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H2+O2->H2O

Input interpretation

H_2 hydrogen + O_2 oxygen ⟶ H_2O water
H_2 hydrogen + O_2 oxygen ⟶ H_2O water

Balanced equation

Balance the chemical equation algebraically: H_2 + O_2 ⟶ H_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 O_2 ⟶ c_3 H_2O Set the number of atoms in the reactants equal to the number of atoms in the products for H and O: H: | 2 c_1 = 2 c_3 O: | 2 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 2 H_2 + O_2 ⟶ 2 H_2O
Balance the chemical equation algebraically: H_2 + O_2 ⟶ H_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 O_2 ⟶ c_3 H_2O Set the number of atoms in the reactants equal to the number of atoms in the products for H and O: H: | 2 c_1 = 2 c_3 O: | 2 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2 + O_2 ⟶ 2 H_2O

Structures

 + ⟶
+ ⟶

Names

hydrogen + oxygen ⟶ water
hydrogen + oxygen ⟶ water

Reaction thermodynamics

Enthalpy

 | hydrogen | oxygen | water molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -285.8 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -571.7 kJ/mol  | H_initial = 0 kJ/mol | | H_final = -571.7 kJ/mol ΔH_rxn^0 | -571.7 kJ/mol - 0 kJ/mol = -571.7 kJ/mol (exothermic) | |
| hydrogen | oxygen | water molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -285.8 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -571.7 kJ/mol | H_initial = 0 kJ/mol | | H_final = -571.7 kJ/mol ΔH_rxn^0 | -571.7 kJ/mol - 0 kJ/mol = -571.7 kJ/mol (exothermic) | |

Gibbs free energy

 | hydrogen | oxygen | water molecular free energy | 0 kJ/mol | 231.7 kJ/mol | -237.1 kJ/mol total free energy | 0 kJ/mol | 231.7 kJ/mol | -474.2 kJ/mol  | G_initial = 231.7 kJ/mol | | G_final = -474.2 kJ/mol ΔG_rxn^0 | -474.2 kJ/mol - 231.7 kJ/mol = -705.9 kJ/mol (exergonic) | |
| hydrogen | oxygen | water molecular free energy | 0 kJ/mol | 231.7 kJ/mol | -237.1 kJ/mol total free energy | 0 kJ/mol | 231.7 kJ/mol | -474.2 kJ/mol | G_initial = 231.7 kJ/mol | | G_final = -474.2 kJ/mol ΔG_rxn^0 | -474.2 kJ/mol - 231.7 kJ/mol = -705.9 kJ/mol (exergonic) | |

Entropy

 | hydrogen | oxygen | water molecular entropy | 115 J/(mol K) | 205 J/(mol K) | 69.91 J/(mol K) total entropy | 230 J/(mol K) | 205 J/(mol K) | 139.8 J/(mol K)  | S_initial = 435 J/(mol K) | | S_final = 139.8 J/(mol K) ΔS_rxn^0 | 139.8 J/(mol K) - 435 J/(mol K) = -295.2 J/(mol K) (exoentropic) | |
| hydrogen | oxygen | water molecular entropy | 115 J/(mol K) | 205 J/(mol K) | 69.91 J/(mol K) total entropy | 230 J/(mol K) | 205 J/(mol K) | 139.8 J/(mol K) | S_initial = 435 J/(mol K) | | S_final = 139.8 J/(mol K) ΔS_rxn^0 | 139.8 J/(mol K) - 435 J/(mol K) = -295.2 J/(mol K) (exoentropic) | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2 + O_2 ⟶ H_2O 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: 2 H_2 + O_2 ⟶ 2 H_2O 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_2 | 2 | -2 O_2 | 1 | -1 H_2O | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2 | 2 | -2 | ([H2])^(-2) O_2 | 1 | -1 | ([O2])^(-1) H_2O | 2 | 2 | ([H2O])^2 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 = ([H2])^(-2) ([O2])^(-1) ([H2O])^2 = ([H2O])^2/(([H2])^2 [O2])
Construct the equilibrium constant, K, expression for: H_2 + O_2 ⟶ H_2O 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: 2 H_2 + O_2 ⟶ 2 H_2O 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_2 | 2 | -2 O_2 | 1 | -1 H_2O | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2 | 2 | -2 | ([H2])^(-2) O_2 | 1 | -1 | ([O2])^(-1) H_2O | 2 | 2 | ([H2O])^2 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 = ([H2])^(-2) ([O2])^(-1) ([H2O])^2 = ([H2O])^2/(([H2])^2 [O2])

Rate of reaction

Construct the rate of reaction expression for: H_2 + O_2 ⟶ H_2O 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: 2 H_2 + O_2 ⟶ 2 H_2O 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_2 | 2 | -2 O_2 | 1 | -1 H_2O | 2 | 2 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_2 | 2 | -2 | -1/2 (Δ[H2])/(Δt) O_2 | 1 | -1 | -(Δ[O2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(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 = -1/2 (Δ[H2])/(Δt) = -(Δ[O2])/(Δt) = 1/2 (Δ[H2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2 + O_2 ⟶ H_2O 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: 2 H_2 + O_2 ⟶ 2 H_2O 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_2 | 2 | -2 O_2 | 1 | -1 H_2O | 2 | 2 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_2 | 2 | -2 | -1/2 (Δ[H2])/(Δt) O_2 | 1 | -1 | -(Δ[O2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(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 = -1/2 (Δ[H2])/(Δt) = -(Δ[O2])/(Δt) = 1/2 (Δ[H2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | hydrogen | oxygen | water formula | H_2 | O_2 | H_2O name | hydrogen | oxygen | water IUPAC name | molecular hydrogen | molecular oxygen | water
| hydrogen | oxygen | water formula | H_2 | O_2 | H_2O name | hydrogen | oxygen | water IUPAC name | molecular hydrogen | molecular oxygen | water

Substance properties

 | hydrogen | oxygen | water molar mass | 2.016 g/mol | 31.998 g/mol | 18.015 g/mol phase | gas (at STP) | gas (at STP) | liquid (at STP) melting point | -259.2 °C | -218 °C | 0 °C boiling point | -252.8 °C | -183 °C | 99.9839 °C density | 8.99×10^-5 g/cm^3 (at 0 °C) | 0.001429 g/cm^3 (at 0 °C) | 1 g/cm^3 surface tension | | 0.01347 N/m | 0.0728 N/m dynamic viscosity | 8.9×10^-6 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) odor | odorless | odorless | odorless
| hydrogen | oxygen | water molar mass | 2.016 g/mol | 31.998 g/mol | 18.015 g/mol phase | gas (at STP) | gas (at STP) | liquid (at STP) melting point | -259.2 °C | -218 °C | 0 °C boiling point | -252.8 °C | -183 °C | 99.9839 °C density | 8.99×10^-5 g/cm^3 (at 0 °C) | 0.001429 g/cm^3 (at 0 °C) | 1 g/cm^3 surface tension | | 0.01347 N/m | 0.0728 N/m dynamic viscosity | 8.9×10^-6 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) odor | odorless | odorless | odorless

Units