Question

Which is true for the combustion of sucrose (\[{{C}_{12}}{{H}_{22}}{{O}_{11}}\]) at 25°C?
A. \[\Delta H>\Delta E\]
B. \[\Delta H<\Delta E\]
C. \[\Delta H=\Delta E\]
D. None

Answer 1

Hint: \[\Delta H>\Delta E\] when the number of moles of products is greater than the number of moles of reactants. If \[\Delta H<\Delta E\], then the number of moles of gaseous reactants is greater than the number of moles of gaseous products and if \[\Delta H=\Delta E\], then the number of moles of gases do not change or are equal.

Complete step by step solution:
First let’s know about Heat of combustion (\[\Delta H\]) and change in Internal energy (\[\Delta E\]). Heat of combustion is the change in enthalpy when one mole of substance burns under standard conditions.
Internal energy is the total energy of the system (potential + kinetic). If there is a change in the internal energy of a system, then energy must have been exchanged between the system and the surroundings.
The relation between enthalpy (\[\Delta H\]) and internal energy (\[\Delta E\]) can be given by
\[\Delta H=\Delta E+P\Delta V\] (i)
We can also write \[P\Delta V\] as
\[P\Delta V=nRT\]
Therefore, \[P\Delta V=\Delta ngRT\]
Now putting this in equation (i),
\[\Delta H=\Delta E+\Delta ngRT\]
Where, \[\Delta H\] = enthalpy of the reaction
\[\Delta E\] = Internal energy of the reaction
\[\Delta ng\] = Change in the gaseous moles of the reaction, it can be calculated by subtracting the number of moles reactant from the number of moles of reactant. (no. of moles of product - no. of moles of reactant.)
\[R\] = gas constant
\[T\] = temperature
Sucrose is also called common sugar. It's a disaccharide made up of two monosaccharide namely glucose and fructose.
Now in the given question combustion of sucrose is taking place at 25℃.
Let’s write the chemical reaction for it:
\[{{C}_{12}}{{H}_{22}}{{O}_{11}}(s)\,+\,12{{O}_{2}}(g)\,\to \,12C{{O}_{2}}(g)\,+\,11{{H}_{2}}O(l)\]
Now change in the gaseous moles of the reaction:
Ang = No. of moles of \[C{{O}_{2}}\] - No. of moles of \[{{O}_{2}}\]
        = 12 - 12 = 0
Now \[\Delta H=\Delta E+\Delta ngRT\]
\[\Delta H=\Delta E+O\times RT\]
Therefore, \[\Delta H=\Delta E\], i.e., enthalpy is equal to internal energy.

Therefore, the correct answer to the question is option (c).

Note: Three factors that can affect the enthalpy and internal energy are:
 (1) the concentrations of the reactants and products involved in the reaction
 (2) the temperature of the system
 (3) the partial pressure of any gases involved in the reaction.