PH3 + HClO3 = HCl + H3PO4

Input interpretation

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PH_3 (phosphine) + HClO3 ⟶ HCl (hydrogen chloride) + H_3PO_4 (phosphoric acid)

Balanced equation

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$Balance the chemical equation algebraically: PH_3 + HClO3 ⟶ HCl + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 PH_3 + c_2 HClO3 ⟶ c_3 HCl + c_4 H_3PO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, P, Cl and O: H: | 3 c_1 + c_2 = c_3 + 3 c_4 P: | c_1 = c_4 Cl: | c_2 = c_3 O: | 3 c_2 = 4 c_4 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 4/3 c_3 = 4/3 c_4 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 3 c_2 = 4 c_3 = 4 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 PH_3 + 4 HClO3 ⟶ 4 HCl + 3 H_3PO_4$

Balance the chemical equation algebraically: PH_3 + HClO3 ⟶ HCl + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 PH_3 + c_2 HClO3 ⟶ c_3 HCl + c_4 H_3PO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, P, Cl and O: H: | 3 c_1 + c_2 = c_3 + 3 c_4 P: | c_1 = c_4 Cl: | c_2 = c_3 O: | 3 c_2 = 4 c_4 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 = 4/3 c_3 = 4/3 c_4 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 3 c_2 = 4 c_3 = 4 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 PH_3 + 4 HClO3 ⟶ 4 HCl + 3 H_3PO_4

+ HClO3 ⟶ +

Names

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phosphine + HClO3 ⟶ hydrogen chloride + phosphoric acid

Equilibrium constant

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Construct the equilibrium constant, K, expression for: PH_3 + HClO3 ⟶ HCl + H_3PO_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: 3 PH_3 + 4 HClO3 ⟶ 4 HCl + 3 H_3PO_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 PH_3 | 3 | -3 HClO3 | 4 | -4 HCl | 4 | 4 H_3PO_4 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression PH_3 | 3 | -3 | ([PH3])^(-3) HClO3 | 4 | -4 | ([HClO3])^(-4) HCl | 4 | 4 | ([HCl])^4 H_3PO_4 | 3 | 3 | ([H3PO4])^3 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 = ([PH3])^(-3) ([HClO3])^(-4) ([HCl])^4 ([H3PO4])^3 = (([HCl])^4 ([H3PO4])^3)/(([PH3])^3 ([HClO3])^4)

Rate of reaction

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Construct the rate of reaction expression for: PH_3 + HClO3 ⟶ HCl + H_3PO_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: 3 PH_3 + 4 HClO3 ⟶ 4 HCl + 3 H_3PO_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 PH_3 | 3 | -3 HClO3 | 4 | -4 HCl | 4 | 4 H_3PO_4 | 3 | 3 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 PH_3 | 3 | -3 | -1/3 (Δ[PH3])/(Δt) HClO3 | 4 | -4 | -1/4 (Δ[HClO3])/(Δt) HCl | 4 | 4 | 1/4 (Δ[HCl])/(Δt) H_3PO_4 | 3 | 3 | 1/3 (Δ[H3PO4])/(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 = -1/3 (Δ[PH3])/(Δt) = -1/4 (Δ[HClO3])/(Δt) = 1/4 (Δ[HCl])/(Δt) = 1/3 (Δ[H3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)