | BaSO_4 | BaS Hill formula | S | BaH_2O_2 | H_2O | BaO_4S | BaS name | mixed sulfur | barium hydroxide | water | barium sulfate | barium sulfide IUPAC name | sulfur | barium(+2) cation dihydroxide | water | barium(+2) cation sulfate | thioxobarium
| mixed sulfur | barium hydroxide | water | barium sulfate | barium sulfide formula | S | Ba(OH)_2 | H_2O | BaSO_4 | BaS Hill formula | S | BaH_2O_2 | H_2O | BaO_4S | BaS name | mixed sulfur | barium hydroxide | water | barium sulfate | barium sulfide IUPAC name | sulfur | barium(+2) cation dihydroxide | water | barium(+2) cation sulfate | thioxobarium

Substance properties

| mixed sulfur | barium hydroxide | water | barium sulfate | barium sulfide molar mass | 32.06 g/mol | 171.34 g/mol | 18.015 g/mol | 233.38 g/mol | 169.39 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | solid (at STP) melting point | 112.8 °C | 300 °C | 0 °C | 1345 °C | 1999.85 °C boiling point | 444.7 °C | | 99.9839 °C | | density | 2.07 g/cm^3 | 2.2 g/cm^3 | 1 g/cm^3 | 4.5 g/cm^3 | 4.25 g/cm^3 solubility in water | | | | insoluble | surface tension | | | 0.0728 N/m | | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |
| mixed sulfur | barium hydroxide | water | barium sulfate | barium sulfide molar mass | 32.06 g/mol | 171.34 g/mol | 18.015 g/mol | 233.38 g/mol | 169.39 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | solid (at STP) melting point | 112.8 °C | 300 °C | 0 °C | 1345 °C | 1999.85 °C boiling point | 444.7 °C | | 99.9839 °C | | density | 2.07 g/cm^3 | 2.2 g/cm^3 | 1 g/cm^3 | 4.5 g/cm^3 | 4.25 g/cm^3 solubility in water | | | | insoluble | surface tension | | | 0.0728 N/m | | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |

Units

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