Question

Heat of combustion of ${C_2}{H_4}$ is $ - 337K.cal.$ If $11.35litre$ ${O_2}$ is used at STP, in the combustion heat liberated is __________K.cal.
A. $56.17$
B. $14.04$
C. $42.06$
D. $27.7$

Answer 1

Hint: The heat of combustion means the heat evolved when one mole of a substance is burnt in oxygen at constant volume. It depends on the number of a carbon atom.

Complete step by step answer:
The heat of combustion is used as a basis for comparing the heating value of fuels. The fuel that produces a greater amount of heat for a given cost has a more economical value. It is also used in comparing the stabilities of chemical compounds. The standard enthalpy of formation of ethene is equal to $52.4kJ/mol.$ The heat of combustion is calculated from the standard enthalpy of formation ( $\Delta {H_f}^\circ $) of the substances included in the reaction.
Combustion of ${C_2}{H_4}$:

${C_2}{H_4}\left( g \right) + 3{O_2}\left( g \right) \to 2C{O_2}\left( g \right) + 2{H_2}O\left( l \right)$

Where, ${C_2}{H_4}$ is ethene, ${O_2}$ is oxygen, $C{O_2}$ is carbon dioxide and ${H_2}O$ is water.

$\Delta H = - 337kcal$

At STP, 1 mole of gas = $22.4$L

Thus, 3 moles of a gas = $3 \times 22.4 = 67.2L$

Amount of oxygen = 11.35litre (Given)

According to the reaction, the heat liberated when $67.2$L of oxygen is used = $337$kcal.

When 11.35litre of oxygen is used, heat liberated = $\dfrac{337}{67.2} \times 11.35 = 56.91 \approx 56.17kcal.$

Therefore, the amount of heat liberation when a given amount of oxygen is used is 56.17kcal.

So, the correct answer is Option A.

Note: Heat of combustion is determined by burning a known amount of the material in a bomb calorimeter with an excess of oxygen and it can also be determined by measuring the temperature change. Remember the volume of a mole of gas is equal to $22.4$L at standard temperature and pressure.