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KOH + KBrO2 + CCl3NO2 = H2O + KCl + KNO3 + KBr + K2CO3

Input interpretation

KOH potassium hydroxide + KBrO2 + Cl_3CNO_2 trichloronitromethane ⟶ H_2O water + KCl potassium chloride + KNO_3 potassium nitrate + KBr potassium bromide + K_2CO_3 pearl ash
KOH potassium hydroxide + KBrO2 + Cl_3CNO_2 trichloronitromethane ⟶ H_2O water + KCl potassium chloride + KNO_3 potassium nitrate + KBr potassium bromide + K_2CO_3 pearl ash

Balanced equation

Balance the chemical equation algebraically: KOH + KBrO2 + Cl_3CNO_2 ⟶ H_2O + KCl + KNO_3 + KBr + K_2CO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 KBrO2 + c_3 Cl_3CNO_2 ⟶ c_4 H_2O + c_5 KCl + c_6 KNO_3 + c_7 KBr + c_8 K_2CO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Br, C, Cl and N: H: | c_1 = 2 c_4 K: | c_1 + c_2 = c_5 + c_6 + c_7 + 2 c_8 O: | c_1 + 2 c_2 + 2 c_3 = c_4 + 3 c_6 + 3 c_8 Br: | c_2 = c_7 C: | c_3 = c_8 Cl: | 3 c_3 = c_5 N: | c_3 = c_6 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 = 12 c_2 = 1 c_3 = 2 c_4 = 6 c_5 = 6 c_6 = 2 c_7 = 1 c_8 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 KOH + KBrO2 + 2 Cl_3CNO_2 ⟶ 6 H_2O + 6 KCl + 2 KNO_3 + KBr + 2 K_2CO_3
Balance the chemical equation algebraically: KOH + KBrO2 + Cl_3CNO_2 ⟶ H_2O + KCl + KNO_3 + KBr + K_2CO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 KBrO2 + c_3 Cl_3CNO_2 ⟶ c_4 H_2O + c_5 KCl + c_6 KNO_3 + c_7 KBr + c_8 K_2CO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Br, C, Cl and N: H: | c_1 = 2 c_4 K: | c_1 + c_2 = c_5 + c_6 + c_7 + 2 c_8 O: | c_1 + 2 c_2 + 2 c_3 = c_4 + 3 c_6 + 3 c_8 Br: | c_2 = c_7 C: | c_3 = c_8 Cl: | 3 c_3 = c_5 N: | c_3 = c_6 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 = 12 c_2 = 1 c_3 = 2 c_4 = 6 c_5 = 6 c_6 = 2 c_7 = 1 c_8 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 KOH + KBrO2 + 2 Cl_3CNO_2 ⟶ 6 H_2O + 6 KCl + 2 KNO_3 + KBr + 2 K_2CO_3

Structures

+ KBrO2 + ⟶ + + + +
+ KBrO2 + ⟶ + + + +

Names

potassium hydroxide + KBrO2 + trichloronitromethane ⟶ water + potassium chloride + potassium nitrate + potassium bromide + pearl ash
potassium hydroxide + KBrO2 + trichloronitromethane ⟶ water + potassium chloride + potassium nitrate + potassium bromide + pearl ash

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + KBrO2 + Cl_3CNO_2 ⟶ H_2O + KCl + KNO_3 + KBr + K_2CO_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: 12 KOH + KBrO2 + 2 Cl_3CNO_2 ⟶ 6 H_2O + 6 KCl + 2 KNO_3 + KBr + 2 K_2CO_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 KOH | 12 | -12 KBrO2 | 1 | -1 Cl_3CNO_2 | 2 | -2 H_2O | 6 | 6 KCl | 6 | 6 KNO_3 | 2 | 2 KBr | 1 | 1 K_2CO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 12 | -12 | ([KOH])^(-12) KBrO2 | 1 | -1 | ([KBrO2])^(-1) Cl_3CNO_2 | 2 | -2 | ([Cl3CNO2])^(-2) H_2O | 6 | 6 | ([H2O])^6 KCl | 6 | 6 | ([KCl])^6 KNO_3 | 2 | 2 | ([KNO3])^2 KBr | 1 | 1 | [KBr] K_2CO_3 | 2 | 2 | ([K2CO3])^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 = ([KOH])^(-12) ([KBrO2])^(-1) ([Cl3CNO2])^(-2) ([H2O])^6 ([KCl])^6 ([KNO3])^2 [KBr] ([K2CO3])^2 = (([H2O])^6 ([KCl])^6 ([KNO3])^2 [KBr] ([K2CO3])^2)/(([KOH])^12 [KBrO2] ([Cl3CNO2])^2)
Construct the equilibrium constant, K, expression for: KOH + KBrO2 + Cl_3CNO_2 ⟶ H_2O + KCl + KNO_3 + KBr + K_2CO_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: 12 KOH + KBrO2 + 2 Cl_3CNO_2 ⟶ 6 H_2O + 6 KCl + 2 KNO_3 + KBr + 2 K_2CO_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 KOH | 12 | -12 KBrO2 | 1 | -1 Cl_3CNO_2 | 2 | -2 H_2O | 6 | 6 KCl | 6 | 6 KNO_3 | 2 | 2 KBr | 1 | 1 K_2CO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 12 | -12 | ([KOH])^(-12) KBrO2 | 1 | -1 | ([KBrO2])^(-1) Cl_3CNO_2 | 2 | -2 | ([Cl3CNO2])^(-2) H_2O | 6 | 6 | ([H2O])^6 KCl | 6 | 6 | ([KCl])^6 KNO_3 | 2 | 2 | ([KNO3])^2 KBr | 1 | 1 | [KBr] K_2CO_3 | 2 | 2 | ([K2CO3])^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 = ([KOH])^(-12) ([KBrO2])^(-1) ([Cl3CNO2])^(-2) ([H2O])^6 ([KCl])^6 ([KNO3])^2 [KBr] ([K2CO3])^2 = (([H2O])^6 ([KCl])^6 ([KNO3])^2 [KBr] ([K2CO3])^2)/(([KOH])^12 [KBrO2] ([Cl3CNO2])^2)

