Ncert Exercises & Solutions

Q.4. $∆ U^{θ}$ of combustion of methane is – X kJ $m o l^{- 1}$ . The value of $∆ H^{θ}$

(i) = $∆ U^{θ}$

(ii) > $∆ U^{θ}$

(iii) < $∆ U^{θ}$

(iv) = 0

NEETprep Answer:

Since

∆ H^{θ} = ∆ U^{θ} + ∆ n_{g} RT and ∆ U^{θ} = - X {kJmol}^{- 1},

∆ H^{θ} = (- X) + ∆ n_{g} RT . \Rightarrow ∆ H^{θ} < ∆ U^{θ}

Therefore, alternative (iii) is correct.