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Question: Heat of combustion of \({C_2}{H_4}\) is \( - 337K.cal.\) If \(11.35litre\) \({O_2}\) is used at STP,...

Heat of combustion of C2H4{C_2}{H_4} is −337K.cal. - 337K.cal. If 11.35litre11.35litre O2{O_2} is used at STP, in the combustion heat liberated is __________K.cal.
A. 56.1756.17
B. 14.0414.04
C. 42.0642.06
D. 27.727.7

Explanation

Solution

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.52.4kJ/mol. The heat of combustion is calculated from the standard enthalpy of formation ( ΔHf∘\Delta {H_f}^\circ ) of the substances included in the reaction.
Combustion of C2H4{C_2}{H_4}:

C2H4(g)+3O2(g)→2CO2(g)+2H2O(l){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, C2H4{C_2}{H_4} is ethene, O2{O_2} is oxygen, CO2C{O_2} is carbon dioxide and H2O{H_2}O is water.

ΔH=−337kcal\Delta H = - 337kcal

At STP, 1 mole of gas = 22.422.4L

Thus, 3 moles of a gas = 3×22.4=67.2L3 \times 22.4 = 67.2L

Amount of oxygen = 11.35litre (Given)

According to the reaction, the heat liberated when 67.267.2L of oxygen is used = 337337kcal.

When 11.35litre of oxygen is used, heat liberated = 33767.2×11.35=56.91≈56.17kcal.\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 22.422.4L at standard temperature and pressure.