Rate of reaction

Construct the rate of reaction expression for: KOH + KBrO2 + Cl_3CNO_2 ⟶ H_2O + KCl + KNO_3 + KBr + K_2CO_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: 12 KOH + KBrO2 + 2 Cl_3CNO_2 ⟶ 6 H_2O + 6 KCl + 2 KNO_3 + KBr + 2 K_2CO_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 KOH | 12 | -12 KBrO2 | 1 | -1 Cl_3CNO_2 | 2 | -2 H_2O | 6 | 6 KCl | 6 | 6 KNO_3 | 2 | 2 KBr | 1 | 1 K_2CO_3 | 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 KOH | 12 | -12 | -1/12 (Δ[KOH])/(Δt) KBrO2 | 1 | -1 | -(Δ[KBrO2])/(Δt) Cl_3CNO_2 | 2 | -2 | -1/2 (Δ[Cl3CNO2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) KCl | 6 | 6 | 1/6 (Δ[KCl])/(Δt) KNO_3 | 2 | 2 | 1/2 (Δ[KNO3])/(Δt) KBr | 1 | 1 | (Δ[KBr])/(Δt) K_2CO_3 | 2 | 2 | 1/2 (Δ[K2CO3])/(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 = -1/12 (Δ[KOH])/(Δt) = -(Δ[KBrO2])/(Δt) = -1/2 (Δ[Cl3CNO2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/6 (Δ[KCl])/(Δt) = 1/2 (Δ[KNO3])/(Δt) = (Δ[KBr])/(Δt) = 1/2 (Δ[K2CO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + KBrO2 + Cl_3CNO_2 ⟶ H_2O + KCl + KNO_3 + KBr + K_2CO_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: 12 KOH + KBrO2 + 2 Cl_3CNO_2 ⟶ 6 H_2O + 6 KCl + 2 KNO_3 + KBr + 2 K_2CO_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 KOH | 12 | -12 KBrO2 | 1 | -1 Cl_3CNO_2 | 2 | -2 H_2O | 6 | 6 KCl | 6 | 6 KNO_3 | 2 | 2 KBr | 1 | 1 K_2CO_3 | 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 KOH | 12 | -12 | -1/12 (Δ[KOH])/(Δt) KBrO2 | 1 | -1 | -(Δ[KBrO2])/(Δt) Cl_3CNO_2 | 2 | -2 | -1/2 (Δ[Cl3CNO2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) KCl | 6 | 6 | 1/6 (Δ[KCl])/(Δt) KNO_3 | 2 | 2 | 1/2 (Δ[KNO3])/(Δt) KBr | 1 | 1 | (Δ[KBr])/(Δt) K_2CO_3 | 2 | 2 | 1/2 (Δ[K2CO3])/(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 = -1/12 (Δ[KOH])/(Δt) = -(Δ[KBrO2])/(Δt) = -1/2 (Δ[Cl3CNO2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/6 (Δ[KCl])/(Δt) = 1/2 (Δ[KNO3])/(Δt) = (Δ[KBr])/(Δt) = 1/2 (Δ[K2CO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | KBrO2 | trichloronitromethane | water | potassium chloride | potassium nitrate | potassium bromide | pearl ash formula | KOH | KBrO2 | Cl_3CNO_2 | H_2O | KCl | KNO_3 | KBr | K_2CO_3 Hill formula | HKO | BrKO2 | CCl_3NO_2 | H_2O | ClK | KNO_3 | BrK | CK_2O_3 name | potassium hydroxide | | trichloronitromethane | water | potassium chloride | potassium nitrate | potassium bromide | pearl ash IUPAC name | potassium hydroxide | | trichloro-nitromethane | water | potassium chloride | potassium nitrate | potassium bromide | dipotassium carbonate
| potassium hydroxide | KBrO2 | trichloronitromethane | water | potassium chloride | potassium nitrate | potassium bromide | pearl ash formula | KOH | KBrO2 | Cl_3CNO_2 | H_2O | KCl | KNO_3 | KBr | K_2CO_3 Hill formula | HKO | BrKO2 | CCl_3NO_2 | H_2O | ClK | KNO_3 | BrK | CK_2O_3 name | potassium hydroxide | | trichloronitromethane | water | potassium chloride | potassium nitrate | potassium bromide | pearl ash IUPAC name | potassium hydroxide | | trichloro-nitromethane | water | potassium chloride | potassium nitrate | potassium bromide | dipotassium carbonate

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