Cs-O2g- CO2- g- - H-94-3 kcal-mol COg-12O2g- CO2g- - H-67-4 kcal-mol O2g- 2Og- - H-117-4 kcal-mol COg- Cg-Og- - H-230-6 kcal-mol-Calculate - H for Cs- Cg in kcal-mol-

The correct option is D 145 To get C(s)→ C(g) C(s)+O_2(g)→ CO_2(g)→ CO_2(g);Δ H = 94 3...(1) CO(g)+12O_2(g)→ CO_2(g);Δ H = 67 4...(2) ...

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Bond Order and Enthalpy

Cs+O2g→ CO2, ...

Question

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

17.1

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154

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145

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Solution

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Similar questions

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); △ 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 = + 94.0 K c a l .

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g), Δ H = - 67.7 K c a l .

^{- 1}

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

C H_{3} C O O H (l)

25^{\circ} C

C H_{3} C O O H (l) + 2 O_{2} (g) \to 2 C O_{2} + 2 H_{2} O (l); Δ H = - 208.34

C (s) + O_{2} (g) \to C O_{2} (g); Δ H = - 94.05

H_{2} (g) + \frac{1}{2} O_{2} \to H_{2} O (l); Δ H = - 68.32

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