MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Heat of Formation

If S + O2⟶ ...

Question

If $S + O_{2} ⟶ S O_{2}; Δ H = - 298.2 k J$
$S O_{2} + 1 / 2 O_{2} ⟶ S O_{3}; Δ H = - 98.7 k J$
$S O_{3} + H_{2} O ⟶ H_{2} S O_{4}; Δ H = - 130.2 k J$
$H_{2} + 1 / 2 O_{2} ⟶ H_{2} O; Δ H = - 287.3 k J$
Then the enthalpy of formation of $H_{2} S O_{4} a t 298 K$ is :

A

- 814.4 k J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

- 650.3 k J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

- 320.5 k J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

- 233.5 k J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- 814.4 k J$
According to Hess law,
enthalpy of formation of $H_{2} S O_{4} a t 298 K$ is sum of all
$S + O_{2} ⟶ S O_{2}; Δ H = - 298.2 k J$

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

$S O_{3} + H_{2} O ⟶ H_{2} S O_{4}; Δ H = - 130.2 k J$
$H_{2} + 1 / 2 O_{2} ⟶ H_{2} O; Δ H = - 287.3 k J$
$Δ H$ = $- 298.2 - 98.7 - 130.2 - 287.3 =$ $- 814.4 k J$

Suggest Corrections

thumbs-up

2

Similar questions

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

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 = - 398.2 k J; S O_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J

S O_{2} + 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 sulphuric acid at

298

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 heat of formation of

H_{2} {S O}_{4}

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}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Thermochemistry

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Heat of Formation

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon