Thermodynamics PYQs Set 11 (PDF)

Set 11 of 14 · Questions 301-330 of 393

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Mere Teacher · Previous Year Questions NEET Chemistry · Thermodynamics · Set 11 (Q301-330)
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301. $For a reaction to occur spontaneously$ (1) $Δ H must be negative$ (2) $Δ S must be negative$ (3) $(Δ H - TS) must be negative$ (4) $(Δ H + TS) must be negative.$ 302. $The value of entropy of solar system is$ (1) $increasing$ (2) $decreasing$ (3) $constant$ (4) $zero.$ 303. $Consider the following reactions: (i) H_{(aq)}^{+} + O H_{(aq)}^{-} = H_{2} O_{(l)}, Δ H = - X_{1} kJmo l^{- 1} (ii) H_{2 (g)} + 1/2 O_{2 (g)} = H_{2} O_{(l)}, Δ H = - X_{2} kJmo l^{- 1} (iii) C O_{2 (g)} + H_{2 (g)} = C O_{(g)} + H_{2} O_{(l)}, Δ H = - X_{3} kJmo k^{- 1} (iv) C_{2} H_{2 (g)} + 5/2 O_{2 (g)} = 2 C O_{2 (g)} + H_{2} O_{(l)}, Δ H = + X_{4} kJmo k^{- 1} Enthalpy of formation of H_{2} O_{(l)} is$ (1) $+ X_{3} kJmo 1^{- 1}$ (2) $- X_{4} kJmo 1^{- 1}$ (3) $+ X_{1} kJmo 1^{- 1}$ (4) $- X_{2} kJmo l^{- 1}$ 304. $The standard enthalpy of formation of CO (g), CO_{2} (g), N_{2} O (g) and N_{2} O_{4} (g) are respectively - 10, - 393, 81 and - 10 kJ mol^{- 1} . Enthalpy change (in kJ) of the following reaction is N_{2} O_{4} (g) + 3 CO (g) ⟶ N_{2} O (g) + 3 CO_{2} (g)$ (1) $-1058$ (2) $+1058$ (3) $-957$ (4) $+957$ 305. $Which reaction is not feasible?$ (1) $2 KI + B r_{2} \to 2 KBr + I_{2}$ (2) $2 KBr + I_{2} \to 2 KI + B r_{2}$ (3) $2 KBr + C l_{2} \to 2 KCl + B r_{2}$ (4) $2 H_{2} O + 2 F_{2} \to 4 HF + O_{2}$ 306. $When ' X' g of graphite is completely burnt in a bomb calorimeter in excess of O_{2} at 298 K and 1 atm pressure as given in the equation C (graphite) + O_{2} (g) \to CO_{2} (g) The temperature of calorimeter raised from 298 K to 302 K . If the heat capacity of the calorimeter and molar enthalpy change for the reaction at 1 atm and 298 K are 20.7 kJ K^{- 1} and - 248.4 kJ mol^{- 1}, ' X' in g is$ (1) $8$ (2) $2$ (3) $3$ (4) $4$ 307. $Identify the invalid equation.$ (1) $Δ H = Σ H_{products} - Σ H_{reactants}$ (2) $Δ H = Δ U + P Δ V$ (3) $Δ H_{(reaction)}^{o} = Δ H_{(product bonds)}^{o} - Δ H_{(reactant bocds)}^{o}$ (4) $Δ H = Δ U + Δ nRT$ 308. $Cell reaction is spontaneous when$ (1) $Δ G^{o} is negative$ (2) $Δ G^{o} is positive$ (3) $Δ E_{red}^{o} is positive$ (4) $Δ E_{red}^{o} is negative.$ 309. $A reaction occurs spontaneously if$ (1) $T Δ S < Δ H and both ΔH and ΔS are +ve$ (2) $T Δ S > Δ H and Δ H is + ve and Δ S is -ve$ (3) $T Δ S > Δ H and both Δ H and Δ S are + ve$ (4) $T Δ S = Δ H and both ΔH and ΔS are +ve$ 310. $Unit of entropy is$ (1) $J K^{- 1} mo l^{- 1}$ (2) $Jmo l^{- 1}$ (3) $J^{- 1} K^{- 1} mo l^{- 1}$ (4) $J Kmo l^{- 1}$ 311. $The following two reactions are known F e_{2} O_{3 (s)} + 3 C O_{(g)} \to 2 F e_{(s)} + 3 C O_{2 (g)}; Δ H = - 26.8 kJ Fe O_{(s)} + C O_{(g)} \to F e_{(s)} + C O_{2 (g)}; Δ H = - 16.5 kJ The value of Δ H for the following reaction F e_{2} O_{3 (s)} + C O_{(g)} \to 2 Fe O_{(s)} + C