The Total solution for NCERT class 6-12
Comment on the thermodynamic stability of NO(g), given
N2(g) + O2(g) → NO(g) ; ΔrHθ = 90 kJ mol–1
NO(g) +O2(g) → NO2(g) : ΔrHθ= –74 kJ mol–1
Answer - 21 : -
The positive value of ΔrH indicates that heatis absorbed during the formation of NO(g). This means that NO(g) has higher energythan the reactants (N2 and O2). Hence, NO(g) is unstable.
The negative value of ΔrH indicates that heatis evolved during the formation of NO2(g) from NO(g) and O2(g). The product, NO2(g) is stabilized withminimum energy.
Hence,unstable NO(g) changes to stable NO2(g).
Calculate the entropy change in surroundingswhen 1.00 mol of H2O(l) is formed understandard conditions. ΔfHθ = –286 kJ mol–1.
Answer - 22 : -
It is given that 286 kJ mol–1 of heat is evolvedon the formation of 1 mol of H2O(l). Thus, an equal amount ofheat will be absorbed by the surroundings.
qsurr = +286 kJ mol–1