Search

H2O2 + KCrO2 = H2O + O2 + KCrO4

Input interpretation

H_2O_2 hydrogen peroxide + KCrO2 ⟶ H_2O water + O_2 oxygen + KCrO4
H_2O_2 hydrogen peroxide + KCrO2 ⟶ H_2O water + O_2 oxygen + KCrO4

Balanced equation

Balance the chemical equation algebraically: H_2O_2 + KCrO2 ⟶ H_2O + O_2 + KCrO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O_2 + c_2 KCrO2 ⟶ c_3 H_2O + c_4 O_2 + c_5 KCrO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, K and Cr: H: | 2 c_1 = 2 c_3 O: | 2 c_1 + 2 c_2 = c_3 + 2 c_4 + 4 c_5 K: | c_2 = c_5 Cr: | c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_2 = 1 c_3 = c_1 c_4 = c_1/2 - 1 c_5 = 1 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 4 and solve for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 4 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2O_2 + KCrO2 ⟶ 4 H_2O + O_2 + KCrO4
Balance the chemical equation algebraically: H_2O_2 + KCrO2 ⟶ H_2O + O_2 + KCrO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O_2 + c_2 KCrO2 ⟶ c_3 H_2O + c_4 O_2 + c_5 KCrO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, K and Cr: H: | 2 c_1 = 2 c_3 O: | 2 c_1 + 2 c_2 = c_3 + 2 c_4 + 4 c_5 K: | c_2 = c_5 Cr: | c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_2 = 1 c_3 = c_1 c_4 = c_1/2 - 1 c_5 = 1 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 4 and solve for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 4 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2O_2 + KCrO2 ⟶ 4 H_2O + O_2 + KCrO4

Structures

+ KCrO2 ⟶ + + KCrO4
+ KCrO2 ⟶ + + KCrO4

Names

hydrogen peroxide + KCrO2 ⟶ water + oxygen + KCrO4
hydrogen peroxide + KCrO2 ⟶ water + oxygen + KCrO4

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O_2 + KCrO2 ⟶ H_2O + O_2 + KCrO4 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: 4 H_2O_2 + KCrO2 ⟶ 4 H_2O + O_2 + KCrO4 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_2 | 4 | -4 KCrO2 | 1 | -1 H_2O | 4 | 4 O_2 | 1 | 1 KCrO4 | 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_2 | 4 | -4 | ([H2O2])^(-4) KCrO2 | 1 | -1 | ([KCrO2])^(-1) H_2O | 4 | 4 | ([H2O])^4 O_2 | 1 | 1 | [O2] KCrO4 | 1 | 1 | [KCrO4] 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 = ([H2O2])^(-4) ([KCrO2])^(-1) ([H2O])^4 [O2] [KCrO4] = (([H2O])^4 [O2] [KCrO4])/(([H2O2])^4 [KCrO2])
Construct the equilibrium constant, K, expression for: H_2O_2 + KCrO2 ⟶ H_2O + O_2 + KCrO4 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: 4 H_2O_2 + KCrO2 ⟶ 4 H_2O + O_2 + KCrO4 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_2 | 4 | -4 KCrO2 | 1 | -1 H_2O | 4 | 4 O_2 | 1 | 1 KCrO4 | 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_2 | 4 | -4 | ([H2O2])^(-4) KCrO2 | 1 | -1 | ([KCrO2])^(-1) H_2O | 4 | 4 | ([H2O])^4 O_2 | 1 | 1 | [O2] KCrO4 | 1 | 1 | [KCrO4] 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 = ([H2O2])^(-4) ([KCrO2])^(-1) ([H2O])^4 [O2] [KCrO4] = (([H2O])^4 [O2] [KCrO4])/(([H2O2])^4 [KCrO2])

