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H2O + Cl2 + CrI3 = HCl + HIO3 + H2CrO4

Input interpretation

H_2O water + Cl_2 chlorine + Cr_1I_3 chromium(III) iodide ⟶ HCl hydrogen chloride + HIO_3 iodic acid + H_2CrO_4 chromic acid
H_2O water + Cl_2 chlorine + Cr_1I_3 chromium(III) iodide ⟶ HCl hydrogen chloride + HIO_3 iodic acid + H_2CrO_4 chromic acid

Balanced equation

Balance the chemical equation algebraically: H_2O + Cl_2 + Cr_1I_3 ⟶ HCl + HIO_3 + H_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Cl_2 + c_3 Cr_1I_3 ⟶ c_4 HCl + c_5 HIO_3 + c_6 H_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Cl, Cr and I: H: | 2 c_1 = c_4 + c_5 + 2 c_6 O: | c_1 = 3 c_5 + 4 c_6 Cl: | 2 c_2 = c_4 Cr: | c_3 = c_6 I: | 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 = 13 c_2 = 21/2 c_3 = 1 c_4 = 21 c_5 = 3 c_6 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 26 c_2 = 21 c_3 = 2 c_4 = 42 c_5 = 6 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 26 H_2O + 21 Cl_2 + 2 Cr_1I_3 ⟶ 42 HCl + 6 HIO_3 + 2 H_2CrO_4
Balance the chemical equation algebraically: H_2O + Cl_2 + Cr_1I_3 ⟶ HCl + HIO_3 + H_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Cl_2 + c_3 Cr_1I_3 ⟶ c_4 HCl + c_5 HIO_3 + c_6 H_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Cl, Cr and I: H: | 2 c_1 = c_4 + c_5 + 2 c_6 O: | c_1 = 3 c_5 + 4 c_6 Cl: | 2 c_2 = c_4 Cr: | c_3 = c_6 I: | 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 = 13 c_2 = 21/2 c_3 = 1 c_4 = 21 c_5 = 3 c_6 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 26 c_2 = 21 c_3 = 2 c_4 = 42 c_5 = 6 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 26 H_2O + 21 Cl_2 + 2 Cr_1I_3 ⟶ 42 HCl + 6 HIO_3 + 2 H_2CrO_4

Structures

+ + ⟶ + +
+ + ⟶ + +

Names

water + chlorine + chromium(III) iodide ⟶ hydrogen chloride + iodic acid + chromic acid
water + chlorine + chromium(III) iodide ⟶ hydrogen chloride + iodic acid + chromic acid

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + Cl_2 + Cr_1I_3 ⟶ HCl + HIO_3 + H_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: 26 H_2O + 21 Cl_2 + 2 Cr_1I_3 ⟶ 42 HCl + 6 HIO_3 + 2 H_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 H_2O | 26 | -26 Cl_2 | 21 | -21 Cr_1I_3 | 2 | -2 HCl | 42 | 42 HIO_3 | 6 | 6 H_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 H_2O | 26 | -26 | ([H2O])^(-26) Cl_2 | 21 | -21 | ([Cl2])^(-21) Cr_1I_3 | 2 | -2 | ([Cr1I3])^(-2) HCl | 42 | 42 | ([HCl])^42 HIO_3 | 6 | 6 | ([HIO3])^6 H_2CrO_4 | 2 | 2 | ([H2CrO4])^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 = ([H2O])^(-26) ([Cl2])^(-21) ([Cr1I3])^(-2) ([HCl])^42 ([HIO3])^6 ([H2CrO4])^2 = (([HCl])^42 ([HIO3])^6 ([H2CrO4])^2)/(([H2O])^26 ([Cl2])^21 ([Cr1I3])^2)
Construct the equilibrium constant, K, expression for: H_2O + Cl_2 + Cr_1I_3 ⟶ HCl + HIO_3 + H_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: 26 H_2O + 21 Cl_2 + 2 Cr_1I_3 ⟶ 42 HCl + 6 HIO_3 + 2 H_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 H_2O | 26 | -26 Cl_2 | 21 | -21 Cr_1I_3 | 2 | -2 HCl | 42 | 42 HIO_3 | 6 | 6 H_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 H_2O | 26 | -26 | ([H2O])^(-26) Cl_2 | 21 | -21 | ([Cl2])^(-21) Cr_1I_3 | 2 | -2 | ([Cr1I3])^(-2) HCl | 42 | 42 | ([HCl])^42 HIO_3 | 6 | 6 | ([HIO3])^6 H_2CrO_4 | 2 | 2 | ([H2CrO4])^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 = ([H2O])^(-26) ([Cl2])^(-21) ([Cr1I3])^(-2) ([HCl])^42 ([HIO3])^6 ([H2CrO4])^2 = (([HCl])^42 ([HIO3])^6 ([H2CrO4])^2)/(([H2O])^26 ([Cl2])^21 ([Cr1I3])^2)

