The standard enthalpy of reaction for the dissolution of silica in aqueous hf is 4.6 kj mol–1 . what is the standard enthalpy of formation of sif4(g)?

sio2(s) + 4 hf(aq) → sif4(g) + 2 h2o(l) ∆hºrxn = 4.6 kj mol–1

compound sio2(s) hf(aq) h2o(l) sif4(g)
∆hºf, kj mol–1 –910.9 –320.1 –285.8

(a) –1624.3 kj mol^–1
(b) –1615.1 kj mol^–1
(c) –949.8 kj mol^–1
(d) –940.6 kj mol^–1

B) ) –1615.1 kJ mol^–1

Explanation:

since

SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l) ∆Hºrxn = 4.6 kJ mol–1

the enhalpy of reaction will be

∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr

where ∆Hºrxn= enthalpy of reaction , ∆Hºfp= standard enthalpy of formation of products , ∆Hºfr = standard enthalpy of formation of reactants , νp=stoichiometric coffficient of products, νr=stoichiometric coffficient of reactants

therefore

∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr

4.6 kJ/mol = [1*∆HºfX + 2*(–285.8 kJ/mol)] - [1*(–910.9kJ/mol) + 4*(–320.1 kJ/mol)]

4.6 kJ/mol =∆HºfX -571.6 kJ/mol + 2191.3 kJ/mol

∆HºfX = 4.6 kJ/mol + 571.6 kJ/mol - 2191.3 kJ/mol = -1615.1 kJ/mol

therefore ∆HºfX (unknown standard enthalpy of formation = standard enthalpy of formation of SiF4(g) ) = -1615.1 kJ/mol

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )