NEET Practice Questions (MCQs) with Answers & Solutions

To find the boiling temperature at 1 atm, use the Gibbs free energy equation at equilibrium: $\Delta G = \Delta H - T\Delta S = 0$. \n\nGiven: $\Delta H = 30\,kJ\,mol^{-1}$ and $\Delta S = 75\,J\,mol^{-1}\,K^{-1} = 0.075\,kJ\,mol^{-1}\,K^{-1}$.\n\nSet $\Delta G$ to zero and solve for $T$: \n$0 = 30\,kJ\,mol^{-1} - T(0.075\,kJ\,mol^{-1}\,K^{-1})$.\n\n$T = \frac{30\,kJ\,mol^{-1}}{0.075\,kJ\,mol^{-1}\,K^{-1}} = 400\,K$.\n\nThus, the boiling temperature at 1 atm is 400 K. The correct option is $o1$.

Spontaneous adsorption of a gas on a solid surface is generally an exothermic process. During adsorption, gas molecules adhere to the surface, resulting in a decrease in disorder (entropy) of the gas molecules because they are more confined. This leads to a decrease in the entropy of the system.\n\nTherefore, the correct option is $o3$, stating that entropy decreases during spontaneous adsorption.

From first  law of Thermodynamics, $ \triangle E  =  q +  w  \Rightarrow  nC_vdT = nCdT – PdV......	(1) $ 

Now according to process, P = V and according to ideal gas equation, PV = nRT We have, V2 = nRT $ On differentiating, 2VdV = nRdT and PdV = VdV = { nRdT \over 2 } $

So from first equation we have, $ nC_vdT = nCdT – { nRdT \over 2 } $ So, $ C_v = C – { R \over 2 } $ $ Hence C = { 4R \over 2 } $

In a bomb calorimeter, the volume is constant, so no work is done (w = 0). For the reaction of zinc with sulfuric acid: $$ ext{Zn(s) + H}_2 ext{SO}_4(aq) ightarrow ext{ZnSO}_4(aq) + ext{H}_2(g) $$, the internal energy change $$ riangle U $$ is negative because the reaction releases energy (exothermic reaction). Hence, the correct option is: $$ riangle U < 0, w = 0 $$.

To find the change in enthalpy for the reaction, we use the enthalpies of formation of the reactants and products. The reaction given is: $$2Al + Cr_2O_3 ightarrow 2Cr + Al_2O_3$$ The enthalpy change of the reaction ($ riangle H$) can be calculated using the formula: $$ riangle H = ext{Sum of enthalpies of formation of products} - ext{Sum of enthalpies of formation of reactants}$$ For the given reaction: Products: $Al_2O_3$ with enthalpy of formation = -1596 kJ Reactants: $Cr_2O_3$ with enthalpy of formation = -1134 kJ So, $$ riangle H = [-1596] - [-1134] = -1596 + 1134 = -462 ext{ kJ}$$ Therefore, the correct answer is -462 kJ.

The internal energy change ($ riangle U$) for a system depends only on the initial and final states and is independent of the path taken. Since the system returns to its initial state (A), the net change in internal energy is zero. This is a fundamental concept in thermodynamics: $$ riangle U_{ ext{net}} = riangle U_{ ext{A to B}} + riangle U_{ ext{B to A}} = 40 ext{ kJ} + (-40 ext{ kJ}) = 0$$ Hence, the net change in internal energy is zero.

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.