Input interpretation
![KI potassium iodide + CH_3CO_2H acetic acid + KNO_2 potassium nitrite ⟶ H_2O water + I_2 iodine + NO nitric oxide + CH_3COOK potassium acetate](../image_source/4c6e5cd466518706080efc3634138ef1.png)
KI potassium iodide + CH_3CO_2H acetic acid + KNO_2 potassium nitrite ⟶ H_2O water + I_2 iodine + NO nitric oxide + CH_3COOK potassium acetate
Balanced equation
![Balance the chemical equation algebraically: KI + CH_3CO_2H + KNO_2 ⟶ H_2O + I_2 + NO + CH_3COOK Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KI + c_2 CH_3CO_2H + c_3 KNO_2 ⟶ c_4 H_2O + c_5 I_2 + c_6 NO + c_7 CH_3COOK Set the number of atoms in the reactants equal to the number of atoms in the products for I, K, C, H, O and N: I: | c_1 = 2 c_5 K: | c_1 + c_3 = c_7 C: | 2 c_2 = 2 c_7 H: | 4 c_2 = 2 c_4 + 3 c_7 O: | 2 c_2 + 2 c_3 = c_4 + c_6 + 2 c_7 N: | c_3 = c_6 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 = 2 c_2 = 4 c_3 = 2 c_4 = 2 c_5 = 1 c_6 = 2 c_7 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KI + 4 CH_3CO_2H + 2 KNO_2 ⟶ 2 H_2O + I_2 + 2 NO + 4 CH_3COOK](../image_source/c234fcd5af87fc9dc3655e3f82d36105.png)
Balance the chemical equation algebraically: KI + CH_3CO_2H + KNO_2 ⟶ H_2O + I_2 + NO + CH_3COOK Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KI + c_2 CH_3CO_2H + c_3 KNO_2 ⟶ c_4 H_2O + c_5 I_2 + c_6 NO + c_7 CH_3COOK Set the number of atoms in the reactants equal to the number of atoms in the products for I, K, C, H, O and N: I: | c_1 = 2 c_5 K: | c_1 + c_3 = c_7 C: | 2 c_2 = 2 c_7 H: | 4 c_2 = 2 c_4 + 3 c_7 O: | 2 c_2 + 2 c_3 = c_4 + c_6 + 2 c_7 N: | c_3 = c_6 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 = 2 c_2 = 4 c_3 = 2 c_4 = 2 c_5 = 1 c_6 = 2 c_7 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KI + 4 CH_3CO_2H + 2 KNO_2 ⟶ 2 H_2O + I_2 + 2 NO + 4 CH_3COOK
Structures
![+ + ⟶ + + +](../image_source/7b5ccead46105908c9d26e4f96825690.png)
+ + ⟶ + + +
Names
![potassium iodide + acetic acid + potassium nitrite ⟶ water + iodine + nitric oxide + potassium acetate](../image_source/52cb12fb90cd360ac4ac1b46af7dcdd8.png)
potassium iodide + acetic acid + potassium nitrite ⟶ water + iodine + nitric oxide + potassium acetate
Equilibrium constant
![Construct the equilibrium constant, K, expression for: KI + CH_3CO_2H + KNO_2 ⟶ H_2O + I_2 + NO + CH_3COOK 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 KI + 4 CH_3CO_2H + 2 KNO_2 ⟶ 2 H_2O + I_2 + 2 NO + 4 CH_3COOK 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 KI | 2 | -2 CH_3CO_2H | 4 | -4 KNO_2 | 2 | -2 H_2O | 2 | 2 I_2 | 1 | 1 NO | 2 | 2 CH_3COOK | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KI | 2 | -2 | ([KI])^(-2) CH_3CO_2H | 4 | -4 | ([CH3CO2H])^(-4) KNO_2 | 2 | -2 | ([KNO2])^(-2) H_2O | 2 | 2 | ([H2O])^2 I_2 | 1 | 1 | [I2] NO | 2 | 2 | ([NO])^2 CH_3COOK | 4 | 4 | ([CH3COOK])^4 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 = ([KI])^(-2) ([CH3CO2H])^(-4) ([KNO2])^(-2) ([H2O])^2 [I2] ([NO])^2 ([CH3COOK])^4 = (([H2O])^2 [I2] ([NO])^2 ([CH3COOK])^4)/(([KI])^2 ([CH3CO2H])^4 ([KNO2])^2)](../image_source/2cf01365d028bd76117c40e862476154.png)
