AgNO3 + HF = HNO3 + AgF

AgNO_3 silver nitrate + HF hydrogen fluoride ⟶ HNO_3 nitric acid + AgF silver fluoride

Balance the chemical equation algebraically: AgNO_3 + HF ⟶ HNO_3 + AgF Add stoichiometric coefficients, c_i, to the reactants and products: c_1 AgNO_3 + c_2 HF ⟶ c_3 HNO_3 + c_4 AgF Set the number of atoms in the reactants equal to the number of atoms in the products for Ag, N, O, F and H: Ag: | c_1 = c_4 N: | c_1 = c_3 O: | 3 c_1 = 3 c_3 F: | c_2 = c_4 H: | 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 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | AgNO_3 + HF ⟶ HNO_3 + AgF

+ ⟶ +

silver nitrate + hydrogen fluoride ⟶ nitric acid + silver fluoride

Construct the equilibrium constant, K, expression for: AgNO_3 + HF ⟶ HNO_3 + AgF 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: AgNO_3 + HF ⟶ HNO_3 + AgF 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 AgNO_3 | 1 | -1 HF | 1 | -1 HNO_3 | 1 | 1 AgF | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression AgNO_3 | 1 | -1 | ([AgNO3])^(-1) HF | 1 | -1 | ([HF])^(-1) HNO_3 | 1 | 1 | [HNO3] AgF | 1 | 1 | [AgF] 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 = ([AgNO3])^(-1) ([HF])^(-1) [HNO3] [AgF] = ([HNO3] [AgF])/([AgNO3] [HF])

Construct the rate of reaction expression for: AgNO_3 + HF ⟶ HNO_3 + AgF 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: AgNO_3 + HF ⟶ HNO_3 + AgF 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 AgNO_3 | 1 | -1 HF | 1 | -1 HNO_3 | 1 | 1 AgF | 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 AgNO_3 | 1 | -1 | -(Δ[AgNO3])/(Δt) HF | 1 | -1 | -(Δ[HF])/(Δt) HNO_3 | 1 | 1 | (Δ[HNO3])/(Δt) AgF | 1 | 1 | (Δ[AgF])/(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[AgNO3])/(Δt) = -(Δ[HF])/(Δt) = (Δ[HNO3])/(Δt) = (Δ[AgF])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| silver nitrate | hydrogen fluoride | nitric acid | silver fluoride formula | AgNO_3 | HF | HNO_3 | AgF Hill formula | AgNO_3 | FH | HNO_3 | AgF name | silver nitrate | hydrogen fluoride | nitric acid | silver fluoride IUPAC name | silver nitrate | hydrogen fluoride | nitric acid | fluorosilver

| silver nitrate | hydrogen fluoride | nitric acid | silver fluoride molar mass | 169.87 g/mol | 20.006 g/mol | 63.012 g/mol | 126.8666 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) | solid (at STP) melting point | 212 °C | -83.36 °C | -41.6 °C | 300 °C boiling point | | 19.5 °C | 83 °C | 1150 °C density | | 8.18×10^-4 g/cm^3 (at 25 °C) | 1.5129 g/cm^3 | 5.852 g/cm^3 solubility in water | soluble | miscible | miscible | dynamic viscosity | | 1.2571×10^-5 Pa s (at 20 °C) | 7.6×10^-4 Pa s (at 25 °C) | odor | odorless | | |