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Cramer's Rule

if S + O2—-> ...

Question

if S + O2----> SO2 ; ( DH= -298.2 KJ ) SO2 +1/2O2------>SO3 ( DH=-98.7 KJ) SO3 + H2O------>H2SO4 (DH=-130.2 KJ) H2 + 1/2H2O-----> H2O ( DH=-287.3KJ) then the enthalpy of formarion of H2SO4 at 298K is 1)-814 KJ 2))-650.3 KJ 3)-320.5 KJ 4)'433.5 KJ

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Similar questions

S + O_{2} \to S O_{2}

Δ H = - 298.2

S O_{2} + \frac{1}{2} O_{2} \to S O_{3}

Δ H = - 98.7

S O_{3} + H_{2} O \to H_{2} S O_{4}

Δ H = - 130.2

H_{2} + \frac{1}{2} O_{2} \to H_{2} O

Δ H = - 287.3

Then the enthalpy of formation of

H_{2} S O_{4}

S + O_{2} ⟶ S O_{2}; Δ H = - 298.2 k J

S O_{2} + 1 / 2 O_{2} ⟶ S O_{3}; Δ H = - 98.7 k J

S O_{3} + H_{2} O ⟶ H_{2} S O_{4}; Δ H = - 130.2 k J

H_{2} + 1 / 2 O_{2} ⟶ H_{2} O; Δ H = - 287.3 k J

Then the enthalpy of formation of

H_{2} S O_{4} a t 298 K

S + O_{2} \to S O_{2}, △ H = - 298.2 k J m o l e^{- 1}

S O_{2} + 1 / 2 O_{2} \to S O_{3}, △ H = - 98.7 k J m o l e^{- 1}

S O_{3} + H_{2} O \to H_{2} S O_{4}, △ H = - 130.2 k J m o l e^{- 1}

H_{2} + 1 / 2 O_{2} \to H_{2} O, △ H = - 287.3 k J m o l e^{- 1}

the enthalpy of formation of

H_{2} S O_{4}

298 K

S + O_{2} ⟶ S O_{2}

Δ H = - 398.2 k J

S O_{2} + \frac{1}{2} O_{2} ⟶ S O_{3}

Δ H = - 98.7 k J

S O_{3} + H_{2} O ⟶ H_{2} S O_{4}

Δ H = - 130.2 k J

H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O

Δ H = - 227.3 k J

The enthalpy of formation of sulphuric acid at

298 K

S + O_{2} \to S O_{2}; Δ H = - 298.2 k J

S O_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J

S O_{3} + H_{2} O \to H_{2} S O_{4}; Δ H = - 130.2 k J

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 227.3 k J

the enthalpy of formation of

H_{2} S O_{4}

298

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