Question

Question 6.4 ΔUθof combustion of methane is – X kJ mol–1. The value of ΔHθ is

(i) = ΔUθ

(ii) > ΔUθ

(iii) < ΔUθ

(iv) = 0

Class XI Thermodynamics Page 182

Answer 1

concept :- according to 1st law of thermodynamics,
∆H°=∆U°+∆ngRT
so, we have to find firstly , ∆ng by writing chemical equation of combustion of methane.
where, ∆ng=total mole of gaseous products - total mole of gaseous reactants.


Combustion of methane:
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l)

∆ng = 1 - (1 + 2) = -2

Now,
∆H°=∆U°+∆ngRT
=-X°-2RT
=-(X°+RT) < -X°

hence, ∆H°< ∆U°

Hence, option (iii) is correct.

Answer 2

ANSWER

Reaction;

CH 4(g)+2O 2(g)⟶CO 2(g)+2H 2O(l)

Now,

Δn g =(n p−n r)=1−3=−2

Again,

ΔH o =ΔUo+Δn g RT

ΔH o=−X−2RT

∴ΔH o <ΔU o

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