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 atm. If the $\Delta H_C^{\circ }\left(C_2{H}_4\right)$ and $\Delta H_C^{\circ }\left(C{H}_4\right)$ are -1400 and -900 kJ/mol find heat evolved on burning this mixture

• A. 20.91 kJ
• B. 50.88 kJ
• C. 185 kJ
• D. 160 kJ

Answer 1

The correct option is C 185 kJ

The total number of moles of ethene can be obtained by the ideal gas equation.

$n_{C_2{H}_4}=\frac{PV}{RT}$......(1)

The proportion of the number of moles of ethene to methane is 2:1. The volume proportion will be the same as the proportion of the number of moles. It will be 2:1.

$V_{C_2{H}_4}=\frac{2}{3}\times 3.67$ $V_{C{H}_4}=\frac{1}{3}\times 3.67$

Substitute values in equation (i) for ethene and for methane.

$n_{C_2{H}_4}=\frac{1\times 2\times 3.67}{0.082\times 3\times 298}$ $n_{C{H}_4}=\frac{1\times 3.67}{3\times 0.082\times 298}$

Heat evolved during combustion of ethene is $=\frac{2\times 3.67}{3\times 0.082\times 298}\times \left(1400\right)$

Heat evolved during combustion of methane is $=\frac{1\times 3.67}{3\times 0.082\times 298}\times 900$

Total heat evolved on burning this mixture is $=140+45=185kJ$

Note: Heat evolved during combustion of n moles of the substance is the product of the number of moles of the substance and the heat evolved during combustion of 1 mole of the substance.

Hence, the correct option is $\text{C}$