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Question: The difference in \(\Delta {\rm H}\) and \(\Delta {\rm E}\) for the combustion of methane at \({25^ ...

The difference in ΔH\Delta {\rm H} and ΔE\Delta {\rm E} for the combustion of methane at 25C{25^ \circ }C would be :
A.zero
B.2×298×2cal - 2 \times 298 \times 2cal
C.2×298×3cal - 2 \times 298 \times 3cal
D.2×25×3cal - 2 \times 25 \times 3cal

Explanation

Solution

ΔH\Delta {\rm H} is the change in enthalpy and ΔE\Delta {\rm E} 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)C{H_4}(g) + 2{O_2}(g) \to C{O_2}(g) + 2{H_2}O(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\Delta {\rm E} = {q_v} and ΔH=qp\Delta {\rm H} = {q_p} .
Now to calculate the difference between the enthalpy change and internal energy we can use the following formula :
ΔHΔE=ΔngRT\Delta {\rm H} - \Delta {\rm E} = \Delta {n_g}RT (equation 1)
where , ΔH\Delta {\rm H} = change in enthalpy
ΔE\Delta {\rm E} = change in internal energy
Δng\Delta {n_g} = difference between the number of moles of the gaseous products and those of the gaseous reactants .
R = universal gas constant ( r=2cal/degree/mole )\left( {{\text{ }}r = 2cal/degree/mole{\text{ }}} \right)
T = temperature
Now , first we will take out the value of Δng\Delta {n_g}
As we can see from the equation there are 3 moles of gaseous reactants and 1 mole of gaseous product .
Therefore , Δng=13=2\Delta {n_g} = 1 - 3 = - 2
value of R is taken in cal/degree/molecal/\deg ree/mole
Temperature = 25C=(273+25)=298K{25^ \circ }C = (273 + 25) = 298K
So , now on substituting the values in equation 1 , we get
ΔHΔE=2×298×2cal\Delta {\rm H} - \Delta {\rm E} = - 2 \times 298 \times 2cal
Hence , option B is correct .

Note:
We should be very careful while calculating the value of Δng\Delta {n_g} 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\Delta {n_g} .