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KOH + Br2 + CrSO4 = H2O + K2SO4 + KBr + K2CrO4

Input interpretation

KOH potassium hydroxide + Br_2 bromine + CrSO4 ⟶ H_2O water + K_2SO_4 potassium sulfate + KBr potassium bromide + K_2CrO_4 potassium chromate
KOH potassium hydroxide + Br_2 bromine + CrSO4 ⟶ H_2O water + K_2SO_4 potassium sulfate + KBr potassium bromide + K_2CrO_4 potassium chromate

Balanced equation

Balance the chemical equation algebraically: KOH + Br_2 + CrSO4 ⟶ H_2O + K_2SO_4 + KBr + K_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Br_2 + c_3 CrSO4 ⟶ c_4 H_2O + c_5 K_2SO_4 + c_6 KBr + c_7 K_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Br, Cr and S: H: | c_1 = 2 c_4 K: | c_1 = 2 c_5 + c_6 + 2 c_7 O: | c_1 + 4 c_3 = c_4 + 4 c_5 + 4 c_7 Br: | 2 c_2 = c_6 Cr: | c_3 = c_7 S: | 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 = 8 c_2 = 2 c_3 = 1 c_4 = 4 c_5 = 1 c_6 = 4 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 KOH + 2 Br_2 + CrSO4 ⟶ 4 H_2O + K_2SO_4 + 4 KBr + K_2CrO_4
Balance the chemical equation algebraically: KOH + Br_2 + CrSO4 ⟶ H_2O + K_2SO_4 + KBr + K_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Br_2 + c_3 CrSO4 ⟶ c_4 H_2O + c_5 K_2SO_4 + c_6 KBr + c_7 K_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Br, Cr and S: H: | c_1 = 2 c_4 K: | c_1 = 2 c_5 + c_6 + 2 c_7 O: | c_1 + 4 c_3 = c_4 + 4 c_5 + 4 c_7 Br: | 2 c_2 = c_6 Cr: | c_3 = c_7 S: | 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 = 8 c_2 = 2 c_3 = 1 c_4 = 4 c_5 = 1 c_6 = 4 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 KOH + 2 Br_2 + CrSO4 ⟶ 4 H_2O + K_2SO_4 + 4 KBr + K_2CrO_4

Structures

+ + CrSO4 ⟶ + + +
+ + CrSO4 ⟶ + + +

Names

potassium hydroxide + bromine + CrSO4 ⟶ water + potassium sulfate + potassium bromide + potassium chromate
potassium hydroxide + bromine + CrSO4 ⟶ water + potassium sulfate + potassium bromide + potassium chromate

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + Br_2 + CrSO4 ⟶ H_2O + K_2SO_4 + KBr + 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: 8 KOH + 2 Br_2 + CrSO4 ⟶ 4 H_2O + K_2SO_4 + 4 KBr + 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 KOH | 8 | -8 Br_2 | 2 | -2 CrSO4 | 1 | -1 H_2O | 4 | 4 K_2SO_4 | 1 | 1 KBr | 4 | 4 K_2CrO_4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 8 | -8 | ([KOH])^(-8) Br_2 | 2 | -2 | ([Br2])^(-2) CrSO4 | 1 | -1 | ([CrSO4])^(-1) H_2O | 4 | 4 | ([H2O])^4 K_2SO_4 | 1 | 1 | [K2SO4] KBr | 4 | 4 | ([KBr])^4 K_2CrO_4 | 1 | 1 | [K2CrO4] 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])^(-8) ([Br2])^(-2) ([CrSO4])^(-1) ([H2O])^4 [K2SO4] ([KBr])^4 [K2CrO4] = (([H2O])^4 [K2SO4] ([KBr])^4 [K2CrO4])/(([KOH])^8 ([Br2])^2 [CrSO4])
Construct the equilibrium constant, K, expression for: KOH + Br_2 + CrSO4 ⟶ H_2O + K_2SO_4 + KBr + 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: 8 KOH + 2 Br_2 + CrSO4 ⟶ 4 H_2O + K_2SO_4 + 4 KBr + 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 KOH | 8 | -8 Br_2 | 2 | -2 CrSO4 | 1 | -1 H_2O | 4 | 4 K_2SO_4 | 1 | 1 KBr | 4 | 4 K_2CrO_4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 8 | -8 | ([KOH])^(-8) Br_2 | 2 | -2 | ([Br2])^(-2) CrSO4 | 1 | -1 | ([CrSO4])^(-1) H_2O | 4 | 4 | ([H2O])^4 K_2SO_4 | 1 | 1 | [K2SO4] KBr | 4 | 4 | ([KBr])^4 K_2CrO_4 | 1 | 1 | [K2CrO4] 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])^(-8) ([Br2])^(-2) ([CrSO4])^(-1) ([H2O])^4 [K2SO4] ([KBr])^4 [K2CrO4] = (([H2O])^4 [K2SO4] ([KBr])^4 [K2CrO4])/(([KOH])^8 ([Br2])^2 [CrSO4])

