If S - O2- SO2- H - -298-2 kJ mole-1 SO2 - 1-2 O2- SO3- H - -98-7 kJ mole-1 SO3 - H2O - H2SO4- H - -130-2 kJ mole-1 H2 - 1-2O2- H2O- H - -287-3 kJ mole-1the enthalpy of formation of H2SO4 at 298 K will be -

The correct option is A 814.4 kJ mole^ 1 S+O_2→ SO_2 Δ H= 298.2 kJ/mole SO_2+12O_2→ SO_3 Δ H= 98.7 kJ/mole SO_3+H_2O→ H_2SO_4 Δ H= 130. ...

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Heat of Formation

If S + O2→ ...

Question

If $S + O_{2} \to S O_{2}, △ H = - 298.2 k J m o l e^{- 1}$
$S O_{2} + 1 / 2 O_{2} \to S O_{3}, △ H = - 98.7 k J m o l e^{- 1}$
$S O_{3} + H_{2} O \to H_{2} S O_{4}, △ H = - 130.2 k J m o l e^{- 1}$
$H_{2} + 1 / 2 O_{2} \to H_{2} O, △ H = - 287.3 k J m o l e^{- 1}$
the enthalpy of formation of $H_{2} S O_{4}$ at $298 K$ will be ?

- 814.4 k J m o l e^{- 1}

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+ 814.4 k J m o l e^{- 1}

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- 650.3 k J m o l e^{- 1}

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- 433.7 k J m o l e^{- 1}

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Solution

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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

H_{2} S O_{4}

S + O_{2} \to {S O}_{2}; △ H = - 398.2 k J : {S O}_{2} + \frac{1}{2} O_{2} \to {S O}_{3}; △ H = - 98.7 k J

{S O}_{3} + H_{2} O \to H_{2} {S O}_{4}; △ H = - 130.2 k J : H_{2} + \frac{1}{2} O_{2} \to H_{2} O, △ H = - 227.3 k J

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

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

H_{2} S O_{4}

298

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

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

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