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Hess' Law

Cs + O2g → CO...

Question

$C (s) + O_{2} (g) \to C O_{2} (g), Δ H = + 94.0 K c a l .$

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

From the above reactions find how much heat (Kcal mole $^{- 1}$ ) would be produced in the following reaction:

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

A

20.6

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B

26.3

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C

44.2

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D

161.7

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Solution

The correct option is D 161.7
The thermochemical reactions are as given below.

$C (s) + O_{2} (g) \to C O_{2} (g) Δ H = + 94.0 K c a l .$ ......(1)

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

The reaction (2) is reversed and added to the reaction (1) to obtain the reaction $C (s) + \frac{1}{2} O_{2} (g) \to C O (g)$ .

The heat produced in this reaction is $94.0 - (- 67.7) = 161.7$ Kcal/mole

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Similar questions

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)

Q. Consider the following reaction

C (s) + O_{2} (g) ⟶ {C O}_{2} (g) + x k J

C O (g) + \frac{1}{2} O_{2} (g) ⟶ {C O}_{2} (g) + y k J

The heat of formation of

C O (g)

Δ H

k J

for the following reaction:

C (g) + O_{2} (g) \to C O_{2} (g)

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

Δ H = + 131 k J

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

Δ H = - 282 k J

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

Δ H = - 242 k J

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)

Q. Which of the following equations has have enthalpy changes equal to

Δ H_{c o m b} C

I

C (s) + O_{2} (g) ⟶ C O_{2} (g)

I I

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

I I I

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

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