O_{2 (g)} is$ (1) $+ 10.3 kJ$ (2) $- 43.3 kJ$ (3) $- 10.3 kJ$ (4) $+ 6.2 kJ$ 312. $Given H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 285 kJ N_{2} O_{5} (g) + H_{2} O (l) \to 2 HNO_{3} (l); Δ H = - 76.6 kJ N_{2} (g) + 3 O_{2} (g) + H_{2} (g) \to 2 HNO_{3} (l); Δ H = - 348.2 kJ Calculate the Δ H of 2 N_{2} (g) + 5 O_{2} (g) \to 2 N_{2} O_{5} (g) . [TS / July 19, 2022 (I)]$ (1) $572 kJ$ (2) $419 kJ$ (3) $14.5 kJ$ (4) $26.8 kJ$ 313. $Entropy of the universe is$ (1) $constant$ (2) $zero$ (3) $continuously decreasing$ (4) $continuously increasing.$ 314. $Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) : Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always - 57 kJ mol^{- 1} . Reason (R) : Enthalpy of neutralisation is the amount of heat liberated when one mole of H^{+} ions furnished by acid combine with one mole of OH^{-} ions furnished by base to form one mole of water. In the light of the above statements, choose the correct answer from the options given below.$ (1) $(A) is true but (R) is false$ (2) $Both (A) and (R) are true but (R) is not the correct explanation of (A)$ (3) $Both (A) and (R) are true and (R) is the correct explanation of (A)$ (4) $(A) is false but (R) is true$ 315. $C_{2} H_{6} (g) + 3.5 O_{2} (g) \to 2 CO_{2} (g) + 3 H_{2} O (g) Δ S_{vap} (H_{2} O, l) = x_{1} cal K^{- 1} (boiling point = T_{1}) Δ H_{f} (H_{2} O, l) = x_{2} Δ H_{f} (CO_{2}) = x_{3} Δ_{H_{f}} (C_{2} H_{6}) = x_{4} Hence, l H for the reaction is -$ (1) $2 x_{3} + 3 x_{2} - x_{4}$ (2) $2 x_{3} + 3 x_{2} - x_{4} + 3 x_{1} T_{1}$ (3) $2 x_{3} + 3 x_{2} - x_{4} - 3 x_{1} T_{1}$ (4) $x_{1} T_{1} + X_{2} + X_{3} - x_{4}$ 316. $Given : (i) C (graphite) + O_{2} (g) \to CO_{2} (g); Δ_{r} H^{0} = xkJ mol^{- 1} (ii) C (graphite) + \frac{1}{2} O_{2} (g) \to CO (g); Δ_{r} H^{0} = ykJ mol^{- 1} (iii) CO (g) + \frac{1}{2} O_{2} (g) \to CO_{2} (g); Δ_{r} H^{0} = zkJ mol^{- 1} Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct?$ (1) $x = y + z$ (2) $z = x + y$ (3) $y = 2 z - x$ (4) $x = y - z$ 317. $If for the reaction CCl_{4} (g) \to C (g) + 4 Cl (g) the following data is given Δ_{vap} H^{\circ} CCl_{4} (l) = 30 kJ mol^{- 1}, vap = vaporization Δ_{f} H^{\circ} CCl_{4} = - 136.0 kJ mol^{- 1}, f = formation Δ_{a} H_{C} = 714.0 kJ mol^{- 1}, a = atomization Δ_{a} H^{\circ} Cl_{2} = 242.0 kJ mol^{- 1}, a = atomization the bond mean enthalpy of C - Cl in CCl_{4} (1) is$ (1) $-319$ (2) $326$ (3) $-326$ (4) $292$ 318. $From the following data at 25^{\circ} C, calculate the Δ_{r} H^{0} for H_{2} O (g) \to 2 H (g) + O (g) . reaction: Δ_{r} H^{0} (kJ mol^{- 1}) \frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) \to OH (g) 42.09 H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g) - 242 H_{2} (g) \to 2 H (g) 436 O_{2} (g) \to 2 O (g) 496$ (1) $1174 kJ mol^{- 1}$ (2) $742 kJ mol^{- 1}$ (3) $926 kJ mol^{- 1}$ (4) $690 kJ mol^{- 1}$ 319. $Which of the following is not correct?