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Enthalpy

Given:Cs + O2...

Question

Given:
$C (s) + O_{2} (g) \to C O_{2}, (g); △ H = - 94.3 k c a l / m o l$
$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); △ H = - 67.4 k c a l / m o l$
$O_{2} (g) \to 2 O (g); △ H = 117.4 k c a l / m o l$
$C O (g) \to C (g) + O (g); △ H = 230.6 k c a l / m o l$
Calculate $△ H$ for $C (s) \to C (g)$ in $k c a l / m o l$ .

A

171

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B

154

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C

117

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D

145

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Solution

The correct option is D 145
Given reactions are,
$e q . (1)$
$C (s) + O_{2} (g) \to C O_{2}, (g); △ H_{1} = - 94.3 k c a l / m o l$

$e q . (2)$
$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); △ H_{2} = - 67.4 k c a l / m o l$

$e q . (3)$
$O_{2} (g) \to 2 O (g); △ H_{3} = 117.4 k c a l / m o l$

$e q . (4)$
$C O (g) \to C (g) + O (g); △ H_{4} = 230.6 k c a l / m o l$

The required reaction,
$C (s) \to C (g)$ can be obtained by the combination,
Target equation = $e q . (1) - e q . (2) - \frac{1}{2} e q . (3) + e q . (4)$
So, enthalpy change will be:
$△ H = △ H_{1} - △ H_{2} - \frac{1}{2} △ H_{3} + △ H_{4}$
putting all the values,
$= - 94.3 + 67.4 - 58.7 + 230.6 = 145 k c a l / m o l .$

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Similar questions

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

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

O_{2} (g) \to 2 O (g); △ H = 117.4 k c a l / m o l

C O (g) \to C (g) + O (g); △ H = 230.6 k c a l / m o l

△ H

C (s) \to C (g)

k c a l / m o l

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

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

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

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

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

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

Δ H = - 67.4

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

Δ H = 117.4

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

Δ H = 230.6

kcal/mol.
Calculate

Δ H

C (s) \to C (g)

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

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

O_{2} (g) \to 2 O (g); Δ H = 117.4

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

Δ H

C (s)

\to

C (g)

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

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

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

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

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

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