Question

$28.0\, L$ of $CO _2$ is produced on complete combustion of $168 \, L$ gaseous mixture of ethene and methane at $25^{\circ} C$ and $1 \, atm$ Heat evolved during the combustion process is ______$kJ$ Given : $\Delta H _{ c }\left( CH _4\right)=-900 \, kJ\, mol ^{-1}$ $\Delta H _{ c }\left( C _2 H _4\right)=-1400\, kJ \, mol ^{-1}$

Answer 1

Correct Answer: 847 - 848

The correct answer is 925.

$\Rightarrow 28=16.8+x$
$x=11.2L$
$n_{C{H}_4}=\frac{PV}{RT}=\frac{1\times 5.6}{0.082\times 298}=0.229\text{mole}$
$n_{C_2{H}_2}=\frac{11.2}{0.082\times 298}=0.458\text{mole}$
$\therefore \text{Heat evolved}=0.229\times 900+0.458\times 1400$
$=206.1+641.2$
$=847.3kJ$

Answer 2

The volume of \(\mathrm{CO_2}\) produced enables us to calculate the moles of \(\mathrm{CH_4}\) and \(\mathrm{C_2H_4}\) combusted, from which we can determine the total heat evolved using their respective combustion enthalpies.
Let, the volume of \[ C_2H_4 \text{ be } x \text{ litres.} \]
\[ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \]
Initial volume: \(x\)
Final volume: \(2x\)
For \(\mathrm{CH_4}\), the reaction is \[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \]
Initial volume: \(16.8 - x\)
Final volume: \(16.8 - x\)
The total volume of \(\mathrm{CO_2}\) produced is: \[ \text{Total } CO_2 = 2x + (16.8 - x) \] Thus, \[ 28 = 16.8 + x \] Solving for \(x\), we get \[ x = 11.2 \text{ L} \] Now, calculate the moles of the gases: \[ n_{CH_4} = \frac{PV}{RT} = \frac{1 \times 5.6}{0.082 \times 298} = 0.229 \text{ moles} \] \[ n_{C_2H_4} = \frac{11.2}{0.082 \times 298} = 0.458 \text{ moles} \] Therefore, the heat evolved is: \[ \text{Heat evolved} = 0.229 \times 900 + 0.458 \times 1400 \] \[ = 206.1 + 641.2 = 847.3 \text{ kJ} \]