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H2SO4 + C2H5OH + H2OK2CrO4 = H2O + K2SO4 + Cr2(SO4)3 + CH3CHO

Input interpretation

H_2SO_4 sulfuric acid + CH_3CH_2OH ethanol + H2OK2CrO4 ⟶ H_2O water + K_2SO_4 potassium sulfate + Cr_2(SO_4)_3 chromium sulfate + CH_3CHO acetaldehyde
H_2SO_4 sulfuric acid + CH_3CH_2OH ethanol + H2OK2CrO4 ⟶ H_2O water + K_2SO_4 potassium sulfate + Cr_2(SO_4)_3 chromium sulfate + CH_3CHO acetaldehyde

Balanced equation

Balance the chemical equation algebraically: H_2SO_4 + CH_3CH_2OH + H2OK2CrO4 ⟶ H_2O + K_2SO_4 + Cr_2(SO_4)_3 + CH_3CHO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2SO_4 + c_2 CH_3CH_2OH + c_3 H2OK2CrO4 ⟶ c_4 H_2O + c_5 K_2SO_4 + c_6 Cr_2(SO_4)_3 + c_7 CH_3CHO Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, S, C, K and Cr: H: | 2 c_1 + 6 c_2 + 2 c_3 = 2 c_4 + 4 c_7 O: | 4 c_1 + c_2 + 5 c_3 = c_4 + 4 c_5 + 12 c_6 + c_7 S: | c_1 = c_5 + 3 c_6 C: | 2 c_2 = 2 c_7 K: | 2 c_3 = 2 c_5 Cr: | c_3 = 2 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_6 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5 c_2 = 3 c_3 = 2 c_4 = 10 c_5 = 2 c_6 = 1 c_7 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 H_2SO_4 + 3 CH_3CH_2OH + 2 H2OK2CrO4 ⟶ 10 H_2O + 2 K_2SO_4 + Cr_2(SO_4)_3 + 3 CH_3CHO
Balance the chemical equation algebraically: H_2SO_4 + CH_3CH_2OH + H2OK2CrO4 ⟶ H_2O + K_2SO_4 + Cr_2(SO_4)_3 + CH_3CHO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2SO_4 + c_2 CH_3CH_2OH + c_3 H2OK2CrO4 ⟶ c_4 H_2O + c_5 K_2SO_4 + c_6 Cr_2(SO_4)_3 + c_7 CH_3CHO Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, S, C, K and Cr: H: | 2 c_1 + 6 c_2 + 2 c_3 = 2 c_4 + 4 c_7 O: | 4 c_1 + c_2 + 5 c_3 = c_4 + 4 c_5 + 12 c_6 + c_7 S: | c_1 = c_5 + 3 c_6 C: | 2 c_2 = 2 c_7 K: | 2 c_3 = 2 c_5 Cr: | c_3 = 2 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_6 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5 c_2 = 3 c_3 = 2 c_4 = 10 c_5 = 2 c_6 = 1 c_7 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 H_2SO_4 + 3 CH_3CH_2OH + 2 H2OK2CrO4 ⟶ 10 H_2O + 2 K_2SO_4 + Cr_2(SO_4)_3 + 3 CH_3CHO

Structures

+ + H2OK2CrO4 ⟶ + + +
+ + H2OK2CrO4 ⟶ + + +

Names

sulfuric acid + ethanol + H2OK2CrO4 ⟶ water + potassium sulfate + chromium sulfate + acetaldehyde
sulfuric acid + ethanol + H2OK2CrO4 ⟶ water + potassium sulfate + chromium sulfate + acetaldehyde