Construct the equilibrium constant, K, expression for: KI + CH_3CO_2H + KNO_2 ⟶ H_2O + I_2 + NO + CH_3COOK 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 KI + 4 CH_3CO_2H + 2 KNO_2 ⟶ 2 H_2O + I_2 + 2 NO + 4 CH_3COOK 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 KI | 2 | -2 CH_3CO_2H | 4 | -4 KNO_2 | 2 | -2 H_2O | 2 | 2 I_2 | 1 | 1 NO | 2 | 2 CH_3COOK | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KI | 2 | -2 | ([KI])^(-2) CH_3CO_2H | 4 | -4 | ([CH3CO2H])^(-4) KNO_2 | 2 | -2 | ([KNO2])^(-2) H_2O | 2 | 2 | ([H2O])^2 I_2 | 1 | 1 | [I2] NO | 2 | 2 | ([NO])^2 CH_3COOK | 4 | 4 | ([CH3COOK])^4 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 = ([KI])^(-2) ([CH3CO2H])^(-4) ([KNO2])^(-2) ([H2O])^2 [I2] ([NO])^2 ([CH3COOK])^4 = (([H2O])^2 [I2] ([NO])^2 ([CH3COOK])^4)/(([KI])^2 ([CH3CO2H])^4 ([KNO2])^2)
Rate of reaction
![Construct the rate of reaction expression for: KI + CH_3CO_2H + KNO_2 ⟶ H_2O + I_2 + NO + CH_3COOK 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 KI + 4 CH_3CO_2H + 2 KNO_2 ⟶ 2 H_2O + I_2 + 2 NO + 4 CH_3COOK 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 KI | 2 | -2 CH_3CO_2H | 4 | -4 KNO_2 | 2 | -2 H_2O | 2 | 2 I_2 | 1 | 1 NO | 2 | 2 CH_3COOK | 4 | 4 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 KI | 2 | -2 | -1/2 (Δ[KI])/(Δt) CH_3CO_2H | 4 | -4 | -1/4 (Δ[CH3CO2H])/(Δt) KNO_2 | 2 | -2 | -1/2 (Δ[KNO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) I_2 | 1 | 1 | (Δ[I2])/(Δt) NO | 2 | 2 | 1/2 (Δ[NO])/(Δt) CH_3COOK | 4 | 4 | 1/4 (Δ[CH3COOK])/(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[KI])/(Δt) = -1/4 (Δ[CH3CO2H])/(Δt) = -1/2 (Δ[KNO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[I2])/(Δt) = 1/2 (Δ[NO])/(Δt) = 1/4 (Δ[CH3COOK])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/798986c2da077a7aae78c8e87b7a8b4a.png)
Construct the rate of reaction expression for: KI + CH_3CO_2H + KNO_2 ⟶ H_2O + I_2 + NO + CH_3COOK 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 KI + 4 CH_3CO_2H + 2 KNO_2 ⟶ 2 H_2O + I_2 + 2 NO + 4 CH_3COOK 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 KI | 2 | -2 CH_3CO_2H | 4 | -4 KNO_2 | 2 | -2 H_2O | 2 | 2 I_2 | 1 | 1 NO | 2 | 2 CH_3COOK | 4 | 4 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 KI | 2 | -2 | -1/2 (Δ[KI])/(Δt) CH_3CO_2H | 4 | -4 | -1/4 (Δ[CH3CO2H])/(Δt) KNO_2 | 2 | -2 | -1/2 (Δ[KNO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) I_2 | 1 | 1 | (Δ[I2])/(Δt) NO | 2 | 2 | 1/2 (Δ[NO])/(Δt) CH_3COOK | 4 | 4 | 1/4 (Δ[CH3COOK])/(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[KI])/(Δt) = -1/4 (Δ[CH3CO2H])/(Δt) = -1/2 (Δ[KNO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[I2])/(Δt) = 1/2 (Δ[NO])/(Δt) = 1/4 (Δ[CH3COOK])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| potassium iodide | acetic acid | potassium nitrite | water | iodine | nitric oxide | potassium acetate formula | KI | CH_3CO_2H | KNO_2 | H_2O | I_2 | NO | CH_3COOK Hill formula | IK | C_2H_4O_2 | KNO_2 | H_2O | I_2 | NO | C_2H_3KO_2 name | potassium iodide | acetic acid | potassium nitrite | water | iodine | nitric oxide | potassium acetate IUPAC name | potassium iodide | acetic acid | potassium nitrite | water | molecular iodine | nitric oxide | potassium acetate](../image_source/4332c5b87eb0c5c41057b57c71f811e5.png)
| potassium iodide | acetic acid | potassium nitrite | water | iodine | nitric oxide | potassium acetate formula | KI | CH_3CO_2H | KNO_2 | H_2O | I_2 | NO | CH_3COOK Hill formula | IK | C_2H_4O_2 | KNO_2 | H_2O | I_2 | NO | C_2H_3KO_2 name | potassium iodide | acetic acid | potassium nitrite | water | iodine | nitric oxide | potassium acetate IUPAC name | potassium iodide | acetic acid | potassium nitrite | water | molecular iodine | nitric oxide | potassium acetate