If H2S-3-2O2- H2O-SO2- - H-136-7Kcal H2-1-2O2- H2O- H-68-4Kcal S-O2- SO2- H-71Kcalwhat is - Hf of H2S

The correct option is B 2.7 Kcal(i) H_2S+3/2O_2→ H_2O+SO_2 136.7 Kcal(ii) H_2+1/2O_2→ H_2O 68.4 nbsp;Kcal(iii) S+O_2→ SO_2 71 nbsp;Kcal(i ...

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Byju's Answer

Standard IX

Mathematics

Symmetric Difference of a Set

If H2S+3/2O...

Question

If $H_{2} S + \frac{3}{2} O_{2} \to H_{2} O + S O_{2},$
$Δ H = - 136.7 K c a l$
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O, Δ H = - 68.4 K c a l$
$S + O_{2} \to S O_{2}, Δ H = - 71 K c a l$
what is $Δ H_{f} o f H_{2} S$

2.7 Kcal

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-2.7 Kcal

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5.4 Kcal

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-5.4 Kcal

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Solution

The correct option is B -2.7 Kcal
(i) $H_{2} S + \frac{3}{2} O_{2} \to H_{2} O + S O_{2} - 136.7$ Kcal
(ii) $H_{2} + \frac{1}{2} O_{2} \to H_{2} O - 68.4$ Kcal
(iii) $S + O_{2} \to S O_{2} - 71$ Kcal
(ii) + (iii) - (i)
$H_{2} + S \to H_{2} S$
$Δ H =$ -68.4 - 71 + 136.7
$=$ - 2.7 Kcal

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Similar questions

Q. 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 :

Q. If

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

K will be?

Q. If

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}

at 298 K is

Q. If :

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

K will be.

Q. If

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

will be?

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If H2S+32O2→H2O+SO2,ΔH=−136.7KcalH2+12O2→H2O,ΔH=−68.4KcalS+O2→SO2,ΔH=−71Kcalwhat is ΔHfofH2S

The correct option is B -2.7 Kcal(i) H2S+32O2→H2O+SO2−136.7 Kcal(ii) H2+12O2→H2O−68.4 Kcal(iii) S+O2→SO2−71 Kcal(ii) + (iii) - (i)H2+S→H2SΔH= -68.4 - 71 + 136.7= - 2.7 Kcal

If $H_{2} S + \frac{3}{2} O_{2} \to H_{2} O + S O_{2},$
$Δ H = - 136.7 K c a l$
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O, Δ H = - 68.4 K c a l$
$S + O_{2} \to S O_{2}, Δ H = - 71 K c a l$
what is $Δ H_{f} o f H_{2} S$

The correct option is B -2.7 Kcal
(i) $H_{2} S + \frac{3}{2} O_{2} \to H_{2} O + S O_{2} - 136.7$ Kcal
(ii) $H_{2} + \frac{1}{2} O_{2} \to H_{2} O - 68.4$ Kcal
(iii) $S + O_{2} \to S O_{2} - 71$ Kcal
(ii) + (iii) - (i)
$H_{2} + S \to H_{2} S$
$Δ H =$ -68.4 - 71 + 136.7
$=$ - 2.7 Kcal