NEET Practice Questions (MCQs) with Answers & Solutions

The given reaction is H_2(g) + I_2(g) → 2HI with ΔH = 12.40 kcal. This means that the formation of 2 moles of HI releases 12.40 kcal. Therefore, the heat of formation of 1 mole of HI is half of this value, which is 12.40 kcal / 2 = 6.20 kcal. Hence, the correct option is 6.20 kcal.

The heat of combustion of carbon monoxide (CO) to carbon dioxide (CO_2) can be determined by subtracting the heat of formation of CO from the heat of formation of CO_2: ext{ΔH}_{combustion} = -94.0 ext{ kcal} - (-26.4 ext{ kcal}) = -67.6 ext{ kcal}. Therefore, the correct option is -67.6 kcal.

The heat of conversion ( ext{ΔH}{conversion}) from rhombic sulphur to monoclinic sulphur is calculated by subtracting the heat of combustion of rhombic sulphur from the heat of combustion of monoclinic sulphur: ext{ΔH}{conversion} = -71030 ext{ cal} - (-70960 ext{ cal}) = 70 ext{ cal}. Therefore, the correct option is 70 cal.

$ W = – P \triangle V $ $ = – 1 \times 105 (1 \times 10^ {–2} – 1 \times 10 ^ {–3} ) = – 1 \times 10 ^ 5 \times 9 \times 10 ^ {–3} = – 900 J $ .

$ 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