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Question: For complete combustion of ethanol: The amount of heat produced as measured in a bomb calorimeter ...

For complete combustion of ethanol:
The amount of heat produced as measured in a bomb calorimeter is 1364.47kJ/mol1364.47\,kJ/mol at 250C{25^0}C .
Assuming ideality, the enthalpy of combustion, ΔcH{\Delta _c}H of the reaction will be:
C2H5OH(l)+3O2(g)2CO2(g)+3H2O(l){C_2}{H_5}OH(l) + 3{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l)
[R=8.314J/mol][R = 8.314\,J/mol]
(A) 1460.50kJ/mol - 1460.50\,kJ/mol
(B) 1350.50kJ/mol - 1350.50\,kJ/mol
(C) 1366.95kJ/mol - 1366.95\,kJ/mol
(D) 1361.95kJ/mol - 1361.95\,kJ/mol

Explanation

Solution

Combustion is a chemical process in which a substance is burned in presence of oxygen and released heat. In combustion, the heat energy that is produced is mainly used to enhance the energy of the system or you can say some useful work.

Complete step by step answer:
-The enthalpy of combustion of a substance is given as the amount of heat energy released out when one mole of a substance is burned completely in presence of oxygen.
-The equation of reaction of combustion of ethanol is-
C2H5OH(l)+3O2(g)2CO2(g)+3H2O(l){C_2}{H_5}OH(l) + 3{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l)
Given, amount of heat measured is 1364.47kJ/mol1364.47\,kJ/mol and [R=8.314J/mol][R = 8.314\,J/mol]
We have to calculate the enthalpy of combustion.
-In a thermodynamic system, the change in enthalpy is given as:
ΔH=ΔE+PΔV\Delta H = \Delta E + P\Delta V
Where, ΔE\Delta E ==change in internal energy and PΔV=P\Delta V = amount of useful work.

Here T=T = Temperature, which is 250C{25^0}C
Temperature in Kelvin=25+273=298KKelvin = 25 + 273 = 298K
Δng{\Delta _n}g is the change in number of moles of gaseous molecules, which is calculated by subtracting the number of moles of products from the number of moles of reactants. It will be;
Δng=23=1{\Delta _n}g = 2 - 3 = - 1
-From the given formula calculate the enthalpy of combustion is given as;
ΔH=ΔE+PΔV\Delta H = \Delta E + P\Delta V
ΔH=ΔE+ΔngRT\Delta H = \Delta E + {\Delta _n}gRT
ΔH=1364.47+(1×8.3141000×298)\Delta H = - 1364.47\, + \left( { - 1 \times \dfrac{{8.314}}{{1000}} \times 298} \right)
ΔH=1366.95kJ/mol\Delta H = - 1366.95\,kJ/mol
Hence, the enthalpy of combustion of ethanol will be 1366.95kJ/mol - 1366.95\,kJ/mol.
So, the correct answer is “Option A”.

ADDITIONAL INFORMATION:
In thermodynamics, Enthalpy is used as a measurement of energy in a system. The amount of enthalpy is given as the total content of heat present in the system which is equal to the internal energy of the system and the product of volume and pressure.

Note:
Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burnt in oxygen with all the reactants and products are in their standard state under standard conditions.