
The difference in $\Delta {\rm H}$ and $\Delta {\rm E}$ for the combustion of methane at ${25^ \circ }C$ would be :
A.zero
B.$ - 2 \times 298 \times 2cal$
C.$ - 2 \times 298 \times 3cal$
D.$ - 2 \times 25 \times 3cal$
Answer
575.7k+ views
Hint: $\Delta {\rm H}$ is the change in enthalpy and $\Delta {\rm E}$ 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
$C{H_4}(g) + 2{O_2}(g) \to C{O_2}(g) + 2{H_2}O(l)$
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 , $\Delta {\rm E} = {q_v}$ and $\Delta {\rm H} = {q_p}$ .
Now to calculate the difference between the enthalpy change and internal energy we can use the following formula :
$\Delta {\rm H} - \Delta {\rm E} = \Delta {n_g}RT$ (equation 1)
where , $\Delta {\rm H}$ = change in enthalpy
$\Delta {\rm E}$ = change in internal energy
$\Delta {n_g}$ = difference between the number of moles of the gaseous products and those of the gaseous reactants .
R = universal gas constant \[\left( {{\text{ }}r = 2cal/degree/mole{\text{ }}} \right)\]
T = temperature
Now , first we will take out the value of $\Delta {n_g}$
As we can see from the equation there are 3 moles of gaseous reactants and 1 mole of gaseous product .
Therefore , $\Delta {n_g} = 1 - 3 = - 2$
value of R is taken in $cal/\deg ree/mole$
Temperature = ${25^ \circ }C = (273 + 25) = 298K$
So , now on substituting the values in equation 1 , we get
$\Delta {\rm H} - \Delta {\rm E} = - 2 \times 298 \times 2cal$
Hence , option B is correct .
Note:
We should be very careful while calculating the value of $\Delta {n_g}$ 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 $\Delta {n_g}$ .
Complete answer:
When combustion of methane takes place the following reaction is observed
$C{H_4}(g) + 2{O_2}(g) \to C{O_2}(g) + 2{H_2}O(l)$
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 , $\Delta {\rm E} = {q_v}$ and $\Delta {\rm H} = {q_p}$ .
Now to calculate the difference between the enthalpy change and internal energy we can use the following formula :
$\Delta {\rm H} - \Delta {\rm E} = \Delta {n_g}RT$ (equation 1)
where , $\Delta {\rm H}$ = change in enthalpy
$\Delta {\rm E}$ = change in internal energy
$\Delta {n_g}$ = difference between the number of moles of the gaseous products and those of the gaseous reactants .
R = universal gas constant \[\left( {{\text{ }}r = 2cal/degree/mole{\text{ }}} \right)\]
T = temperature
Now , first we will take out the value of $\Delta {n_g}$
As we can see from the equation there are 3 moles of gaseous reactants and 1 mole of gaseous product .
Therefore , $\Delta {n_g} = 1 - 3 = - 2$
value of R is taken in $cal/\deg ree/mole$
Temperature = ${25^ \circ }C = (273 + 25) = 298K$
So , now on substituting the values in equation 1 , we get
$\Delta {\rm H} - \Delta {\rm E} = - 2 \times 298 \times 2cal$
Hence , option B is correct .
Note:
We should be very careful while calculating the value of $\Delta {n_g}$ 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 $\Delta {n_g}$ .
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