Question

For the reaction \[{{C}_{3}}{{H}_{8}}(g)+5{{O}_{2}}(g)\to 3C{{O}_{2}}(g)+4{{H}_{2}}O(l)\] at a constant temperature, \[\Delta H-\Delta E\] is:
(a)+3RT
(b)-RT
(c)+RT
(d)-3RT

Answer 1

Hint: We know that the First Law of thermodynamics states that energy can neither be created nor destroyed. It is only converted from one form to another form. It is based on the internal energy of the system, heat transferred and the work done by the system. Enthalpy of the system is also involved.

Complete step by step solution:
According to first law of thermodynamics \[\Delta H-\Delta E=\Delta {{n}_{g}}RT\]
In the above equation, \[\Delta {{n}_{g}}\]= no. of moles gaseous of products-no. of moles gaseous of reactants.
\[\Delta E\] is the change in internal energy, \[\Delta H\] is the change in enthalpy. R is Universal Gas constant and T is Temperature in Kelvin scale.
In the reaction \[{{C}_{3}}{{H}_{8}}(g)+5{{O}_{2}}(g)\to 3C{{O}_{2}}(g)+4{{H}_{2}}O(l)\]
From the reaction we know that the Number of moles of gaseous products = 3(carbon dioxide) and
Number moles of gaseous reactants = 1(propane)+5(oxygen) = 6
Therefore, we get \[\Delta {{n}_{g}}\]= 3-6 = -3
After substituting the above value of \[\Delta {{n}_{g}}\] in \[\Delta H-\Delta E=\Delta {{n}_{g}}RT\], we get
\[\Delta H-\Delta E=-3RT\]

So, the correct answer is “Option D”.

Additional Information:
The main equation for First Law of Thermodynamics is \[\Delta E=q+w\], where q is heat transferred and w is work done. If \[\Delta H\] is known we can easily calculate \[\Delta E\]using the above equation. In case of reactions involving solids and liquids, or any combination of these, there are no appreciable changes in their volumes. In such cases, \[\Delta H=\Delta E\]. But in gases they are different. We know, \[\Delta H=\Delta E+\Delta (PV)\] here for gases, \[PV=nRT\]. Substituting the value of PV, we get \[\Delta H=\Delta E+\Delta nRT\]. For reactions that result in a net production of gas, \[\Delta {{n}_{g}}\]> 0, so \[\Delta E\]< \[\Delta H\]. In endothermic reactions, change in enthalpy is greater than 0(\[\Delta H\]> 0) so the net consumption of gas has\[\Delta {{n}_{g}}\]< 0 and \[\Delta E\]> ΔH.

Note: We have to keep in mind that the equation \[\Delta H-\Delta E=\Delta {{n}_{g}}RT\]is only applicable in the cases of gases. For solids and liquids it is \[\Delta H=\Delta E\].