Question

43. The difference in delta H and delta E for the
combustion of methane at 27°C would be
1)-1800 Cal
2) -162 Cal
3) -1200 Cal
4) 0



please answer me only dont ask thanks, follow ,​

Answer 1

Answer:

c option is correct –1200 Cal

Answer 2

-1200 Cal. Is the right answer.

The balanced reaction for combustion of methane is :ch4+2o2=co2+2h2o

In such reactions, the change in the number of moles is calculated for gaseous products only.

The number of moles of CH 4,o2

ans CO are 1, 1 and 2 respectively as seen from the balanced reaction given above.

Therefore, the change in the number of moles is given by =Δn

g

=n

products

−n

reactants

=1−(2+1) =−2

We know the relation: ΔH=ΔE+Δn

g

RT, where ΔH is the enthalpy change, ΔE is the internal energy change and Δn

g

is the change in the number of moles (as per the law of thermodynamics).

The value of gas constant (R) is 2 cal and temperature is 27

o

C or 300 K (given).

∴ΔH−ΔE=Δn

g

RT=−2×2×300 =−1200 calories

Hence, the difference in ΔH and ΔU for the combustion of methane at 27

o

C is −1200 cal.