Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

To find the equilibrium constant (K) for the reaction, we use the relationship between the standard Gibbs free energy change ($ riangle G^ ext{°}$) and the equilibrium constant (K): $$ riangle G^ ext{°} = -RT ext{ln}(K)$$ Given: $$ riangle G^ ext{°} = -4.606 ext{ kcal} = -4606 ext{ cal}$$ $$T = 227^ ext{°C} = 227 + 273 = 500 ext{ K}$$ $$R = 2.0 ext{ cal K}^{-1} ext{ mol}^{-1}$$ Substituting these values into the equation, we get: $$-4606 = -2.0 imes 500 imes ext{ln}(K)$$ Solving for K: $$ ext{ln}(K) = rac{4606}{2.0 imes 500} = 4.606$$ $$K = e^{4.606} hickapprox 100$$ Therefore, the equilibrium constant (K) is approximately 100.

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 $