Test Index

NCERT Class XI Chemistry Thermodynamics Solutions

© examsnet.com
Question : 19 of 22
Marks: +1, -0
For the reaction, 2A(g)+B(g)2A_{(g)} + B_{(g)} → 2D(g)2D_{(g)}, ΔU° = –10.5 kJ and ΔS° = – 44.1 J K−1\mathrm{J\,K^{-1}}. Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously.
Solution:  
ΔH° = ΔU° + ΔngRT\Delta n_{g}RT, ΔU° = –10.5 kJ mol−1\mathrm{kJ\,mol^{-1}}, R = 8.314 J K−1 mol−1\mathrm{J\,K^{-1}\,mol^{-1}}, T = 298 K Δng\Delta n_{g} = 2 – (2 + 1) = –1
∴ ΔH° = –10.5 – (1) × 8.314 × 10−310^{-3} × 298 = –10.5 – 2.48 = –12.98 kJ mol−1\mathrm{kJ\,mol^{-1}}
Now ΔG° = ΔH° – TΔS° = –12.98 – 298 × (–44.1 × 10−310^{-3})
= –12.98 +13.14 = 0.16 kJ mol−1\mathrm{kJ\,mol^{-1}}
Since DG° is positive, the reaction is not spontaneous.
© examsnet.com
Go to Question:
Prev Question
Next Question