Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

In isolated system, the expansion of gas is carried out adiabatically. Since heat exchange between system and surrounding is not possible i.e. q = 0 and secondary wrev is always greater than wirr- therefore for reversible process there must be comparatively higher decreases in internal energy i.e. $ \triangle U $ for reversible process will be more negative. Hence, final temperature in reversible process will be smaller than irreversible process. $ \therefore( T_f ) _{irrev} \gt (T_f)_{rev} $

In an isolated system, there is no exchange of energy or matter between the system and surrounding. For a spontaneous process in an isolated system, the change in entropy is positive, i.e. S > 0. Most of the spontaneous chemical reactions are exothermic. A number of endothermic reaction are spontaneous e.g melting of ice (an endothermic process) is a spontaneous reaction. The two factors which are responsible for the spontaneity of process are (i) tendency to acquire minimum energy (ii) tendency to acquire maximum randomness

$ \triangle G ^\circ = \triangle H ^\circ – T \triangle S ^ \circ $ $ for a spontaneous process \triangle G ^ \circ \lt 0 $ $ \triangle H ^ \circ – T \triangle S ^\circ \lt 0$ $ T \triangle S ^ \circ \gt \triangle H ^ \circ $ $ T \gt { \triangle H ^ \circ \over \triangle S ^ \circ } , T \gt { 179.1 \times 1000 \over 160.2 } $ $ T \gt 1117.9 , K \approx 1118 K $ .

For an isothermal process involving an ideal gas, the change in enthalpy (ΔH) is zero. This is because enthalpy is a function of temperature for an ideal gas, and in an isothermal process, the temperature remains constant. Therefore, ΔH = 0.

$ \triangle U = \triangle H – \triangle nRT $ $ = 41000 – 1 \triangle 8.314 \triangle 373 = 41000 – 3101.122 = 37898.878 J mol ^ {–1} = 37.9 kJ mol ^ { –1} $ .

$ C + 2H_2 \rightarrow CH_4; \triangle H ^\circ = – 74.8 kJ mol ^ {–1} $ In order to calculate average energy for C – H bond formation we should know the followng data. $ C(graphite) \rightarrow C(g); \triangle H f ^ \circ = enthalpy of sublimation of carbon$ $ H_2 (g) \rightarrow 2H(g) ; \triangle H ^\circ bond dissociation energy of H_2 $ .

$ \triangle S ^\circ reaction = 50 – 1/2 (60) – 3/2 (40) = –40 JK ^ {–1} $ For reaction to be at equilibrium $ \triangle G = 0 $ $ \triangle H – T \triangle S = 0 \Rightarrow T = { \triangle H \over \triangle S } = { 30000 \over 40 } = 750 K $

$ H_2(g) + O_2(g) \rightarrow H_2O (l) \triangle H = –286.20 kJ $ $ \triangle H_r = \triangle H_f (H_2O, l ) – \triangle H_f (H_2 , g) \triangle H_f (O_2 , g) –286.20 = \triangle H_f (H_2O ( l )) $ $ So \triangle H_f (H_2O, l) = –286.20 KJ/mole $ $ H_2O (l) \rightarrow H^+ (aq) + OH^– (aq) \triangle H = 57.32 kJ $ $ \triangle H_r = \triangle H ^\circ f (H^+, aq) + \triangle H ^\circ f(OH–, aq) – \triangle H ^\circ f (H2O, l ) $ $ 57.32 = 0 + \triangle H ^ \circ f (OH^–, aq) – (–286.20) $ $ \triangle H ^\circ f (OH ^ –, aq) = 57.32 – 286.20 = –228.88 kJ. $

The average bond enthalpy of N-H bonds in NH3 can be calculated using the formula: ΔHf(NH3) = Σ(Bond enthalpies of reactants) - Σ(Bond enthalpies of products) Given: ΔHf(NH3) = -46 kJ/mol ΔHf(H2 from atoms) = -436 kJ/mol ΔHf(N2 from atoms) = -712 kJ/mol Let x be the bond enthalpy of N-H. The formation of NH3 involves breaking 1/2 N2 and 3/2 H2: ΔH = 1/2(-712) + 3/2(-436) - 3x -46 = -356 - 654 - 3x 3x = -1000 + 46 3x = -1000 + 46 x = 352 kJ/mol.