Question
Question: The difference in \(\Delta {\rm H}\) and \(\Delta {\rm E}\) for the combustion of methane at \({25^ ...
The difference in ΔH and ΔE for the combustion of methane at 25∘C would be :
A.zero
B.−2×298×2cal
C.−2×298×3cal
D.−2×25×3cal
Solution
ΔH is the change in enthalpy and ΔE is the change in internal energy of a particular gas . Both internal energy and enthalpy change are state functions , also they both are extensive properties , that is , they both depend upon the quantity of matter contained in the system .
Complete answer:
When combustion of methane takes place the following reaction is observed
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
We cannot determine the absolute value of neither internal energy nor enthalpy . The internal energy change is the heat evolved or absorbed at constant volume whereas the enthalpy change is the heat evolved or absorbed at constant pressure , that is , ΔE=qv and ΔH=qp .
Now to calculate the difference between the enthalpy change and internal energy we can use the following formula :
ΔH−ΔE=ΔngRT (equation 1)
where , ΔH = change in enthalpy
ΔE = change in internal energy
Δng = difference between the number of moles of the gaseous products and those of the gaseous reactants .
R = universal gas constant ( r=2cal/degree/mole )
T = temperature
Now , first we will take out the value of Δng
As we can see from the equation there are 3 moles of gaseous reactants and 1 mole of gaseous product .
Therefore , Δng=1−3=−2
value of R is taken in cal/degree/mole
Temperature = 25∘C=(273+25)=298K
So , now on substituting the values in equation 1 , we get
ΔH−ΔE=−2×298×2cal
Hence , option B is correct .
Note:
We should be very careful while calculating the value of Δng as it is the difference between the number of moles of gaseous products and the number of moles of gaseous reactants . The reactants and products which are in solid or liquid state should be considered while calculating Δng .