Search

H2O + RhF6 = HF + HRhO4 + RhO2

Input interpretation

H_2O water + RhF_6 rhodium(VI) fluoride ⟶ HF hydrogen fluoride + HRhO4 + RhO_2 rhodium dioxide
H_2O water + RhF_6 rhodium(VI) fluoride ⟶ HF hydrogen fluoride + HRhO4 + RhO_2 rhodium dioxide

Balanced equation

Balance the chemical equation algebraically: H_2O + RhF_6 ⟶ HF + HRhO4 + RhO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 RhF_6 ⟶ c_3 HF + c_4 HRhO4 + c_5 RhO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, F and Rh: H: | 2 c_1 = c_3 + c_4 O: | c_1 = 4 c_4 + 2 c_5 F: | 6 c_2 = c_3 Rh: | c_2 = c_4 + 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_5 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 3 c_3 = 18 c_4 = 2 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 H_2O + 3 RhF_6 ⟶ 18 HF + 2 HRhO4 + RhO_2
Balance the chemical equation algebraically: H_2O + RhF_6 ⟶ HF + HRhO4 + RhO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 RhF_6 ⟶ c_3 HF + c_4 HRhO4 + c_5 RhO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, F and Rh: H: | 2 c_1 = c_3 + c_4 O: | c_1 = 4 c_4 + 2 c_5 F: | 6 c_2 = c_3 Rh: | c_2 = c_4 + 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_5 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 3 c_3 = 18 c_4 = 2 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 H_2O + 3 RhF_6 ⟶ 18 HF + 2 HRhO4 + RhO_2

Structures

+ ⟶ + HRhO4 +
+ ⟶ + HRhO4 +

Names

water + rhodium(VI) fluoride ⟶ hydrogen fluoride + HRhO4 + rhodium dioxide
water + rhodium(VI) fluoride ⟶ hydrogen fluoride + HRhO4 + rhodium dioxide

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + RhF_6 ⟶ HF + HRhO4 + RhO_2 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: 10 H_2O + 3 RhF_6 ⟶ 18 HF + 2 HRhO4 + RhO_2 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 | 10 | -10 RhF_6 | 3 | -3 HF | 18 | 18 HRhO4 | 2 | 2 RhO_2 | 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 | 10 | -10 | ([H2O])^(-10) RhF_6 | 3 | -3 | ([RhF6])^(-3) HF | 18 | 18 | ([HF])^18 HRhO4 | 2 | 2 | ([HRhO4])^2 RhO_2 | 1 | 1 | [RhO2] 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])^(-10) ([RhF6])^(-3) ([HF])^18 ([HRhO4])^2 [RhO2] = (([HF])^18 ([HRhO4])^2 [RhO2])/(([H2O])^10 ([RhF6])^3)
Construct the equilibrium constant, K, expression for: H_2O + RhF_6 ⟶ HF + HRhO4 + RhO_2 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: 10 H_2O + 3 RhF_6 ⟶ 18 HF + 2 HRhO4 + RhO_2 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 | 10 | -10 RhF_6 | 3 | -3 HF | 18 | 18 HRhO4 | 2 | 2 RhO_2 | 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 | 10 | -10 | ([H2O])^(-10) RhF_6 | 3 | -3 | ([RhF6])^(-3) HF | 18 | 18 | ([HF])^18 HRhO4 | 2 | 2 | ([HRhO4])^2 RhO_2 | 1 | 1 | [RhO2] 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])^(-10) ([RhF6])^(-3) ([HF])^18 ([HRhO4])^2 [RhO2] = (([HF])^18 ([HRhO4])^2 [RhO2])/(([H2O])^10 ([RhF6])^3)

