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Spontaneity

Only N2, CO...

Question

Only $N_{2}$ , CO and $C O_{2}$ gases remain after 0.72 gm of carbon is treated with one litre of air at $27^{0} C$ and 4.92 atm pressure. Assume air composition $O_{2}$ = 20%, $N_{2}$ = 79% and $C O_{2}$ = 1% (by volume). The heat evolved (in Kcal) under constant pressure is:
Given : $C + O_{2} \to C O_{2}, Δ H = - 100 K c a l / m o l$
$C + \frac{1}{2} O_{2} \to C O, Δ H = - 25 K c a l / m o l$

1 kcal

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2 kcal

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3 kcal

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4 kcal

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Solution

The correct option is C 3 kcal
$n_{a i r} = \frac{4.92 \times 1}{24.6}$ = 0.2 mol.
$n_{O_{2}}$ $= 0.2 \times 0.2 =$ 0.04 mol, $n_{c} = \frac{0.72}{12}$
$C + \frac{1}{2} O_{2} \to C O$
initial mole 0.06 0.04 0
final mole 0 0.01 0.06
$C + O_{2} \to C O_{2}$
initial mole 0.06 0.01 0
final mole 0.04 0 0.02

Heat evolved $= 0.04 \times 25 + 0.02 \times 100 = 1 + 2 = 3 K c a l$

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From the thermochemical reactions,

C + 1/2O₂ → CO; ∆H = -110.5 kJ

CO + 1/2O₂→ CO₂; ∆H = -283.2 kJ

The heat of reaction C + O₂ → CO₂ is:

C (g r a p h i t e) + \frac{1}{2} O_{2} \to C O Δ H_{1} = - 110.5 k J / m o l

C O + \frac{1}{2} O_{2} \to C O_{2} Δ H = - 283.2 k J / m o l

k J / m o l

C (g r a p h i t e) + O_{2} \to C O_{2} Δ H = ?

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