H2O + CrO3 = H2CrO4

H_2O (water) + CrO_3 (chromium trioxide) ⟶ H_2CrO_4 (chromic acid)

Balance the chemical equation algebraically: H_2O + CrO_3 ⟶ H_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 CrO_3 ⟶ c_3 H_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and Cr: H: | 2 c_1 = 2 c_3 O: | c_1 + 3 c_2 = 4 c_3 Cr: | c_2 = c_3 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 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + CrO_3 ⟶ H_2CrO_4

+ ⟶

water + chromium trioxide ⟶ chromic acid

Construct the equilibrium constant, K, expression for: H_2O + CrO_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: H_2O + CrO_3 ⟶ 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 | 1 | -1 CrO_3 | 1 | -1 H_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 H_2O | 1 | -1 | ([H2O])^(-1) CrO_3 | 1 | -1 | ([CrO3])^(-1) H_2CrO_4 | 1 | 1 | [H2CrO4] 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])^(-1) ([CrO3])^(-1) [H2CrO4] = ([H2CrO4])/([H2O] [CrO3])

Construct the rate of reaction expression for: H_2O + CrO_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: H_2O + CrO_3 ⟶ 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 | 1 | -1 CrO_3 | 1 | -1 H_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 H_2O | 1 | -1 | -(Δ[H2O])/(Δt) CrO_3 | 1 | -1 | -(Δ[CrO3])/(Δt) H_2CrO_4 | 1 | 1 | (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 = -(Δ[H2O])/(Δt) = -(Δ[CrO3])/(Δt) = (Δ[H2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | chromium trioxide | chromic acid formula | H_2O | CrO_3 | H_2CrO_4 Hill formula | H_2O | CrO_3 | CrH_2O_4 name | water | chromium trioxide | chromic acid IUPAC name | water | trioxochromium | dihydroxy-dioxo-chromium

| water | chromium trioxide | chromic acid molar mass | 18.015 g/mol | 99.993 g/mol | 118.01 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) melting point | 0 °C | 196 °C | 196 °C boiling point | 99.9839 °C | | density | 1 g/cm^3 | | 2.7 g/cm^3 solubility in water | | very soluble | very soluble surface tension | 0.0728 N/m | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | odor | odorless | odorless |