Chemistry MCQs for NEET — Practice Questions with Answers

$ Let the bond dissociation energy of XY, X_2 and Y_2 be x,x and x, KJ/mol respectively,$ $ { 1 \over 2 } X_2 + { 1 \over 2 } Y_2 \rightarrow XY ; \triangle Hf = -200 KJ mol ^ { -1} $ $ \triangle Hreaction = [(sum of bond dissociation energy of all reactants) – (sum of bond dissociation energy ofproduct)] $ $ = \left [ { { 1 \over 2 } \triangle H_ {x2} + { 1 \over 2} \triangle H_ {y2} - \triangle H_ {xy} } \right] = { x \over 2 } + { 0.5 x \over 2 } - x = - 200 $ $ \therefore x = 800 KJ mol ^ { -1} $ Second Method $ XY \rightarrow X_{(g) } + Y \triangle H _{(g)} = a + kJ / mole ; ...(i) $ $ X_2 \rightarrow 2 X \triangle H = a+kJ / mole ......(ii) $ $ Y_2 \rightarrow 2 Y \triangle H = 0.5 a kJ / mole ; ......(iii) $ $ { 1 \over 2} \times (ii) + {1 \over 2} \times (iii) - (i) , gives { 1 \over 2} X_2 + { 1 \over 2 } Y_2 \rightarrow XY ; $

$ N_2 + 3H_2 \rightarrow 2 NH_3 $ $ \triangle n = 2 -4 = -2 $ $ \triangle H = \triangle U + \trianglw n RT = \triangle U - 2 RT $ $ \triangle H \lt \triangle U $

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. $