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In conversion of lime–stone to lime, $ CaCO_3(s) \rightarrow CaO(s) + CO_2(g) $ the values of $OH ^\circ $ and $ OS ^\circ $ are $ +179.1 kJ mol ^ {–1} $ and 160.2 J/K respectively at 298 K and 1 bar. Assuming that $OH ^\circ $ and $OS ^ \circ $ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is :

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Explanation

$ \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 a reversible process at T = 300 K, the volume is increased from $ V_i = 1 L to V_f = 10 L$ . Calculate H if the process is isothermal

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Explanation

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.

Assuming that water vapour is an ideal gas, the internal energy change (OU) when 1 mol of water is vapourisedat 1 bar pressure and $ 100 ^\circ C $ , (Given : Molar enthalpy of vapourization of water at 1 bar and $ 373 K = 41 kJ mol ^ {–1} and R = 8.3 J mol ^ {–1} K ^{–1}) $ will be :

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Explanation

$ \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} $ .

The standard enthalpy of formation $ (OHf ^ \circ ) $ at 398 K for methane, $ CH_4(g) is 74.8 kJ mol ^ {–1} $ . The additional information required to determine the average energy for C – H bond formation would be.

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Explanation

$ 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 $ .

Standard entropy of $X_2, Y_2 and XY_3$ are $60, 40 and 50 JK^{–1} mol^{–1}$ , respectively. For the reaction,$ 1/2 X_2 + 3/2 Y_2 \rightarrow XY_3 OH = – 30 kJ $ . To be at equilibrium the temperature will be :

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Explanation

$ \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 $

On the basis of the following thermochemical data : $ (O_ƒG ^\circ H ^+_{(aq)} = 0) $ $H_2O(l) \rightarrow H+ (aq) + OH^– (aq.) ; \triangle H = 57.32 kJ $ $ H_2(g) + O_2(g) \rightarrow H_2O( l); \triangle H = – 286.20 kJ $ The value of enthalpy of formation of $OH^ - ion at 25 ^\circ C $ is :

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Explanation

$ 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. $

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $CH_3OH(l) + 3/2O2 (g) \rightarrow CO_2(g) + 2H_2O(l) $ . At 298 K, standard Gibb’s energies of formation for $ CH_3OH(l), H_2O(l) and CO_2 (g) are –166.2,–237.2 and –394.4 kJ mol^{–1} $ respectively. If standard enthalpy of combustion of methanol is $ –726kJ mol ^ {–1} $ , efficiency of the fuel cell will be :

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The standard enthalpy of formation of $ NH_3 is – 46.0 kJ mol^{–1} $ . If the enthalpy of formation of $H_2$ from its atoms is $ –436 kJ mol^{–1} $ and that of $ N_2 is –712 kJ mol^{–1} $ , the average bond enthalpy of N – H bond in $NH_3$ is

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Explanation

The average bond enthalpy of N-H bonds in NH3 can be calculated using the formula: ΔHf(NH3) = Σ(Bond enthalpies of reactants) - Σ(Bond enthalpies of products) Given: ΔHf(NH3) = -46 kJ/mol ΔHf(H2 from atoms) = -436 kJ/mol ΔHf(N2 from atoms) = -712 kJ/mol Let x be the bond enthalpy of N-H. The formation of NH3 involves breaking 1/2 N2 and 3/2 H2: ΔH = 1/2(-712) + 3/2(-436) - 3x -46 = -356 - 654 - 3x 3x = -1000 + 46 3x = -1000 + 46 x = 352 kJ/mol.

Choose the correct option about the following sentnences [T= True , F =False]

(i) Ice in contact with water constitutes a homogeneous system.

(ii) The process is known as isochoric in which the pressure remains constant throughout the change, i.e., dP = 0.

(iii) A spontaneous process is reversible in nature.

(iv) In an isolated system, one form of energy cannot be converted into another, i.e., internal energy remains constant.

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Choose the correct option about the following sentnences [T= True , F =False] (i)Molar heat capacity at constant pressure = Molar heat capacity at constant volume + POV. (ii)A spontaneous process is accompanied by a decrease in entropy. (iii) $ \triangle Hsub = \triangle Hfusion + \triangle Hvap $ . (iv)The standard heat of formation represents the formation of the compound from its elements at $ 25 ^\circ C $ and one atmospheric pressure. (v)Whenever an acid is neutralised by a base, the net reaction is

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