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Question: The value of enthalpy change \[\left( {\Delta H} \right)\] for the reaction\[{C_2}{H_5}OH\left( l \r...

The value of enthalpy change (ΔH)\left( {\Delta H} \right) for the reactionC2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l){C_2}{H_5}OH\left( l \right) + 3{O_2}\left( g \right) \to 2C{O_2}\left( g \right) + 3{H_2}O\left( l \right) at 270C{27^0}C is −1366.5kJ(mol)−1 - 1366.5kJ{\left( {mol} \right)^{ - 1}} The value of internal energy for the above reaction at this temperature will be:
A.−1369.0kJ - 1369.0kJ
B.−1364.0kJ - 1364.0kJ
C.−1361.5kJ - 1361.5kJ
D.−1371.5kJ - 1371.5kJ

Explanation

Solution

The change in internal energy can be calculated from the change in the enthalpy of a reaction, the change in number of moles of gaseous reactants, universal gas constant and temperature in kelvins. The above all the terms are related as follows in the below equation.
Formula used:
ΔH=ΔE+ΔngRT\Delta H = \Delta E + \Delta {n_g}RT
ΔH\Delta H is change in enthalpy
ΔE\Delta E is change in internal energy
Δng\Delta {n_g} is change in number of moles of gaseous products and gaseous reactants
R is universal gas constant
T is temperature in kelvin

Complete answer:
Given that the change in enthalpy is −1366.5kJ(mol)−1 - 1366.5kJ{\left( {mol} \right)^{ - 1}}
The change in number of moles of gaseous products and reactants Δng\Delta {n_g} is 2−3=−12 - 3 = - 1
R is universal gas constant which is 8.31×10−3kJ(mol−K)−18.31 \times {10^{ - 3}}kJ{\left( {mol - K} \right)^{ - 1}}
T is temperature in kelvins which is equal to 27+273=300K27 + 273 = 300K
Substitute all the values in the above formula
−1366.5=ΔE+(−1)×8.31×10−3×300- 1366.5 = \Delta E + \left( { - 1} \right) \times 8.31 \times {10^{ - 3}} \times 300
By further simplification, the value of internal energy will be
ΔE=−1366.5+2.4942=−1364.0kJ\Delta E = - 1366.5 + 2.4942 = - 1364.0kJ
Thus, the value of enthalpy change (ΔH)\left( {\Delta H} \right) for the reaction C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l){C_2}{H_5}OH\left( l \right) + 3{O_2}\left( g \right) \to 2C{O_2}\left( g \right) + 3{H_2}O\left( l \right) at 270C{27^0}C is −1366.5kJ(mol)−1 - 1366.5kJ{\left( {mol} \right)^{ - 1}} The value of internal energy for the above reaction at this temperature will be −1364.0kJ - 1364.0kJ
Hence, option (B) is the correct answer.

Note:
While considering the value of universal gas constant, it should be in the kilojoules but not joules. The change in the number of moles of gaseous reactants and gaseous products only must be considered. The water and ethanol moles should not be considered as these are not gases reactants and gaseous products.