Input interpretation
![Cl_2 chlorine + KOH potassium hydroxide + Cr_2(SO_4)_3 chromium sulfate ⟶ H_2O water + K_2SO_4 potassium sulfate + KCl potassium chloride + K_2CrO_4 potassium chromate](../image_source/40cd334ddad6c54aa975b082adac2851.png)
Cl_2 chlorine + KOH potassium hydroxide + Cr_2(SO_4)_3 chromium sulfate ⟶ H_2O water + K_2SO_4 potassium sulfate + KCl potassium chloride + K_2CrO_4 potassium chromate
Balanced equation
![Balance the chemical equation algebraically: Cl_2 + KOH + Cr_2(SO_4)_3 ⟶ H_2O + K_2SO_4 + KCl + K_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Cl_2 + c_2 KOH + c_3 Cr_2(SO_4)_3 ⟶ c_4 H_2O + c_5 K_2SO_4 + c_6 KCl + c_7 K_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, K, O, Cr and S: Cl: | 2 c_1 = c_6 H: | c_2 = 2 c_4 K: | c_2 = 2 c_5 + c_6 + 2 c_7 O: | c_2 + 12 c_3 = c_4 + 4 c_5 + 4 c_7 Cr: | 2 c_3 = c_7 S: | 3 c_3 = c_5 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 16 c_3 = 1 c_4 = 8 c_5 = 3 c_6 = 6 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 Cl_2 + 16 KOH + Cr_2(SO_4)_3 ⟶ 8 H_2O + 3 K_2SO_4 + 6 KCl + 2 K_2CrO_4](../image_source/a20145c1af4981bde61227a738e42747.png)
Balance the chemical equation algebraically: Cl_2 + KOH + Cr_2(SO_4)_3 ⟶ H_2O + K_2SO_4 + KCl + K_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Cl_2 + c_2 KOH + c_3 Cr_2(SO_4)_3 ⟶ c_4 H_2O + c_5 K_2SO_4 + c_6 KCl + c_7 K_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, K, O, Cr and S: Cl: | 2 c_1 = c_6 H: | c_2 = 2 c_4 K: | c_2 = 2 c_5 + c_6 + 2 c_7 O: | c_2 + 12 c_3 = c_4 + 4 c_5 + 4 c_7 Cr: | 2 c_3 = c_7 S: | 3 c_3 = c_5 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 16 c_3 = 1 c_4 = 8 c_5 = 3 c_6 = 6 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 Cl_2 + 16 KOH + Cr_2(SO_4)_3 ⟶ 8 H_2O + 3 K_2SO_4 + 6 KCl + 2 K_2CrO_4
Structures
![+ + ⟶ + + +](../image_source/824bddc4391ffd1982411c2d8e9f252d.png)
+ + ⟶ + + +
Names
![chlorine + potassium hydroxide + chromium sulfate ⟶ water + potassium sulfate + potassium chloride + potassium chromate](../image_source/eaab7ea038d171ace6824a553417383c.png)
chlorine + potassium hydroxide + chromium sulfate ⟶ water + potassium sulfate + potassium chloride + potassium chromate
Equilibrium constant
![Construct the equilibrium constant, K, expression for: Cl_2 + KOH + Cr_2(SO_4)_3 ⟶ H_2O + K_2SO_4 + KCl + K_2CrO_4 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: 3 Cl_2 + 16 KOH + Cr_2(SO_4)_3 ⟶ 8 H_2O + 3 K_2SO_4 + 6 KCl + 2 K_2CrO_4 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 Cl_2 | 3 | -3 KOH | 16 | -16 Cr_2(SO_4)_3 | 1 | -1 H_2O | 8 | 8 K_2SO_4 | 3 | 3 KCl | 6 | 6 K_2CrO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Cl_2 | 3 | -3 | ([Cl2])^(-3) KOH | 16 | -16 | ([KOH])^(-16) Cr_2(SO_4)_3 | 1 | -1 | ([Cr2(SO4)3])^(-1) H_2O | 8 | 8 | ([H2O])^8 K_2SO_4 | 3 | 3 | ([K2SO4])^3 KCl | 6 | 6 | ([KCl])^6 K_2CrO_4 | 2 | 2 | ([K2CrO4])^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 = ([Cl2])^(-3) ([KOH])^(-16) ([Cr2(SO4)3])^(-1) ([H2O])^8 ([K2SO4])^3 ([KCl])^6 ([K2CrO4])^2 = (([H2O])^8 ([K2SO4])^3 ([KCl])^6 ([K2CrO4])^2)/(([Cl2])^3 ([KOH])^16 [Cr2(SO4)3])](../image_source/e1fe20bbdefbaf4e50a6de84ff10fd03.png)
Construct the equilibrium constant, K, expression for: Cl_2 + KOH + Cr_2(SO_4)_3 ⟶ H_2O + K_2SO_4 + KCl + K_2CrO_4 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: 3 Cl_2 + 16 KOH + Cr_2(SO_4)_3 ⟶ 8 H_2O + 3 K_2SO_4 + 6 KCl + 2 K_2CrO_4 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 Cl_2 | 3 | -3 KOH | 16 | -16 Cr_2(SO_4)_3 | 1 | -1 H_2O | 8 | 8 K_2SO_4 | 3 | 3 KCl | 6 | 6 K_2CrO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Cl_2 | 3 | -3 | ([Cl2])^(-3) KOH | 16 | -16 | ([KOH])^(-16) Cr_2(SO_4)_3 | 1 | -1 | ([Cr2(SO4)3])^(-1) H_2O | 8 | 8 | ([H2O])^8 K_2SO_4 | 3 | 3 | ([K2SO4])^3 KCl | 6 | 6 | ([KCl])^6 K_2CrO_4 | 2 | 2 | ([K2CrO4])^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 = ([Cl2])^(-3) ([KOH])^(-16) ([Cr2(SO4)3])^(-1) ([H2O])^8 ([K2SO4])^3 ([KCl])^6 ([K2CrO4])^2 = (([H2O])^8 ([K2SO4])^3 ([KCl])^6 ([K2CrO4])^2)/(([Cl2])^3 ([KOH])^16 [Cr2(SO4)3])