Rate of reaction

Construct the rate of reaction expression for: H_2O_2 + KCrO2 ⟶ H_2O + O_2 + KCrO4 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: 4 H_2O_2 + KCrO2 ⟶ 4 H_2O + O_2 + KCrO4 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_2 | 4 | -4 KCrO2 | 1 | -1 H_2O | 4 | 4 O_2 | 1 | 1 KCrO4 | 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_2 | 4 | -4 | -1/4 (Δ[H2O2])/(Δt) KCrO2 | 1 | -1 | -(Δ[KCrO2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) O_2 | 1 | 1 | (Δ[O2])/(Δt) KCrO4 | 1 | 1 | (Δ[KCrO4])/(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 = -1/4 (Δ[H2O2])/(Δt) = -(Δ[KCrO2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = (Δ[O2])/(Δt) = (Δ[KCrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O_2 + KCrO2 ⟶ H_2O + O_2 + KCrO4 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: 4 H_2O_2 + KCrO2 ⟶ 4 H_2O + O_2 + KCrO4 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_2 | 4 | -4 KCrO2 | 1 | -1 H_2O | 4 | 4 O_2 | 1 | 1 KCrO4 | 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_2 | 4 | -4 | -1/4 (Δ[H2O2])/(Δt) KCrO2 | 1 | -1 | -(Δ[KCrO2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) O_2 | 1 | 1 | (Δ[O2])/(Δt) KCrO4 | 1 | 1 | (Δ[KCrO4])/(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 = -1/4 (Δ[H2O2])/(Δt) = -(Δ[KCrO2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = (Δ[O2])/(Δt) = (Δ[KCrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| hydrogen peroxide | KCrO2 | water | oxygen | KCrO4 formula | H_2O_2 | KCrO2 | H_2O | O_2 | KCrO4 Hill formula | H_2O_2 | CrKO2 | H_2O | O_2 | CrKO4 name | hydrogen peroxide | | water | oxygen | IUPAC name | hydrogen peroxide | | water | molecular oxygen |
| hydrogen peroxide | KCrO2 | water | oxygen | KCrO4 formula | H_2O_2 | KCrO2 | H_2O | O_2 | KCrO4 Hill formula | H_2O_2 | CrKO2 | H_2O | O_2 | CrKO4 name | hydrogen peroxide | | water | oxygen | IUPAC name | hydrogen peroxide | | water | molecular oxygen |

Substance properties

| hydrogen peroxide | KCrO2 | water | oxygen | KCrO4 molar mass | 34.014 g/mol | 123.09 g/mol | 18.015 g/mol | 31.998 g/mol | 155.09 g/mol phase | liquid (at STP) | | liquid (at STP) | gas (at STP) | melting point | -0.43 °C | | 0 °C | -218 °C | boiling point | 150.2 °C | | 99.9839 °C | -183 °C | density | 1.44 g/cm^3 | | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | solubility in water | miscible | | | | surface tension | 0.0804 N/m | | 0.0728 N/m | 0.01347 N/m | dynamic viscosity | 0.001249 Pa s (at 20 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | odor | | | odorless | odorless |
| hydrogen peroxide | KCrO2 | water | oxygen | KCrO4 molar mass | 34.014 g/mol | 123.09 g/mol | 18.015 g/mol | 31.998 g/mol | 155.09 g/mol phase | liquid (at STP) | | liquid (at STP) | gas (at STP) | melting point | -0.43 °C | | 0 °C | -218 °C | boiling point | 150.2 °C | | 99.9839 °C | -183 °C | density | 1.44 g/cm^3 | | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | solubility in water | miscible | | | | surface tension | 0.0804 N/m | | 0.0728 N/m | 0.01347 N/m | dynamic viscosity | 0.001249 Pa s (at 20 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | odor | | | odorless | odorless |

Units

Random Queries

density of octane
ion class of gallium(I) cation vs thallium(II) cation
formula of 2-ethylhexanal vs 1,3-dibromo-5-(trifluoromethoxy)benzene
formula of dioxygenyl cation vs oxoantimony(III) cation
half-life of Md-250 vs Bk-246
melting point of silver nitrate vs potassium hexacyanomanganate(III)
chlorine trifluoride
DOT numbers of thorium(IV) iodide
name of 2, 6-dimethyl-5-heptenal
C3H6 = H2 + C