A gas mixture 3-67L in volume contain C 2 H 4 and CH 4 is proportion of 2-1 by moles and is at 25oC and 1 atm- If the - H C C 2 H 4 and - H C CH 4 are -1400 and -900kJ-mol- find heat evolved on burning this mixture-

The correct option is C 201kJ Ratio of moles = ratio of volumeSo, [ 2 / 3 of 3.67L= C_ 2 H_ 4 amp; 1 / 3 of 3.67L fo ...

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Byju's Answer

Standard XII

Chemistry

Enthalpy of Combustion

A gas mixture...

Question

A gas mixture $3.67 L$ in volume contain $C_{2} H_{4}$ and ${C H}_{4}$ is proportion of $2 : 1$ by moles and is at $25^{o} C$ and $1$ atm. 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:

20.91 k J

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50.88 k J

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201 k J

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160 k J

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Solution

The correct option is C $201 k J$
Ratio of moles = ratio of volume
So,
$\begin{matrix} \frac{2}{3} o f 3.67 L = C_{2} H_{4} & \frac{1}{3} o f 3.67 L f o r C H_{4} C_{2} H_{4} = 2.45 L & C H_{4} = 1.22 L N o . o f a t S T P n C_{2} H_{4} = \frac{2.45}{22.4} & n C H_{4} = \frac{1.22}{22.4} \Rightarrow 0.109 m o l, & \Rightarrow 0.054 m o l H e a t e v o l v e d = 0.109 \times (- 1400) + 0.054 (900) \Rightarrow - 201.2 K J / m o l \end{matrix}$
So, closest option is option C.

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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