$ (1) $Δ G is negative for a spontaneous reaction$ (2) $Δ G is positive for a spontaneous reaction$ (3) $Δ G is zero for a reversible reaction$ (4) $Δ G is positive for a non-spontaneous reaction$ 320. $Three thermochemical equations are given below (i) C_{(graphite)} + O_{2 (g)} \to C O_{2 (g)}; Δ_{r} H^{O} = x kJ mo l^{- 1} (ii) C_{(graphite)} +^{1} /_{2} O_{2 (g)} \to C O_{(g)}; Δ_{r} H^{O} = y kJ mo l^{- 1} (iii) C O_{(g)} + 1 /_{2} O_{2 (g)} \to C O_{2 (g)}; Δ_{r} H^{O} = z kJ mo l^{- 1} Based on the above equations, find out which of the relationship given below is correct.$ (1) $z = x + y$ (2) $x = y + z$ (3) $y = 2 z - x$ (4) $x = y - z$ 321. $The law formulated by Dr. Nernst is$ (1) $First law of thermodynamics$ (2) $Second law of thermodynamics$ (3) $Third law of thermodynamics$ (4) $Both (a) and (b)$ 322. $At standard conditions, if the change in the enthalpy for the following reaction is - 109 kJ mol^{- 1} H_{2} (g) + Br_{2} (g) \to 2 HBr (g) Given that bond energy of H_{2} and Br_{2} is 435 kJ mol^{- 1} and 192 kJ mol^{- 1}, respectively, what is the bond energy (in kJ mol^{- 1}) of HBr?$ (1) $736$ (2) $518$ (3) $259$ (4) $368$ 323. $Isentropic process is$ (1) $adiabatic and irreversible process$ (2) $isothermal and reversible process$ (3) $adiabatic and reversible process$ (4) $isothermal and reversible for which Q = 0$ 324. $The entropy change when two moles of ideal monoatomic gas is heated from 200 to 30 0^{0} C reversibly and isochoric ally?$ (1) $\frac{3}{2} R ln (\frac{300}{200})$ (2) $\frac{5}{2} R ln (\frac{573}{273})$ (3) $3 R ln (\frac{573}{473})$ (4) $\frac{3}{2} R ln (\frac{573}{473})$ 325. $When reaction in standard state is at equilibrium, then:$ (1) $DHº = 0$ (2) $DSº = 0$ (3) $equilibrium constant K = 0$ (4) $equilibrium constant K = 1$ 326. $The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are Δ_{f} Gº[C(graphite)] = 0 kJ m o l^{-1} Δ_{f} Gº[C(diamond)] = 2.9 kJ m o l^{-1} The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 \times 1 0^{-6} m^{3} m o l^{-1} . If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is : [Useful information: 1 J = 1 kg m^{2} s^{-2}, 1 Pa = 1 kg m^{-1} s^{-2}; 1 bar = 1 0^{5} Pa]$ (1) $58001 bar$ (2) $1450 bar$ (3) $14501 bar$ (4) $29001 bar$ 327. $What will be the melting point of KCl if enthalpy change for the reaction is 7.25 J m o l^{- 1} and entropy change is 0.007 J K^{- 1} m o l^{- 1} ?$ (1) $1835.2 K$ (2) $173 K$ (3) $1035.7 K$ (4) $1285.2 K$ 328. $Mixing of non-reacting gases is generally accompanied by (non-reacting)$ (1) $Decrease in entropy$ (2) $Increase in entropy$ (3) $Change in enthalpy$ (4) $Change in free energy$ 329. $The true statement amongst the following is:$ (1) $Both S and AS are not functions of temperature.$ (2) $S is not a function of temperature but AS is a function of temperature.$ (3) $Both AS and S are functions of temperature.$ (4) $S is a function of temperature but AS is not a function of temperature.$ 330. $For a spontaneous reaction the ∆ G, equilibrium constant (K) and E_{ce l l}^{0} will be respectively$ (1) $- ve, > 1, + ve$ (2) $+ ve, > 1, - ve$ (3) $- ve, < 1, -ve$ (4) $- ve, > 1, - ve$