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2SO_4 + CH_3CH_2OH + H2OK2CrO4 ⟶ H_2O + K_2SO_4 + Cr_2(SO_4)_3 + CH_3CHO 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: 5 H_2SO_4 + 3 CH_3CH_2OH + 2 H2OK2CrO4 ⟶ 10 H_2O + 2 K_2SO_4 + Cr_2(SO_4)_3 + 3 CH_3CHO 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_2SO_4 | 5 | -5 CH_3CH_2OH | 3 | -3 H2OK2CrO4 | 2 | -2 H_2O | 10 | 10 K_2SO_4 | 2 | 2 Cr_2(SO_4)_3 | 1 | 1 CH_3CHO | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2SO_4 | 5 | -5 | ([H2SO4])^(-5) CH_3CH_2OH | 3 | -3 | ([CH3CH2OH])^(-3) H2OK2CrO4 | 2 | -2 | ([H2OK2CrO4])^(-2) H_2O | 10 | 10 | ([H2O])^10 K_2SO_4 | 2 | 2 | ([K2SO4])^2 Cr_2(SO_4)_3 | 1 | 1 | [Cr2(SO4)3] CH_3CHO | 3 | 3 | ([CH3CHO])^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 = ([H2SO4])^(-5) ([CH3CH2OH])^(-3) ([H2OK2CrO4])^(-2) ([H2O])^10 ([K2SO4])^2 [Cr2(SO4)3] ([CH3CHO])^3 = (([H2O])^10 ([K2SO4])^2 [Cr2(SO4)3] ([CH3CHO])^3)/(([H2SO4])^5 ([CH3CH2OH])^3 ([H2OK2CrO4])^2)
Construct the equilibrium constant, K, expression for: H_2SO_4 + CH_3CH_2OH + H2OK2CrO4 ⟶ H_2O + K_2SO_4 + Cr_2(SO_4)_3 + CH_3CHO 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: 5 H_2SO_4 + 3 CH_3CH_2OH + 2 H2OK2CrO4 ⟶ 10 H_2O + 2 K_2SO_4 + Cr_2(SO_4)_3 + 3 CH_3CHO 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_2SO_4 | 5 | -5 CH_3CH_2OH | 3 | -3 H2OK2CrO4 | 2 | -2 H_2O | 10 | 10 K_2SO_4 | 2 | 2 Cr_2(SO_4)_3 | 1 | 1 CH_3CHO | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2SO_4 | 5 | -5 | ([H2SO4])^(-5) CH_3CH_2OH | 3 | -3 | ([CH3CH2OH])^(-3) H2OK2CrO4 | 2 | -2 | ([H2OK2CrO4])^(-2) H_2O | 10 | 10 | ([H2O])^10 K_2SO_4 | 2 | 2 | ([K2SO4])^2 Cr_2(SO_4)_3 | 1 | 1 | [Cr2(SO4)3] CH_3CHO | 3 | 3 | ([CH3CHO])^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 = ([H2SO4])^(-5) ([CH3CH2OH])^(-3) ([H2OK2CrO4])^(-2) ([H2O])^10 ([K2SO4])^2 [Cr2(SO4)3] ([CH3CHO])^3 = (([H2O])^10 ([K2SO4])^2 [Cr2(SO4)3] ([CH3CHO])^3)/(([H2SO4])^5 ([CH3CH2OH])^3 ([H2OK2CrO4])^2)

