A gas mixture 3.67 L in volume contains C2H4 and CH4 is proportion of 2:1 b

Question

A gas mixture

3.67 L

in volume contains

C_{2} H_{4}

and

C H_{4}

is proportion of

2 : 1

by moles and is at

25^{\circ} C

and

1 a t m

. If the

Δ H_{C}

(

C_{2} H_{4})

and

Δ H_{C}

(C H_{4})

are

- 1400

and

- 900 k J / m o l

find heat evolved on burning this mixture

Options:

A . 20.91 kJ

B . 50.88 kJ

C . 185 kJ

D . 160 kJ

Answer: Option C
:
C
Volume of

C_{2} H_{4}

\frac{3.67 \times 2}{3} = 2.45 L i t

Volume of

C H_{4}

1.22 L i t

Mole of

C_{2} H_{5}

P V

n R T n

\frac{P V}{R T}

\frac{1 \times 2.45}{0.082 \times 298} = n = 0.1 m o l e

Mole of

C H_{4}

^{n} C H_{4}

\frac{1 \times 1.22}{0.082 \times 298} = 0.05 m o l e

∵

Energy released due to combusting 1 mole

C_{2} H_{4} = 1400 K J

∴

Energy released due to combusting 1 mole

C_{2} H_{4}

1400 \times 0.1 = 140 k J

Similar energy released in combusting 0.05 mole

C H_{4} = - 900 \times 0.05 = 45

So that heat realized =

140 + 45

185 k J / m o l e

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