From the following data at 25-C-Reaction -rH- kJ-mol 1-2H2g-1-2O2g -42 H2g-1-2O2gH2Og -242 H2g - 2Hg O2g- 2Og -436 -495

The correct options are A Δ _rH^∘ for the reaction H_2O(g)→ 2H(g)+O(g) is 925.5 kJ/mol D Enthalpy of formation of H(g) is +218 kJ/m ...

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Standard XII

Chemistry

Enthalpy

From the foll...

Question

From the following data at $25^{\circ} C$ .
Reaction $Δ_{r} H^{\circ}$ kJ/mol
$\frac{1}{2} H_{2} (g) + \to \frac{1}{2} O_{2} (g)$
$+ 42$
$H_{2} (g) + \frac{1}{2} O_{2} (g) H_{2} O (g)$ $- 242$
$H_{2} (g) \to 2 H (g)$
$O_{2} (g) \to 2 O (g)$
$+ 436$
$+ 495$

Δ_{r} H^{\circ}

for the reaction

H_{2} O (g) \to 2 H (g) + O (g)

925.5 k J / m o l

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Δ_{r} H^{\circ}

for the reaction

O H (g) \to H (g) + O (g)

502 k J / m o l

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Enthalpy of formation of

H (g)

- 218 k J / m o l

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Enthalpy of formation of

H (g)

+ 218 k J / m o l

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Solution

The correct options are
A $Δ_{r} H^{\circ}$ for the reaction $H_{2} O (g) \to 2 H (g) + O (g)$ is $925.5 k J / m o l$
D Enthalpy of formation of $H (g)$ is $+ 218 k J / m o l$
A. $H_{2} O \to 2 H + O Δ H_{r}^{0} = 242 k J / m o l$ ......(1)
$H_{2} \to 2 H Δ H_{r}^{0} = 436 k J / m o l$ ......(2)
$\frac{1}{2} \to O Δ H_{r}^{0} = \frac{495}{2} = 247.5 k J / m o l$ ......(3)
Add reactions (1) to (3)
$H_{2} O \to= 2 H + O$
$Δ H_{r}^{0} = 242 + 436 + 247.5 = 925.5 k J / m o l$
Thus, the option $A$ is correct.
$H_{2} (g) \to 2 H (g) Δ H_{r}^{0} = 436 k J / m o l$ ......(1)
The expression for the enthalpy of formation of $H (g)$ is
$\frac{1}{2} H_{2} (g) \to H (g)$
It is obtained by dividing equation (1) with 2.
Enthalpy of formation of $H (g)$ is $\frac{436}{2} = 218 k J / m o l$
Thus, the option $D$ is correct.

Suggest Corrections

Similar questions

Q. The enthalpy changes of the following reactions at

27^{\circ} C

are

$N a (s) + \frac{1}{2} C l_{2} (g) \to N a C l (s)$	$Δ_{r} H = - 411 k J / m o l$
$H_{2} (g) + S (s) + 2 O_{2} (g) \to H_{2} S O_{4} (l)$	$Δ_{r} H = - 811 k J / m o l$
$2 N a (s) + S (s) + 2 O_{2} (g) \to N a_{2} S O_{4} (s)$	$Δ_{r} H = - 1382 k J / m o l$
$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g)$	$Δ_{r} H = - 92 k J / m o l;$

from these data, the heat change of reaction at constant volume (in kJ/mol) at

27^{\circ} C

for the process

2 N a C l (s) + H_{2} S O_{4} (l) \to N a_{2} S O_{4} (s) + 2 H C l (g)

Q. The enthalpies of formation of

O H (g)

H (g)

and

O (g)

are 21

k J m o l^{- 1}

, 109

k J m o l^{- 1}

and 124

k J m o l^{- 1} .

The value of bond enthalpy

O - H

$C (s) + O_{2} (g) \to C O_{2} (g), Δ_{r} H = - 94.3 k c a l / m o l$

$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g), Δ_{r} H = - 67.4 k c a l / m o l$

$O_{2} (g) \to 2 O (g), Δ_{r} H = 117.4 K c a l / m o l$

$C O (g) \to C (g) + O (g), Δ_{r} H = 230.6 K c a l / m o l$

Calculate: $Δ_{r} H f o r C (s) \to C (g) i n K c a l / m o l$

Q. The enthalpy changes at

298

K in successive breaking of

O - H

bonds of

H O H

are:

H_{2} O (g) \to H (g) + O H (g), Δ H = 498

m o l^{- 1}

O H (g) \to H (g) + O (g), Δ H = 428

m o l^{- 1}

The bond enthalpy of the

O - H

bond is:

Q. The enthalpy changes at

298 K

in successive breaking of

O - H

bonds of water are

H_{2} O (g) \to H (g) + O H (g); Δ H = 498 k J

{m o l}^{- 1}

O H (g) \to H (g) + O (g); Δ H = 428 k J

{m o l}^{- 1}

The bond enthalpy of

O - H

bond is:

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Reaction	$Δ_{r} H^{\circ}$ kJ/mol
$\frac{1}{2} H_{2} (g) + \to \frac{1}{2} O_{2} (g)$	$+ 42$
$H_{2} (g) + \frac{1}{2} O_{2} (g) H_{2} O (g)$	$- 242$
$H_{2} (g) \to 2 H (g)$ $O_{2} (g) \to 2 O (g)$	$+ 436$ $+ 495$

From the following data at 25∘C.ReactionΔrH∘ kJ/mol12H2(g)+→12O2(g)+42H2(g)+12O2(g)H2O(g)−242H2(g)→2H(g)O2(g)→2O(g)+436+495

From the following data at $25^{\circ} C$ .
Reaction $Δ_{r} H^{\circ}$ kJ/mol
$\frac{1}{2} H_{2} (g) + \to \frac{1}{2} O_{2} (g)$
$+ 42$
$H_{2} (g) + \frac{1}{2} O_{2} (g) H_{2} O (g)$ $- 242$
$H_{2} (g) \to 2 H (g)$
$O_{2} (g) \to 2 O (g)$
$+ 436$
$+ 495$