CrO3 + C4H10O = H2O + Cr + C4H8O

CrO_3 chromium trioxide + CH_3(CH_2)_3OH 1-butanol ⟶ H_2O water + Cr chromium + C_4H_8O tetrahydrofuran

Balance the chemical equation algebraically: CrO_3 + CH_3(CH_2)_3OH ⟶ H_2O + Cr + C_4H_8O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CrO_3 + c_2 CH_3(CH_2)_3OH ⟶ c_3 H_2O + c_4 Cr + c_5 C_4H_8O Set the number of atoms in the reactants equal to the number of atoms in the products for Cr, O, C and H: Cr: | c_1 = c_4 O: | 3 c_1 + c_2 = c_3 + c_5 C: | 4 c_2 = 4 c_5 H: | 10 c_2 = 2 c_3 + 8 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3 c_3 = 3 c_4 = 1 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | CrO_3 + 3 CH_3(CH_2)_3OH ⟶ 3 H_2O + Cr + 3 C_4H_8O

+ ⟶ + +

chromium trioxide + 1-butanol ⟶ water + chromium + tetrahydrofuran

Construct the equilibrium constant, K, expression for: CrO_3 + CH_3(CH_2)_3OH ⟶ H_2O + Cr + C_4H_8O 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: CrO_3 + 3 CH_3(CH_2)_3OH ⟶ 3 H_2O + Cr + 3 C_4H_8O 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 CrO_3 | 1 | -1 CH_3(CH_2)_3OH | 3 | -3 H_2O | 3 | 3 Cr | 1 | 1 C_4H_8O | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CrO_3 | 1 | -1 | ([CrO3])^(-1) CH_3(CH_2)_3OH | 3 | -3 | ([CH3(CH2)3OH])^(-3) H_2O | 3 | 3 | ([H2O])^3 Cr | 1 | 1 | [Cr] C_4H_8O | 3 | 3 | ([C4H8O])^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 = ([CrO3])^(-1) ([CH3(CH2)3OH])^(-3) ([H2O])^3 [Cr] ([C4H8O])^3 = (([H2O])^3 [Cr] ([C4H8O])^3)/([CrO3] ([CH3(CH2)3OH])^3)

Construct the rate of reaction expression for: CrO_3 + CH_3(CH_2)_3OH ⟶ H_2O + Cr + C_4H_8O 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: CrO_3 + 3 CH_3(CH_2)_3OH ⟶ 3 H_2O + Cr + 3 C_4H_8O 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 CrO_3 | 1 | -1 CH_3(CH_2)_3OH | 3 | -3 H_2O | 3 | 3 Cr | 1 | 1 C_4H_8O | 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 CrO_3 | 1 | -1 | -(Δ[CrO3])/(Δt) CH_3(CH_2)_3OH | 3 | -3 | -1/3 (Δ[CH3(CH2)3OH])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Cr | 1 | 1 | (Δ[Cr])/(Δt) C_4H_8O | 3 | 3 | 1/3 (Δ[C4H8O])/(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 = -(Δ[CrO3])/(Δt) = -1/3 (Δ[CH3(CH2)3OH])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[Cr])/(Δt) = 1/3 (Δ[C4H8O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| chromium trioxide | 1-butanol | water | chromium | tetrahydrofuran formula | CrO_3 | CH_3(CH_2)_3OH | H_2O | Cr | C_4H_8O Hill formula | CrO_3 | C_4H_10O | H_2O | Cr | C_4H_8O name | chromium trioxide | 1-butanol | water | chromium | tetrahydrofuran IUPAC name | trioxochromium | butan-1-ol | water | chromium | tetrahydrofuran