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Standard XII

Chemistry

Enthalpy of Combustion

One litre sam...

Question

One litre sample of a mixture of methane gas and $O_{2}$ measured at $25^{\circ} C$ and 740 torr was allowed to react at constant pressure in a calorimeter which together with its content has a heat capacity of 1260 cal/degree. The complete combustion of methane to $C O_{2}$ and $H_{2} O$ caused a temperature rise in calorimeter of 0.667 K. The percentage of $C H_{4}$ in original mixture was:
Given: $Δ H_{c o m b u s t i o n}$ of $C H_{4}$ is $- 210.8 k c a l m o l^{- 1}$ .

10%

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20%

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15%

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30%

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Solution

The correct option is A 10%
Heat generated $Q = m s △ T = 120 \times 0.667 = 840.4 c a l$

Number of moles of $C H_{3}$ in the mixture is = $\frac{840}{210.8 \times 10^{3}} = 3.98 \times 10^{- 3}$

Total moles, $n = \frac{P V}{R T} = \frac{740 \times 1}{760 \times 0.0821 \times 298} = 3.98 \times 10^{- 2}$

Mole $%$ of $C H_{4}$ is $\frac{3.98 \times 10^{- 3}}{3.98 \times 10^{- 2}} \times 100 = 10 %$

Hence, option A is correct.

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Similar questions

Q. 1.00 sample of a mixture of

C H_{4} (g)

O_{2} (g)

measured at

25^{\circ} C

& 740 torr was allowed to react at constant pressure in a calorimeter which together with its contents had a heat capacity of

1260 c a l / K

. The complete combustion of methane to

C O_{2}

H_{2} O

caused a temperature rise, in the calorimeter, of 0.667 K. Then the mole precent of

C H_{4}

in the original mixture was:
Given that :

Δ H_{c o m b}^{\circ} (C H_{4}) = - 215

k c a l . m o l^{- 1}; v o l u m e = 1 l i t .

Q. 0.16 g of methane was subjected to combustion at

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( i ) constant volume and ( ii ) constant pressure.

[Given: The thermal capacity of calorimeter system is 17.7 kJ

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(R = 8.314 J m o l^{- 1} K^{- 1})]

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was subjected to the combustion at

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in a bomb calorimeter. The temperature of the calorimeter. The temperature of the calorimeter system (including water) was found to rise by

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. Calculate the heat of combustion of methane at
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Q. 0.16 g of methane was subjected to combustion at

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in a bomb calorimeter. The temperature of the calorimeter system (including water) was found to rise by

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The correct option is A 10%Heat generated Q=ms△T=120×0.667=840.4calNumber of moles of CH3 in the mixture is =840210.8×103=3.98×10−3Total moles, n=PVRT=740×1760×0.0821×298=3.98×10−2Mole % of CH4 is 3.98×10−33.98×10−2×100=10%Hence, option A is correct.