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Enthalpy

When 5 litr...

Question

When $5 l i t r e s$ of a gas mixture of methane and propane is perfectly combusted at $0^{\circ} C$ and $1$ atmosphere, $16 l i t e r$ of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in $k J$ is:
$(Δ H_{c o m b} (C H_{4})) = 890 k J m o l^{- 1}, Δ H_{C o m b} (C_{3} H_{8}) = 2220 k J m o l^{- 1}$

32

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317

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Solution

The correct option is D $317$
The balanced equations of combustion reactions are:
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l), Δ H_{C H_{4} (g)} = 890 k J / m o l$
$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l), Δ H_{C_{3} H_{8} (g)} = 2220 k J / m o l$

Let here $x L C H_{4}$ and $(5 - x) L C_{3} H_{8}$ in 5L gas mixture at STP
So total volume of oxygen consumed as per stoichiometry of the reaction involved will be
$\Rightarrow 2 x + 5 (5 - x) = 16$
$\Rightarrow 2 x + 25 - 5 x = 16$
$\Rightarrow 3 x = 9$
$\Rightarrow x = 3 L$

So volume of $C H_{4} (g) = 3 L$
and volume of $C_{3} H_{8} (g) = 2 L a t S T P$

So corresponding number of moles of gaese present in 5L gas mixture can be obtained by dividing the volumes of gases at STP with 22.4L as any gas of 1mol occupied 22.4L at STP.

$n_{C H_{4} (g)} = \frac{3}{22.4} m o l$

$n_{C_{3} H_{8} (g)} = \frac{2}{22.4} m o l$

So the amount of heat released due to combustion of the 5L gas mixture
$Δ H_{m i x t u r e} = \frac{3}{22.4} Δ H_{C H_{4} (g)} + \frac{2}{22.4} Δ H_{C_{3} H_{8} (g)}$

$Δ H_{m i x t u r e} = \frac{3 \times 890 + 2 \times 2220}{22.4} \approx 317 k J$

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When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0∘C and 1 atmosphere, 16liter of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ is: (ΔHcomb(CH4))=890 kJmol−1,ΔHComb(C3H8)=2220 kJmol−1