Na + SiCl4 + C6H5Cl = NaCl + (C6H5)4Si

Na sodium + SiCl_4 silicon tetrachloride + C_6H_5Cl chlorobenzene ⟶ NaCl sodium chloride + (C6H5)4Si

Balance the chemical equation algebraically: Na + SiCl_4 + C_6H_5Cl ⟶ NaCl + (C6H5)4Si Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Na + c_2 SiCl_4 + c_3 C_6H_5Cl ⟶ c_4 NaCl + c_5 (C6H5)4Si Set the number of atoms in the reactants equal to the number of atoms in the products for Na, Cl, Si, C and H: Na: | c_1 = c_4 Cl: | 4 c_2 + c_3 = c_4 Si: | c_2 = c_5 C: | 6 c_3 = 24 c_5 H: | 5 c_3 = 20 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_1 = 8 c_2 = 1 c_3 = 4 c_4 = 8 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 Na + SiCl_4 + 4 C_6H_5Cl ⟶ 8 NaCl + (C6H5)4Si

+ + ⟶ + (C6H5)4Si

sodium + silicon tetrachloride + chlorobenzene ⟶ sodium chloride + (C6H5)4Si

K_c = ([NaCl]^8 [(C6H5)4Si])/([Na]^8 [SiCl4] [C6H5Cl]^4)

rate = -1/8 (Δ[Na])/(Δt) = -(Δ[SiCl4])/(Δt) = -1/4 (Δ[C6H5Cl])/(Δt) = 1/8 (Δ[NaCl])/(Δt) = (Δ[(C6H5)4Si])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium | silicon tetrachloride | chlorobenzene | sodium chloride | (C6H5)4Si formula | Na | SiCl_4 | C_6H_5Cl | NaCl | (C6H5)4Si Hill formula | Na | Cl_4Si | C_6H_5Cl | ClNa | C24H20Si name | sodium | silicon tetrachloride | chlorobenzene | sodium chloride | IUPAC name | sodium | tetrachlorosilane | chlorobenzene | sodium chloride |