From the given table answer the following questions -COg CO2g H2Og H2g - H- f 298 -kCal - mole -26-42 -94-05 -57-80 - G- f 298 -kCal - mole -32-79 -94-24 -54-640 S-298 -Cal - Kmole 47-351-1-31-2Reaction - H2Og-COg- H2g-CO2g i - rH-298 ii - rG-298 iii - rS-298 iv - rE-298 v S-298 - H2Og -

The correct option is C ( i ) 9.83 kCal/mol; #160;( ii ) 6.81 kCal/mol, #160;( iii ) 10.13 Cal / K mol, #160;( iv ) 9.83 kCal ...

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Byju's Answer

Standard XII

Chemistry

Difference between Internal Energy and Enthalpy

From the give...

Question

From the given table answer the following questions :
CO(g) $C O_{2} (g)$ $H_{2} O (g)$ $H_{2} (g)$
${(Δ H_{f}^{\circ})}_{298}$ ( -kCal / mole ) $-$ 26.42 $-$ 94.05 $-$ 57.8 0
${(Δ G^{\circ}_{f})}_{298}^{\circ}$ ( -kCal / mole ) $-$ 32.79 $-$ 94.24
$-$ 54.64
0
$S_{298}^{\circ}$ ( -Cal / Kmole )
47.3 51.1 ? 31.2
Reaction : $H_{2} O (g) + C O (g) ⇌ H_{2} (g) + C O_{2} (g)$
( i ) $Δ_{r} H_{298}^{\circ}$

( ii ) $Δ_{r} G_{298}^{\circ}$

( iii ) $Δ_{r} S_{298}^{\circ}$

( iv ) $Δ_{r} E_{298}^{\circ}$

( v ) $S_{298}^{\circ} [H_{2} O (g)]$

( i )

- 8.83

kCal/mol;

( ii )

- 6.81

kCal/mol,

( iii )

- 10.13

Cal / K-mol,

( iv )

- 9.83

kCal/mol,

( v )

+ 45.13

Cal / K-mol

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( i )

- 9.83

kCal/mol;

( ii )

- 8.81

kCal/mol,

( iii )

- 10.13

Cal / K-mol,

( iv )

- 9.83

kCal/mol,

( v )

+ 45.13

Cal / K-mol

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( i )

- 9.83

kCal/mol;

( ii )

- 6.81

kCal/mol,

( iii )

- 10.13

Cal / K-mol

( iv )

- 8.83

kCal/mol,

( v )

+ 45.13

Cal / K-mol

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( i )

- 9.83

kCal/mol;

( ii )

- 6.81

kCal/mol,

( iii )

- 10.13

Cal / K-mol,

( iv )

- 9.83

kCal/mol,

( v )

+ 45.13

Cal / K-mol

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Solution

The correct option is C ( i ) $- 9.83$ kCal/mol;
( ii ) $- 6.81$ kCal/mol,
( iii ) $- 10.13$ Cal / K-mol,
( iv ) $- 9.83$ kCal/mol,
( v ) $+ 45.13$ Cal / K-mol

(i) $Δ_{r} H_{298}^{\circ}$ = ${(Δ H_{f}^{\circ})}_{298}$ (Reactants) - ${(Δ H_{f}^{\circ})}_{298}$ (Products),
$Δ_{r} H_{298}^{\circ}$ $= [(- 94.05 + 0) - (- 57.8 - 26.42)]$ $k C a l / m o l$ ,
$Δ_{r} H_{298}^{\circ}$ $= - 9.83 k C a l / m o l$
(ii) $Δ_{r} G_{298}^{\circ}$ = ${(Δ G_{f}^{\circ})}_{298}$ (Reactants) - ${(Δ G_{f}^{\circ})}_{298}$ (Products),
$Δ_{r} G_{298}^{\circ}$ $= [(- 94.24 + 0) - (- 54.64 - 32.79)] k C a l / m o l,$
$Δ_{r} G_{298}^{\circ}$ $= - 6.81 k C a l / m o l$
(iii) $Δ_{r} G_{298}^{\circ}$ = $Δ_{r} H_{298}^{\circ} -$ T $Δ_{r} S_{298}^{\circ}$ ,
Therefore,
$Δ_{r} S_{298}^{\circ}$ $= - (- 6.83 + 9.83) / 298$
$= - 10.13 C a l / K - m o l$
(iv) $Δ_{r} H_{298}^{\circ}$ = $Δ_{r} E_{298}^{\circ}$ $+ n R T$ ,
Now $n = 0$ for this reaction,
Therefore,
$Δ_{r} H_{298}^{\circ}$ = $Δ_{r} E_{298}^{\circ}$ $= - 9.83 k C a l / m o l$

(v) $Δ_{r} S_{298}^{\circ}$ = ${(Δ S_{f}^{\circ})}_{298}$ (Reactants) - ${(Δ S_{f}^{\circ})}_{298}$ (Products),
Therefore,
$- 10.13 = (51.1 + 31.2) - (47.3 +$ $S_{298}^{\circ} [H_{2} O (g)]$ ) $S_{298}^{\circ} [H_{2} O (g)]$
$= 45.13 C a l / K m o l$

Suggest Corrections

Similar questions

I P_{1}

and

I P_{2}

M g

are 178 and 348 Kcal mol

^{- 1}

. The energy required for the reaction

M g \to M g^{2 +} + 2 e^{-}

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

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

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

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

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

C (s) + O_{2} (g) \to C O_{2}, (g);

Δ H = - 94.3

kcal/mol

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g);

Δ H = - 67.4

kcal/mol

O_{2} (g) \to 2 O (g);

Δ H = 117.4

kcal/mol

C O (g) \to C (g) + O (g);

Δ H = 230.6

kcal/mol.
Calculate

Δ H

for

C (s) \to C (g)

in kcal/mol.

C (s) + O_{2} (g) \to C O_{2} (g); Δ H = - 94.3

kcal/mol

C O (g) + 1 / 2 O_{2} (g) \to C O_{2} (g); Δ H = - 67.4

kcal/mol

O_{2} (g) \to 2 O (g); Δ H = 117.4

kcal/mol

C O (g) \to C (g) + O (g); Δ H = 230.6

kcal/mol

Calculate

Δ H

for

C (s)

\to

C (g)

in Kcal/mol.

$I P_{1}$ and $I P_{2}$ of Mg are 178 and 348 k cal ${m o l}^{- 1}$ . The energy required for the reaction Mg $\to$ ${M g}^{2 +}$ + 2 $e^{-}$ is

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	CO(g)	$C O_{2} (g)$	$H_{2} O (g)$	$H_{2} (g)$
${(Δ H_{f}^{\circ})}_{298}$ ( -kCal / mole )	$-$ 26.42	$-$ 94.05	$-$ 57.8	0
${(Δ G^{\circ}_{f})}_{298}^{\circ}$ ( -kCal / mole )	$-$ 32.79	$-$ 94.24	$-$ 54.64	0
$S_{298}^{\circ}$ ( -Cal / Kmole )	47.3	51.1	?	31.2