Rate of reaction

Construct the rate of reaction expression for: H_2O + RhF_6 ⟶ HF + HRhO4 + RhO_2 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: 10 H_2O + 3 RhF_6 ⟶ 18 HF + 2 HRhO4 + RhO_2 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 | 10 | -10 RhF_6 | 3 | -3 HF | 18 | 18 HRhO4 | 2 | 2 RhO_2 | 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 | 10 | -10 | -1/10 (Δ[H2O])/(Δt) RhF_6 | 3 | -3 | -1/3 (Δ[RhF6])/(Δt) HF | 18 | 18 | 1/18 (Δ[HF])/(Δt) HRhO4 | 2 | 2 | 1/2 (Δ[HRhO4])/(Δt) RhO_2 | 1 | 1 | (Δ[RhO2])/(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 = -1/10 (Δ[H2O])/(Δt) = -1/3 (Δ[RhF6])/(Δt) = 1/18 (Δ[HF])/(Δt) = 1/2 (Δ[HRhO4])/(Δt) = (Δ[RhO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + RhF_6 ⟶ HF + HRhO4 + RhO_2 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: 10 H_2O + 3 RhF_6 ⟶ 18 HF + 2 HRhO4 + RhO_2 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 | 10 | -10 RhF_6 | 3 | -3 HF | 18 | 18 HRhO4 | 2 | 2 RhO_2 | 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 | 10 | -10 | -1/10 (Δ[H2O])/(Δt) RhF_6 | 3 | -3 | -1/3 (Δ[RhF6])/(Δt) HF | 18 | 18 | 1/18 (Δ[HF])/(Δt) HRhO4 | 2 | 2 | 1/2 (Δ[HRhO4])/(Δt) RhO_2 | 1 | 1 | (Δ[RhO2])/(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 = -1/10 (Δ[H2O])/(Δt) = -1/3 (Δ[RhF6])/(Δt) = 1/18 (Δ[HF])/(Δt) = 1/2 (Δ[HRhO4])/(Δt) = (Δ[RhO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | rhodium(VI) fluoride | hydrogen fluoride | HRhO4 | rhodium dioxide formula | H_2O | RhF_6 | HF | HRhO4 | RhO_2 Hill formula | H_2O | F_6Rh_1 | FH | HO4Rh | O_2Rh name | water | rhodium(VI) fluoride | hydrogen fluoride | | rhodium dioxide IUPAC name | water | | hydrogen fluoride | | dioxorhodium
| water | rhodium(VI) fluoride | hydrogen fluoride | HRhO4 | rhodium dioxide formula | H_2O | RhF_6 | HF | HRhO4 | RhO_2 Hill formula | H_2O | F_6Rh_1 | FH | HO4Rh | O_2Rh name | water | rhodium(VI) fluoride | hydrogen fluoride | | rhodium dioxide IUPAC name | water | | hydrogen fluoride | | dioxorhodium

Substance properties

| water | rhodium(VI) fluoride | hydrogen fluoride | HRhO4 | rhodium dioxide molar mass | 18.015 g/mol | 216.89592 g/mol | 20.006 g/mol | 167.91 g/mol | 134.903 g/mol phase | liquid (at STP) | | gas (at STP) | | melting point | 0 °C | 70 °C | -83.36 °C | | boiling point | 99.9839 °C | | 19.5 °C | | density | 1 g/cm^3 | 3.1 g/cm^3 | 8.18×10^-4 g/cm^3 (at 25 °C) | | 11.4 g/cm^3 solubility in water | | | miscible | | insoluble surface tension | 0.0728 N/m | | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 1.2571×10^-5 Pa s (at 20 °C) | | odor | odorless | | | |
| water | rhodium(VI) fluoride | hydrogen fluoride | HRhO4 | rhodium dioxide molar mass | 18.015 g/mol | 216.89592 g/mol | 20.006 g/mol | 167.91 g/mol | 134.903 g/mol phase | liquid (at STP) | | gas (at STP) | | melting point | 0 °C | 70 °C | -83.36 °C | | boiling point | 99.9839 °C | | 19.5 °C | | density | 1 g/cm^3 | 3.1 g/cm^3 | 8.18×10^-4 g/cm^3 (at 25 °C) | | 11.4 g/cm^3 solubility in water | | | miscible | | insoluble surface tension | 0.0728 N/m | | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 1.2571×10^-5 Pa s (at 20 °C) | | odor | odorless | | | |

Units

Random Queries

NH3 + (N2H5)HSO4 = (NH4)2SO4 + N2H4
chemical types of 2,5-bis(trifluoromethyl)benzenesulfonyl chloride
net ionic charge of niobium(III) cation vs tantalum(V) cation
H2O + O2 + Pb = Pb(OH)2
ions of barium metasilicate vs lanthanum(III) phosphate hydrate
ion class of hexaamminemanganese(II) cation vs hexaiodoplatinate(IV) anion
DOT numbers of antimony trioxide vs palladium oxide
find morphological branch points image almandine
densities of group 4 elements
PubChem SID number of choline ion vs hexylmagnesium bromide