When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0°C and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ (Δ Hcomb. (CH4)=890 kJ mol-1, DHcomb(C3H8)=2220 kJ mol-1 ) is :-

Question

When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0°C and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ (Δ Hcomb. (CH4)=890 kJ mol-1, DHcomb(C3H8)=2220 kJ mol-1 ) is :- (A) 38 (B) 317 (C) 477 (D) 32

Tardigrade Admin · Accepted Answer

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

5 L

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

at

STP

So corresponding number of moles of gaese present in

5 L

gas mixture can be obtained by dividing the volumes of gases at STP with

22.4 L

as any gas of

1 m o l

occupied

22.4 L

at STP.

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

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

So the amount of heat released due to combustion of the

5 L

gas mixture

Δ H_{mixture} = \frac{3}{22.4} Δ H_{C H_{4} (g)} + \frac{2}{22.4} Δ H_{C_{3} H_{8} (g)}

Δ H_{mixture} = \frac{3 \times 890 + 2 \times 2220}{22.4} \approx 317 k J