Rate of reaction

Construct the rate of reaction expression for: KOH + Br_2 + CrSO4 ⟶ H_2O + K_2SO_4 + KBr + 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: 8 KOH + 2 Br_2 + CrSO4 ⟶ 4 H_2O + K_2SO_4 + 4 KBr + 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 KOH | 8 | -8 Br_2 | 2 | -2 CrSO4 | 1 | -1 H_2O | 4 | 4 K_2SO_4 | 1 | 1 KBr | 4 | 4 K_2CrO_4 | 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 KOH | 8 | -8 | -1/8 (Δ[KOH])/(Δt) Br_2 | 2 | -2 | -1/2 (Δ[Br2])/(Δt) CrSO4 | 1 | -1 | -(Δ[CrSO4])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) K_2SO_4 | 1 | 1 | (Δ[K2SO4])/(Δt) KBr | 4 | 4 | 1/4 (Δ[KBr])/(Δt) K_2CrO_4 | 1 | 1 | (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 = -1/8 (Δ[KOH])/(Δt) = -1/2 (Δ[Br2])/(Δt) = -(Δ[CrSO4])/(Δt) = 1/4 (Δ[H2O])/(Δt) = (Δ[K2SO4])/(Δt) = 1/4 (Δ[KBr])/(Δt) = (Δ[K2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + Br_2 + CrSO4 ⟶ H_2O + K_2SO_4 + KBr + 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: 8 KOH + 2 Br_2 + CrSO4 ⟶ 4 H_2O + K_2SO_4 + 4 KBr + 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 KOH | 8 | -8 Br_2 | 2 | -2 CrSO4 | 1 | -1 H_2O | 4 | 4 K_2SO_4 | 1 | 1 KBr | 4 | 4 K_2CrO_4 | 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 KOH | 8 | -8 | -1/8 (Δ[KOH])/(Δt) Br_2 | 2 | -2 | -1/2 (Δ[Br2])/(Δt) CrSO4 | 1 | -1 | -(Δ[CrSO4])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) K_2SO_4 | 1 | 1 | (Δ[K2SO4])/(Δt) KBr | 4 | 4 | 1/4 (Δ[KBr])/(Δt) K_2CrO_4 | 1 | 1 | (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 = -1/8 (Δ[KOH])/(Δt) = -1/2 (Δ[Br2])/(Δt) = -(Δ[CrSO4])/(Δt) = 1/4 (Δ[H2O])/(Δt) = (Δ[K2SO4])/(Δt) = 1/4 (Δ[KBr])/(Δt) = (Δ[K2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | bromine | CrSO4 | water | potassium sulfate | potassium bromide | potassium chromate formula | KOH | Br_2 | CrSO4 | H_2O | K_2SO_4 | KBr | K_2CrO_4 Hill formula | HKO | Br_2 | CrO4S | H_2O | K_2O_4S | BrK | CrK_2O_4 name | potassium hydroxide | bromine | | water | potassium sulfate | potassium bromide | potassium chromate IUPAC name | potassium hydroxide | molecular bromine | | water | dipotassium sulfate | potassium bromide | dipotassium dioxido-dioxochromium
| potassium hydroxide | bromine | CrSO4 | water | potassium sulfate | potassium bromide | potassium chromate formula | KOH | Br_2 | CrSO4 | H_2O | K_2SO_4 | KBr | K_2CrO_4 Hill formula | HKO | Br_2 | CrO4S | H_2O | K_2O_4S | BrK | CrK_2O_4 name | potassium hydroxide | bromine | | water | potassium sulfate | potassium bromide | potassium chromate IUPAC name | potassium hydroxide | molecular bromine | | water | dipotassium sulfate | potassium bromide | dipotassium dioxido-dioxochromium

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