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Gibbs Free Energy & Spontaneity

When 6.81 L o...

Question

When $6.81 L$ of a gas mixture of methane and propane is perfectly combusted at $0^{\circ} C$ and $1$ bar, $15.89 L$ of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in $kJ is : (Δ H_{comb} (C H_{4}) = 890 {kJ mol}^{- 1}, Δ H_{comb} (C_{3} H_{8}) = 2220 {kJ mol}^{- 1})$

840

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307

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477

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520

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Solution

The correct option is B $307$
Balanced equations for combustion reactions of $C H_{4}$ and $C_{3} H_{8}$ are:
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$
$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)$
Molar volume of gases at $0^{\circ} C$ and $1$ bar is $22.7 L$ . Hence, number of moles in gaseous mixture,
$C H_{4} + C_{3} H_{8} = \frac{6.81}{22.7} = 0.3 mol$
$Number of moles of O_{2} = \frac{15.89}{22.7} = 0.7 mol$
Let $x$ moles of $C H_{4}$ is there in a geseous mixture, so number of moles of $C_{3} H_{8}$ would be $0.3 - x$ . Then, moles of $O_{2}$ consumed,
$2 x + (0.3 - x) 5 = 0.7$
$\Rightarrow x = 0.27$
$Moles of C H_{4} = 0.27$
$Moles of C_{3} H_{8} = 0.3 - 0.27 = 0.03$
$∴ Total amount of heat liberated = 0.27 \times 890 + 0.03 \times 2220 = 306.9 kJ \approx 307 kJ$ .

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When 6.81 L of a gas mixture of methane and propane is perfectly combusted at 0 ∘C and 1 bar, 15.89 L of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ is:(ΔHcomb(CH4)=890 kJ mol−1, ΔHcomb(C3H8)=2220 kJ mol−1)