Rate of reaction
![Construct the rate of reaction expression for: Cl_2 + KOH + Cr_2(SO_4)_3 ⟶ H_2O + K_2SO_4 + KCl + K_2CrO_4 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: 3 Cl_2 + 16 KOH + Cr_2(SO_4)_3 ⟶ 8 H_2O + 3 K_2SO_4 + 6 KCl + 2 K_2CrO_4 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 Cl_2 | 3 | -3 KOH | 16 | -16 Cr_2(SO_4)_3 | 1 | -1 H_2O | 8 | 8 K_2SO_4 | 3 | 3 KCl | 6 | 6 K_2CrO_4 | 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 Cl_2 | 3 | -3 | -1/3 (Δ[Cl2])/(Δt) KOH | 16 | -16 | -1/16 (Δ[KOH])/(Δt) Cr_2(SO_4)_3 | 1 | -1 | -(Δ[Cr2(SO4)3])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) K_2SO_4 | 3 | 3 | 1/3 (Δ[K2SO4])/(Δt) KCl | 6 | 6 | 1/6 (Δ[KCl])/(Δt) K_2CrO_4 | 2 | 2 | 1/2 (Δ[K2CrO4])/(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 = -1/3 (Δ[Cl2])/(Δt) = -1/16 (Δ[KOH])/(Δt) = -(Δ[Cr2(SO4)3])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/3 (Δ[K2SO4])/(Δt) = 1/6 (Δ[KCl])/(Δt) = 1/2 (Δ[K2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/1268337f477bb0f0ed3b46c01e504ac3.png)
Construct the rate of reaction expression for: Cl_2 + KOH + Cr_2(SO_4)_3 ⟶ H_2O + K_2SO_4 + KCl + K_2CrO_4 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: 3 Cl_2 + 16 KOH + Cr_2(SO_4)_3 ⟶ 8 H_2O + 3 K_2SO_4 + 6 KCl + 2 K_2CrO_4 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 Cl_2 | 3 | -3 KOH | 16 | -16 Cr_2(SO_4)_3 | 1 | -1 H_2O | 8 | 8 K_2SO_4 | 3 | 3 KCl | 6 | 6 K_2CrO_4 | 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 Cl_2 | 3 | -3 | -1/3 (Δ[Cl2])/(Δt) KOH | 16 | -16 | -1/16 (Δ[KOH])/(Δt) Cr_2(SO_4)_3 | 1 | -1 | -(Δ[Cr2(SO4)3])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) K_2SO_4 | 3 | 3 | 1/3 (Δ[K2SO4])/(Δt) KCl | 6 | 6 | 1/6 (Δ[KCl])/(Δt) K_2CrO_4 | 2 | 2 | 1/2 (Δ[K2CrO4])/(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 = -1/3 (Δ[Cl2])/(Δt) = -1/16 (Δ[KOH])/(Δt) = -(Δ[Cr2(SO4)3])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/3 (Δ[K2SO4])/(Δt) = 1/6 (Δ[KCl])/(Δt) = 1/2 (Δ[K2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| chlorine | potassium hydroxide | chromium sulfate | water | potassium sulfate | potassium chloride | potassium chromate formula | Cl_2 | KOH | Cr_2(SO_4)_3 | H_2O | K_2SO_4 | KCl | K_2CrO_4 Hill formula | Cl_2 | HKO | Cr_2O_12S_3 | H_2O | K_2O_4S | ClK | CrK_2O_4 name | chlorine | potassium hydroxide | chromium sulfate | water | potassium sulfate | potassium chloride | potassium chromate IUPAC name | molecular chlorine | potassium hydroxide | chromium(+3) cation trisulfate | water | dipotassium sulfate | potassium chloride | dipotassium dioxido-dioxochromium](../image_source/da456cee776fc628bfbe739e7d7c4d21.png)
| chlorine | potassium hydroxide | chromium sulfate | water | potassium sulfate | potassium chloride | potassium chromate formula | Cl_2 | KOH | Cr_2(SO_4)_3 | H_2O | K_2SO_4 | KCl | K_2CrO_4 Hill formula | Cl_2 | HKO | Cr_2O_12S_3 | H_2O | K_2O_4S | ClK | CrK_2O_4 name | chlorine | potassium hydroxide | chromium sulfate | water | potassium sulfate | potassium chloride | potassium chromate IUPAC name | molecular chlorine | potassium hydroxide | chromium(+3) cation trisulfate | water | dipotassium sulfate | potassium chloride | dipotassium dioxido-dioxochromium