KBrO4 = O2 + KBr

KBrO4 ⟶ O_2 oxygen + KBr potassium bromide

Balance the chemical equation algebraically: KBrO4 ⟶ O_2 + KBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KBrO4 ⟶ c_2 O_2 + c_3 KBr Set the number of atoms in the reactants equal to the number of atoms in the products for K, Br and O: K: | c_1 = c_3 Br: | c_1 = c_3 O: | 4 c_1 = 2 c_2 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 = 2 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | KBrO4 ⟶ 2 O_2 + KBr

KBrO4 ⟶ +

KBrO4 ⟶ oxygen + potassium bromide

Construct the equilibrium constant, K, expression for: KBrO4 ⟶ O_2 + KBr 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: KBrO4 ⟶ 2 O_2 + KBr 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 KBrO4 | 1 | -1 O_2 | 2 | 2 KBr | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KBrO4 | 1 | -1 | ([KBrO4])^(-1) O_2 | 2 | 2 | ([O2])^2 KBr | 1 | 1 | [KBr] 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 = ([KBrO4])^(-1) ([O2])^2 [KBr] = (([O2])^2 [KBr])/([KBrO4])

Construct the rate of reaction expression for: KBrO4 ⟶ O_2 + KBr 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: KBrO4 ⟶ 2 O_2 + KBr 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 KBrO4 | 1 | -1 O_2 | 2 | 2 KBr | 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 KBrO4 | 1 | -1 | -(Δ[KBrO4])/(Δt) O_2 | 2 | 2 | 1/2 (Δ[O2])/(Δt) KBr | 1 | 1 | (Δ[KBr])/(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[KBrO4])/(Δt) = 1/2 (Δ[O2])/(Δt) = (Δ[KBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| KBrO4 | oxygen | potassium bromide formula | KBrO4 | O_2 | KBr Hill formula | BrKO4 | O_2 | BrK name | | oxygen | potassium bromide IUPAC name | | molecular oxygen | potassium bromide

| KBrO4 | oxygen | potassium bromide molar mass | 183 g/mol | 31.998 g/mol | 119 g/mol phase | | gas (at STP) | solid (at STP) melting point | | -218 °C | 734 °C boiling point | | -183 °C | 1435 °C density | | 0.001429 g/cm^3 (at 0 °C) | 2.75 g/cm^3 solubility in water | | | soluble surface tension | | 0.01347 N/m | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | odor | | odorless |