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CH4 + ClOH = H2O + CH3Cl

Input interpretation

CH_4 methane + HOCl hypochlorous acid ⟶ H_2O water + CH_3Cl methyl chloride
CH_4 methane + HOCl hypochlorous acid ⟶ H_2O water + CH_3Cl methyl chloride

Balanced equation

Balance the chemical equation algebraically: CH_4 + HOCl ⟶ H_2O + CH_3Cl Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CH_4 + c_2 HOCl ⟶ c_3 H_2O + c_4 CH_3Cl Set the number of atoms in the reactants equal to the number of atoms in the products for C, H, Cl and O: C: | c_1 = c_4 H: | 4 c_1 + c_2 = 2 c_3 + 3 c_4 Cl: | c_2 = c_4 O: | 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: | | CH_4 + HOCl ⟶ H_2O + CH_3Cl
Balance the chemical equation algebraically: CH_4 + HOCl ⟶ H_2O + CH_3Cl Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CH_4 + c_2 HOCl ⟶ c_3 H_2O + c_4 CH_3Cl Set the number of atoms in the reactants equal to the number of atoms in the products for C, H, Cl and O: C: | c_1 = c_4 H: | 4 c_1 + c_2 = 2 c_3 + 3 c_4 Cl: | c_2 = c_4 O: | 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: | | CH_4 + HOCl ⟶ H_2O + CH_3Cl

Structures

+ ⟶ +
+ ⟶ +

Names

methane + hypochlorous acid ⟶ water + methyl chloride
methane + hypochlorous acid ⟶ water + methyl chloride

Equilibrium constant

Construct the equilibrium constant, K, expression for: CH_4 + HOCl ⟶ H_2O + CH_3Cl 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: CH_4 + HOCl ⟶ H_2O + CH_3Cl 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 CH_4 | 1 | -1 HOCl | 1 | -1 H_2O | 1 | 1 CH_3Cl | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CH_4 | 1 | -1 | ([CH4])^(-1) HOCl | 1 | -1 | ([HOCl])^(-1) H_2O | 1 | 1 | [H2O] CH_3Cl | 1 | 1 | [CH3Cl] 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 = ([CH4])^(-1) ([HOCl])^(-1) [H2O] [CH3Cl] = ([H2O] [CH3Cl])/([CH4] [HOCl])
Construct the equilibrium constant, K, expression for: CH_4 + HOCl ⟶ H_2O + CH_3Cl 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: CH_4 + HOCl ⟶ H_2O + CH_3Cl 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 CH_4 | 1 | -1 HOCl | 1 | -1 H_2O | 1 | 1 CH_3Cl | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CH_4 | 1 | -1 | ([CH4])^(-1) HOCl | 1 | -1 | ([HOCl])^(-1) H_2O | 1 | 1 | [H2O] CH_3Cl | 1 | 1 | [CH3Cl] 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 = ([CH4])^(-1) ([HOCl])^(-1) [H2O] [CH3Cl] = ([H2O] [CH3Cl])/([CH4] [HOCl])

Rate of reaction

Construct the rate of reaction expression for: CH_4 + HOCl ⟶ H_2O + CH_3Cl 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: CH_4 + HOCl ⟶ H_2O + CH_3Cl 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 CH_4 | 1 | -1 HOCl | 1 | -1 H_2O | 1 | 1 CH_3Cl | 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 CH_4 | 1 | -1 | -(Δ[CH4])/(Δt) HOCl | 1 | -1 | -(Δ[HOCl])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) CH_3Cl | 1 | 1 | (Δ[CH3Cl])/(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[CH4])/(Δt) = -(Δ[HOCl])/(Δt) = (Δ[H2O])/(Δt) = (Δ[CH3Cl])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: CH_4 + HOCl ⟶ H_2O + CH_3Cl 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: CH_4 + HOCl ⟶ H_2O + CH_3Cl 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 CH_4 | 1 | -1 HOCl | 1 | -1 H_2O | 1 | 1 CH_3Cl | 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 CH_4 | 1 | -1 | -(Δ[CH4])/(Δt) HOCl | 1 | -1 | -(Δ[HOCl])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) CH_3Cl | 1 | 1 | (Δ[CH3Cl])/(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[CH4])/(Δt) = -(Δ[HOCl])/(Δt) = (Δ[H2O])/(Δt) = (Δ[CH3Cl])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| methane | hypochlorous acid | water | methyl chloride formula | CH_4 | HOCl | H_2O | CH_3Cl Hill formula | CH_4 | ClHO | H_2O | CH_3Cl name | methane | hypochlorous acid | water | methyl chloride IUPAC name | methane | hypochlorous acid | water | chloromethane
| methane | hypochlorous acid | water | methyl chloride formula | CH_4 | HOCl | H_2O | CH_3Cl Hill formula | CH_4 | ClHO | H_2O | CH_3Cl name | methane | hypochlorous acid | water | methyl chloride IUPAC name | methane | hypochlorous acid | water | chloromethane

Substance properties

| methane | hypochlorous acid | water | methyl chloride molar mass | 16.04 g/mol | 52.46 g/mol | 18.015 g/mol | 50.48 g/mol phase | gas (at STP) | | liquid (at STP) | gas (at STP) melting point | -182.47 °C | | 0 °C | -97.7 °C boiling point | -161.48 °C | | 99.9839 °C | -24.09 °C density | 6.67151×10^-4 g/cm^3 (at 20 °C) | | 1 g/cm^3 | 0.911 g/cm^3 (at 25 °C) solubility in water | soluble | soluble | | surface tension | 0.0137 N/m | | 0.0728 N/m | 0.0162 N/m dynamic viscosity | 1.114×10^-5 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.834×10^-4 Pa s (at 20 °C) odor | odorless | | odorless | faint | sweet
| methane | hypochlorous acid | water | methyl chloride molar mass | 16.04 g/mol | 52.46 g/mol | 18.015 g/mol | 50.48 g/mol phase | gas (at STP) | | liquid (at STP) | gas (at STP) melting point | -182.47 °C | | 0 °C | -97.7 °C boiling point | -161.48 °C | | 99.9839 °C | -24.09 °C density | 6.67151×10^-4 g/cm^3 (at 20 °C) | | 1 g/cm^3 | 0.911 g/cm^3 (at 25 °C) solubility in water | soluble | soluble | | surface tension | 0.0137 N/m | | 0.0728 N/m | 0.0162 N/m dynamic viscosity | 1.114×10^-5 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.834×10^-4 Pa s (at 20 °C) odor | odorless | | odorless | faint | sweet

Units

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