A key reaction in the production of optical fibers is the reaction: SiCI_4(I) +

Question

A key reaction in the production of optical fibers is the reaction: SiCI_4(I) + 2 H_2O(I) rightarrow SiO_2(s) + 4 HCi(aq) The following standard enthalpies of formation are found in data tables: DeltaH^0_f[SiCI_4(1)] = -687 kJ; DeltaH^0_f[H_2O(I)] = -286 kJ; DeltaH^0_f[SiO_2(s)] = -911 kJ DeltaH^0_f[HCI(g)] = -92 kJ; DeltaH^0_f[H^+ (aq)] = 0 kJ; DeltaH^0_f[CI^-(aq)] = -167 kJ The standard enthalpy of reaction for reaction (1) above is: a. -320 kJ b. -105 kJ c. -30 kJ d. -20 kJ e. +20 kJ f.+105 kJ g. + 320 kJ h. +348 kJ i. None of the above; the correct enthalpy of reaction is kJ

Explanation / Answer

consider the given reaction

SiCl4 + 2H20 ---> SiO2 + 4 HCl

now

HCl is a strong acid

so

it undergoes 100 % dissociation

HCl (aq) --> H+ (aq) + Cl- (aq)

so

the reaction is

SiCl4 + 2H20 ---> SiO2 + 4H+ + 4Cl-

now

we know that

dHrxn = dHfo products - dHfo reactants

so

dHrxn = (dHfo SiO2 ) + 4 x ( dHfo H+ + dHfo Cl- ) - ( dHfo SiCl4) - (2 x dHfo H20)

using given values

we get

dHrxn = ( -911) + 4 x ( 0 -167) - ( -687) - ( 2 x -286)

dHrxn = -320 kJ

so

the answer is a) -320 kJ

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