H 2 g - 1 2 O 2 g - H 2 O l H 2 O l - H 2 O g - - H- x 4Given- E H-H - x 1 E O-O - x 2 E O-H - x 3 - H F of H 2 O vapour is-

The correct option is D x _ 1 + x _ 2 2 2 x _ 3 + x _ 4 H _ 2 + 1 2 O _ 2 ( g ) ⟶ H _ 2 O( l ) ; Δ H ⇒Δ H=2( 0 H ) ( H H ) 1 2 ( 0 ...

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Byju's Answer

Standard XII

Chemistry

Hess' Law

H 2 g + 1 2 O...

Question

$H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l)$
$H_{2} O (l) ⟶ H_{2} O (g); Δ H = x_{4}$
Given,
$E_{H - H} = x_{1}$
$E_{O - O} = x_{2}$
$E_{O - H} = x_{3}$
$Δ H_{F}$ of $H_{2} O$ vapour is:

x_{1} + \frac{x_{2}}{2} - x_{3} + x_{4}

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2 x_{3} - x_{1} - \frac{x_{2}}{2} - x_{4}

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x_{1} + \frac{x_{2}}{2} - 2 x_{3} - x_{4}

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x_{1} + \frac{x_{2}}{2} - 2 x_{3} + x_{4}

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Solution

The correct option is D $x_{1} + \frac{x_{2}}{2} - 2 x_{3} + x_{4}$
$H_{2} + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H$
$\Rightarrow Δ H = 2 (0 - H) - (H - H) - \frac{1}{2} (0 - 0)$
$= 2 x_{3} - x_{1} - \frac{x_{2}}{2}$
for $H_{2} O (l) ⟶ H_{2} O (g) Δ H = x_{4}$
${Δ H}_{f} = x_{4} - [{2 x}_{3} - x_{1} - \frac{x_{2}}{2}]$
$= x_{1} + \frac{x_{2}}{2} - 2 x_{3} + x_{4}$
Ans :- D

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Similar questions

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l)

H_{2} O_{(l)} \to H_{2} O_{(g)}; △ H = x_{4}

Given,

E_{H - H} = x_{1}

E_{O = O} = x_{2}

E_{O - H} = x_{3}

△ H_{f}

H_{2} O

Vapour is:

Q. Five real numbers

x_{1}, x_{2}, x_{3}, x_{4}, x_{5}

are such that:

\sqrt{x_{1} - 1} + 2 \sqrt{x_{2} - 4} + 3 \sqrt{x_{3} - 9} + 4 \sqrt{x_{4} - 16} + 5 \sqrt{x_{5} - 25} = \frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{2}

The value of

\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{2}

If $x_{1}, x_{2}, x_{3}, x_{4}$ are four positive real numbers such that $x_{1} + \frac{1}{x_{2}} = 4, x_{2} + \frac{1}{x_{3}} = 1, x_{3} + \frac{1}{x_{4}} = 4$ and $x_{4} + \frac{1}{x_{1}} = 1,$ then

Q. Determine

Δ H / k J

for the following reaction using the listed enthalpies of reaction:

4 C O (g) + 8 H_{2} (g) ⟶ 3 C H_{4} (g) + C O_{2} (g) + 2 H_{2} O (l)

Given that

C (g r a p h i t e) + 1 / 2 O_{2} (g) ⟶ C O (g); Δ H / k J = - 110.5 k J

C O (g) + 1 / 2 O_{2} (g) ⟶ C O_{2} (g); Δ H / k J = - 282.9 k J

H_{2} (g) + 1 / 2 O_{2} (g) ⟶ H_{2} O (l); Δ H / k J = - 285.8 k J

C (g r a p h i t e) + 2 H_{2} (g) ⟶ C H_{4} (g); Δ H / k J = - 74.8 k J

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H2(g)+12O2(g)⟶H2O(l)H2O(l)⟶H2O(g);ΔH=x4Given,EH−H=x1EO−O=x2EO−H=x3ΔHF of H2O vapour is:

The correct option is D x1+x22−2x3+x4H2+12O2(g)⟶H2O(l);ΔH⇒ΔH=2(0−H)−(H−H)−12(0−0) =2x3−x1−x22for H2O(l)⟶H2O(g)ΔH=x4ΔHf=x4−[2x3−x1−x22] =x1+x22−2x3+x4Ans :- D

$H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l)$
$H_{2} O (l) ⟶ H_{2} O (g); Δ H = x_{4}$
Given,
$E_{H - H} = x_{1}$
$E_{O - O} = x_{2}$
$E_{O - H} = x_{3}$
$Δ H_{F}$ of $H_{2} O$ vapour is: