A gas mixture 3-67 L in volume contain C2H4 and CH4 is proportion of 2 - 1 by moles and is at 25-C and 1 atm- If the HCC2H4 and HC CH4 are -1400 and -900 kJ-mol- Find the heat evolved on burning this mixture-

The correct option is C 185 kJ The total mole of gas, #160; n= PV RT #160; #160; #160; #160; #160; #160; #160; #160; #16 ...

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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 $25^{\circ} C$ and $1 a t m$ . If the $△ H_{C} (C_{2} H_{4})$ and $△ H_{C} (C H_{4})$ are $- 1400$ and $- 900 k J / m o l$ . Find the heat evolved on burning this mixture.

20.91 k J

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50.88 k J

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185 k J

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160 k J

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Solution

The correct option is C $185 k J$
The total mole of gas, $n = \frac{P V}{R T}$
$= \frac{1 a t m \times 3.67 L}{0.0821 L . a t m {m o l}^{- 1} K^{- 1} \times 298.15 K}$
$= 0.15$ moles
Since, $n_{C_{2} H_{4}} : n_{{C H}_{4}} = 2 : 1$
$∴ n_{C_{2} H_{4}} = 0.1 m o l e & n_{{C H}_{4}} = 0.05 m o l e$
$∴$ total heat produced by combusting $C_{2} H_{4}$ and ${C H}_{4}$
$= | {Δ H}_{C} (C_{2} H_{4}) | + | {Δ H}_{C} ({C H}_{4}) |$
$= ∣ ∣ - (1400 k J / m o l) \times n_{C_{2} H_{4}} ∣ ∣ + ∣ ∣ - (900 k J / m o l) \times n_{C_{2} H_{4}} ∣ ∣$
$= 140 k J / m o l e + 45.0 k J / m o l e$
$= 185 k J / m o l e$

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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