In the reaction S -3-2 O 2- SO 3-2 xkJ and SO 2-1-2 O 2- SO 3- ykJHeat of formation of SO 2 isA- x-yB- x-yC- 2 x-yD- 2 x-y

The correct option is C 2x – y S+3/2 o_2 → SO_3 .....(1)Δ H_1 = 2x kJ SO_2 + 1/2 O_2 → SO_3 ....(2) Δ H_2 = y kJ S+O_2 → SO_2 .....(3) Δ H_3 =z kJ From ...

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Byju's Answer

Standard XII

Chemistry

Molar and Specific Heat Capacity

In the reacti...

Question

In the reaction

$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J$ and $S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J$

Heat of formation of $S O_{2}$ is

x + y

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x – y

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2x – y

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2x + y

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Solution

The correct option is C
2x – y

$S + \frac{3}{2} o_{2} \to S O_{3} . . . . . (1) Δ H_{1} = - 2 x - k J$
$S O_{2} + \frac{1}{2} O_{2} \to S O_{3} . . . . (2) Δ H_{2} = - y k J$
$S + O_{2} \to S O_{2} . . . . . (3) Δ H_{3} = z k J$

From Hess's law, $Δ H_{3} + Δ H_{2} = Δ H_{1}$

z - y = -2x $∴$ z = y - 2x

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Similar questions

In the reaction

$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J$ and $S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J$

Heat of formation of $S O_{2}$ is

The heat of formation of $S O_{2}$ is:

$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J$
$S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J$

Q. In the reaction : S + 3/2

O_{2} \to S O_{3}

+ 2x kcal and

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

+ y kcal, heat of formation of

S O_{2}

is:

Q. In the reactions

S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k c a l

and

S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k c a l

, the heat of formation of

S O_{2}

is :

S + \frac{3}{2} O_{2} ⟶ S O_{3} + 2 x k c a l

;

S O_{2} + \frac{1}{2} O_{2} ⟶ S O_{3} + y k c a l

;

The heat of formation of

S O_{2}

is :

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In the reaction S+32O2→SO3+2xkJ and SO2+12O2→SO3+ykJ Heat of formation of SO2 is

The correct option is C 2x – y S+32o2→SO3.....(1)ΔH1=−2x−kJ SO2+12O2→SO3....(2)ΔH2=−ykJ S+O2→SO2.....(3)ΔH3=zkJ From Hess's law, ΔH3+ΔH2=ΔH1 z - y = -2x ∴ z = y - 2x

In the reaction

$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J$ and $S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J$

Heat of formation of $S O_{2}$ is