Rate of reaction

Construct the rate of reaction expression for: H_2SO_4 + CH_3CH_2OH + H2OK2CrO4 ⟶ H_2O + K_2SO_4 + Cr_2(SO_4)_3 + CH_3CHO 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: 5 H_2SO_4 + 3 CH_3CH_2OH + 2 H2OK2CrO4 ⟶ 10 H_2O + 2 K_2SO_4 + Cr_2(SO_4)_3 + 3 CH_3CHO 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_2SO_4 | 5 | -5 CH_3CH_2OH | 3 | -3 H2OK2CrO4 | 2 | -2 H_2O | 10 | 10 K_2SO_4 | 2 | 2 Cr_2(SO_4)_3 | 1 | 1 CH_3CHO | 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 H_2SO_4 | 5 | -5 | -1/5 (Δ[H2SO4])/(Δt) CH_3CH_2OH | 3 | -3 | -1/3 (Δ[CH3CH2OH])/(Δt) H2OK2CrO4 | 2 | -2 | -1/2 (Δ[H2OK2CrO4])/(Δt) H_2O | 10 | 10 | 1/10 (Δ[H2O])/(Δt) K_2SO_4 | 2 | 2 | 1/2 (Δ[K2SO4])/(Δt) Cr_2(SO_4)_3 | 1 | 1 | (Δ[Cr2(SO4)3])/(Δt) CH_3CHO | 3 | 3 | 1/3 (Δ[CH3CHO])/(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 = -1/5 (Δ[H2SO4])/(Δt) = -1/3 (Δ[CH3CH2OH])/(Δt) = -1/2 (Δ[H2OK2CrO4])/(Δt) = 1/10 (Δ[H2O])/(Δt) = 1/2 (Δ[K2SO4])/(Δt) = (Δ[Cr2(SO4)3])/(Δt) = 1/3 (Δ[CH3CHO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2SO_4 + CH_3CH_2OH + H2OK2CrO4 ⟶ H_2O + K_2SO_4 + Cr_2(SO_4)_3 + CH_3CHO 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: 5 H_2SO_4 + 3 CH_3CH_2OH + 2 H2OK2CrO4 ⟶ 10 H_2O + 2 K_2SO_4 + Cr_2(SO_4)_3 + 3 CH_3CHO 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_2SO_4 | 5 | -5 CH_3CH_2OH | 3 | -3 H2OK2CrO4 | 2 | -2 H_2O | 10 | 10 K_2SO_4 | 2 | 2 Cr_2(SO_4)_3 | 1 | 1 CH_3CHO | 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 H_2SO_4 | 5 | -5 | -1/5 (Δ[H2SO4])/(Δt) CH_3CH_2OH | 3 | -3 | -1/3 (Δ[CH3CH2OH])/(Δt) H2OK2CrO4 | 2 | -2 | -1/2 (Δ[H2OK2CrO4])/(Δt) H_2O | 10 | 10 | 1/10 (Δ[H2O])/(Δt) K_2SO_4 | 2 | 2 | 1/2 (Δ[K2SO4])/(Δt) Cr_2(SO_4)_3 | 1 | 1 | (Δ[Cr2(SO4)3])/(Δt) CH_3CHO | 3 | 3 | 1/3 (Δ[CH3CHO])/(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 = -1/5 (Δ[H2SO4])/(Δt) = -1/3 (Δ[CH3CH2OH])/(Δt) = -1/2 (Δ[H2OK2CrO4])/(Δt) = 1/10 (Δ[H2O])/(Δt) = 1/2 (Δ[K2SO4])/(Δt) = (Δ[Cr2(SO4)3])/(Δt) = 1/3 (Δ[CH3CHO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| sulfuric acid | ethanol | H2OK2CrO4 | water | potassium sulfate | chromium sulfate | acetaldehyde formula | H_2SO_4 | CH_3CH_2OH | H2OK2CrO4 | H_2O | K_2SO_4 | Cr_2(SO_4)_3 | CH_3CHO Hill formula | H_2O_4S | C_2H_6O | H2CrK2O5 | H_2O | K_2O_4S | Cr_2O_12S_3 | C_2H_4O name | sulfuric acid | ethanol | | water | potassium sulfate | chromium sulfate | acetaldehyde IUPAC name | sulfuric acid | ethanol | | water | dipotassium sulfate | chromium(+3) cation trisulfate | acetaldehyde
| sulfuric acid | ethanol | H2OK2CrO4 | water | potassium sulfate | chromium sulfate | acetaldehyde formula | H_2SO_4 | CH_3CH_2OH | H2OK2CrO4 | H_2O | K_2SO_4 | Cr_2(SO_4)_3 | CH_3CHO Hill formula | H_2O_4S | C_2H_6O | H2CrK2O5 | H_2O | K_2O_4S | Cr_2O_12S_3 | C_2H_4O name | sulfuric acid | ethanol | | water | potassium sulfate | chromium sulfate | acetaldehyde IUPAC name | sulfuric acid | ethanol | | water | dipotassium sulfate | chromium(+3) cation trisulfate | acetaldehyde

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