Rate of reaction

Construct the rate of reaction expression for: H_2O + Cl_2 + Cr_1I_3 ⟶ HCl + HIO_3 + H_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: 26 H_2O + 21 Cl_2 + 2 Cr_1I_3 ⟶ 42 HCl + 6 HIO_3 + 2 H_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 H_2O | 26 | -26 Cl_2 | 21 | -21 Cr_1I_3 | 2 | -2 HCl | 42 | 42 HIO_3 | 6 | 6 H_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 H_2O | 26 | -26 | -1/26 (Δ[H2O])/(Δt) Cl_2 | 21 | -21 | -1/21 (Δ[Cl2])/(Δt) Cr_1I_3 | 2 | -2 | -1/2 (Δ[Cr1I3])/(Δt) HCl | 42 | 42 | 1/42 (Δ[HCl])/(Δt) HIO_3 | 6 | 6 | 1/6 (Δ[HIO3])/(Δt) H_2CrO_4 | 2 | 2 | 1/2 (Δ[H2CrO4])/(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 = -1/26 (Δ[H2O])/(Δt) = -1/21 (Δ[Cl2])/(Δt) = -1/2 (Δ[Cr1I3])/(Δt) = 1/42 (Δ[HCl])/(Δt) = 1/6 (Δ[HIO3])/(Δt) = 1/2 (Δ[H2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + Cl_2 + Cr_1I_3 ⟶ HCl + HIO_3 + H_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: 26 H_2O + 21 Cl_2 + 2 Cr_1I_3 ⟶ 42 HCl + 6 HIO_3 + 2 H_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 H_2O | 26 | -26 Cl_2 | 21 | -21 Cr_1I_3 | 2 | -2 HCl | 42 | 42 HIO_3 | 6 | 6 H_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 H_2O | 26 | -26 | -1/26 (Δ[H2O])/(Δt) Cl_2 | 21 | -21 | -1/21 (Δ[Cl2])/(Δt) Cr_1I_3 | 2 | -2 | -1/2 (Δ[Cr1I3])/(Δt) HCl | 42 | 42 | 1/42 (Δ[HCl])/(Δt) HIO_3 | 6 | 6 | 1/6 (Δ[HIO3])/(Δt) H_2CrO_4 | 2 | 2 | 1/2 (Δ[H2CrO4])/(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 = -1/26 (Δ[H2O])/(Δt) = -1/21 (Δ[Cl2])/(Δt) = -1/2 (Δ[Cr1I3])/(Δt) = 1/42 (Δ[HCl])/(Δt) = 1/6 (Δ[HIO3])/(Δt) = 1/2 (Δ[H2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | chlorine | chromium(III) iodide | hydrogen chloride | iodic acid | chromic acid formula | H_2O | Cl_2 | Cr_1I_3 | HCl | HIO_3 | H_2CrO_4 Hill formula | H_2O | Cl_2 | CrI_3 | ClH | HIO_3 | CrH_2O_4 name | water | chlorine | chromium(III) iodide | hydrogen chloride | iodic acid | chromic acid IUPAC name | water | molecular chlorine | triiodochromium | hydrogen chloride | iodic acid | dihydroxy-dioxo-chromium
| water | chlorine | chromium(III) iodide | hydrogen chloride | iodic acid | chromic acid formula | H_2O | Cl_2 | Cr_1I_3 | HCl | HIO_3 | H_2CrO_4 Hill formula | H_2O | Cl_2 | CrI_3 | ClH | HIO_3 | CrH_2O_4 name | water | chlorine | chromium(III) iodide | hydrogen chloride | iodic acid | chromic acid IUPAC name | water | molecular chlorine | triiodochromium | hydrogen chloride | iodic acid | dihydroxy-dioxo-chromium

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