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Si + HF = H2 + SiF4

Input interpretation

Si silicon + HF hydrogen fluoride ⟶ H_2 hydrogen + SiF_4 silicon tetrafluoride
Si silicon + HF hydrogen fluoride ⟶ H_2 hydrogen + SiF_4 silicon tetrafluoride

Balanced equation

Balance the chemical equation algebraically: Si + HF ⟶ H_2 + SiF_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Si + c_2 HF ⟶ c_3 H_2 + c_4 SiF_4 Set the number of atoms in the reactants equal to the number of atoms in the products for Si, F and H: Si: | c_1 = c_4 F: | c_2 = 4 c_4 H: | c_2 = 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 = 4 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | Si + 4 HF ⟶ 2 H_2 + SiF_4
Balance the chemical equation algebraically: Si + HF ⟶ H_2 + SiF_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Si + c_2 HF ⟶ c_3 H_2 + c_4 SiF_4 Set the number of atoms in the reactants equal to the number of atoms in the products for Si, F and H: Si: | c_1 = c_4 F: | c_2 = 4 c_4 H: | c_2 = 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 = 4 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Si + 4 HF ⟶ 2 H_2 + SiF_4

Structures

 + ⟶ +
+ ⟶ +

Names

silicon + hydrogen fluoride ⟶ hydrogen + silicon tetrafluoride
silicon + hydrogen fluoride ⟶ hydrogen + silicon tetrafluoride

Equilibrium constant

Construct the equilibrium constant, K, expression for: Si + HF ⟶ H_2 + SiF_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: Si + 4 HF ⟶ 2 H_2 + SiF_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 Si | 1 | -1 HF | 4 | -4 H_2 | 2 | 2 SiF_4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Si | 1 | -1 | ([Si])^(-1) HF | 4 | -4 | ([HF])^(-4) H_2 | 2 | 2 | ([H2])^2 SiF_4 | 1 | 1 | [SiF4] 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 = ([Si])^(-1) ([HF])^(-4) ([H2])^2 [SiF4] = (([H2])^2 [SiF4])/([Si] ([HF])^4)
Construct the equilibrium constant, K, expression for: Si + HF ⟶ H_2 + SiF_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: Si + 4 HF ⟶ 2 H_2 + SiF_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 Si | 1 | -1 HF | 4 | -4 H_2 | 2 | 2 SiF_4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Si | 1 | -1 | ([Si])^(-1) HF | 4 | -4 | ([HF])^(-4) H_2 | 2 | 2 | ([H2])^2 SiF_4 | 1 | 1 | [SiF4] 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 = ([Si])^(-1) ([HF])^(-4) ([H2])^2 [SiF4] = (([H2])^2 [SiF4])/([Si] ([HF])^4)

Rate of reaction

Construct the rate of reaction expression for: Si + HF ⟶ H_2 + SiF_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: Si + 4 HF ⟶ 2 H_2 + SiF_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 Si | 1 | -1 HF | 4 | -4 H_2 | 2 | 2 SiF_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 Si | 1 | -1 | -(Δ[Si])/(Δt) HF | 4 | -4 | -1/4 (Δ[HF])/(Δt) H_2 | 2 | 2 | 1/2 (Δ[H2])/(Δt) SiF_4 | 1 | 1 | (Δ[SiF4])/(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[Si])/(Δt) = -1/4 (Δ[HF])/(Δt) = 1/2 (Δ[H2])/(Δt) = (Δ[SiF4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: Si + HF ⟶ H_2 + SiF_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: Si + 4 HF ⟶ 2 H_2 + SiF_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 Si | 1 | -1 HF | 4 | -4 H_2 | 2 | 2 SiF_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 Si | 1 | -1 | -(Δ[Si])/(Δt) HF | 4 | -4 | -1/4 (Δ[HF])/(Δt) H_2 | 2 | 2 | 1/2 (Δ[H2])/(Δt) SiF_4 | 1 | 1 | (Δ[SiF4])/(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[Si])/(Δt) = -1/4 (Δ[HF])/(Δt) = 1/2 (Δ[H2])/(Δt) = (Δ[SiF4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | silicon | hydrogen fluoride | hydrogen | silicon tetrafluoride formula | Si | HF | H_2 | SiF_4 Hill formula | Si | FH | H_2 | F_4Si name | silicon | hydrogen fluoride | hydrogen | silicon tetrafluoride IUPAC name | silicon | hydrogen fluoride | molecular hydrogen | tetrafluorosilane
| silicon | hydrogen fluoride | hydrogen | silicon tetrafluoride formula | Si | HF | H_2 | SiF_4 Hill formula | Si | FH | H_2 | F_4Si name | silicon | hydrogen fluoride | hydrogen | silicon tetrafluoride IUPAC name | silicon | hydrogen fluoride | molecular hydrogen | tetrafluorosilane

Substance properties

 | silicon | hydrogen fluoride | hydrogen | silicon tetrafluoride molar mass | 28.085 g/mol | 20.006 g/mol | 2.016 g/mol | 104.079 g/mol phase | solid (at STP) | gas (at STP) | gas (at STP) | gas (at STP) melting point | 1410 °C | -83.36 °C | -259.2 °C | -90.2 °C boiling point | 2355 °C | 19.5 °C | -252.8 °C | -86 °C density | 2.33 g/cm^3 | 8.18×10^-4 g/cm^3 (at 25 °C) | 8.99×10^-5 g/cm^3 (at 0 °C) | 0.004254 g/cm^3 (at 25 °C) solubility in water | insoluble | miscible | | decomposes dynamic viscosity | | 1.2571×10^-5 Pa s (at 20 °C) | 8.9×10^-6 Pa s (at 25 °C) |  odor | | | odorless |
| silicon | hydrogen fluoride | hydrogen | silicon tetrafluoride molar mass | 28.085 g/mol | 20.006 g/mol | 2.016 g/mol | 104.079 g/mol phase | solid (at STP) | gas (at STP) | gas (at STP) | gas (at STP) melting point | 1410 °C | -83.36 °C | -259.2 °C | -90.2 °C boiling point | 2355 °C | 19.5 °C | -252.8 °C | -86 °C density | 2.33 g/cm^3 | 8.18×10^-4 g/cm^3 (at 25 °C) | 8.99×10^-5 g/cm^3 (at 0 °C) | 0.004254 g/cm^3 (at 25 °C) solubility in water | insoluble | miscible | | decomposes dynamic viscosity | | 1.2571×10^-5 Pa s (at 20 °C) | 8.9×10^-6 Pa s (at 25 °C) | odor | | | odorless |

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