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If an endothermic reaction is nonspontaneous at freezing point of water and becomes feasible at its boiling point, then
TIPS/Formulae : ΔG = ΔH – TΔS
Since ΔG = ΔH – TΔS for an endothermic reaction, ΔH = +ve and at low temperature ΔS = + ve
Hence ΔG = (+) ΔH – T( + )ΔS and if TΔS < ΔH (at low temp) ΔG = +ve (non spon taneous) But at high temperature, reaction becomes spontaneous i.e. ΔG = –ve. because at higher temperature TΔS > ΔH.
A heat engine abosrbs heat Q_{1} at temperature T_{1} and heat Q_{2} at temperature T_{2}. Work done by the engine is J (Q_{1} + Q_{2}). This data [2002]
According to first law of thermodynamics energy can neither be created nor destroyed although it can be converted from one form to another.
NOTE : Carnot cycle is based upon this principle but ΔUring the conversion of heat into work some mechanical energy is always converted to other form of energy hence this data violates 1st law of thermodynamics.
For the reactions, [2002] 2C + O_{2} → 2CO_{2} ; ΔH = 393 J
2Zn + O_{2} → 2ZnO ; ΔH = 412 J
ΔH negative shows that the reaction is spontaneous.
Higher negative value for Zn shows that the reaction is more feasible.
The heat required to raise the temperature of body by 1 K is called [2002]
The heat required to raise the temperature of body by 1K is called thermal capacity or heat capacity.
The internal energy change when a system goes from state A to B is 40 kJ/mole. If the system goes from A to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy ? [2003]
For a cyclic process then net change in the internal energy is zero because the change in internal energy does not depend on the path.
If at 298 K the bond energies of C — H, C — C, C = C and H — H bonds are respectively 414, 347, 615 and 435 kJ mol^{–1}, the value of enthalpy change for the reaction H_{2}C = CH_{2}(g) + H_{2}(g) → H_{3}C — CH_{3}(g) at 298 K will be [2003]
CH_{2} = CH_{2}(g) + H_{2}(g) → CH_{3 } CH_{3}
Enthalpy change = Bond energy of reactants – Bond energy of products.
ΔH = 1(C = C) + 4 (C – H) + 1 (H  H)  1 (C  C)  6 (C  H)
= 1 (C = C) + 1 ( H – H) – 1 (C – C) – 2 (C– H)
= 615 + 435 – 347 – 2 × 414 = 1050 – 1175 = –125 kJ.
In an irreversible process taking place at constant T and P and in which only pressurevolume work is being done, the change in Gibbs free energy (ΔG) and change in entropy (dS), satisfy the criteria [2003]
For spontaneous reaction, dS > 0 and ΔG should be negative i.e. < 0.
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant K_{c} is [2003]
ΔGº = – RT lnK_{c} or – ΔGº = RT lnK_{c}
The enthalpy change for a reaction does not depend upon [2003]
Enthalpy change for a reaction does not depend upon the nature of intermediate reaction steps.
An ideal gas expands in volume from 1×10^{–3} to 1 × 10^{–2} m^{3} at 300 K against a constant pressure of 1×10^{5} Nm^{–2}. The work done is [2004]
w = PΔV = 10^{5} (1x10^{ 2}  1x10^{ 3}) = 900J
The enthalpies of combustion of carbon and carbon monoxide are – 393.5 and – 283 kJ mol^{–1} respectively. The enthalpy of formation of carbon monoxide per mole is
ΔH = –283.0 kJmol^{–1 }
Operating (i)  (ii), we have
ΔH = 110.5 kJmol^{–1}
Consider an endothermic reaction X → Y with the activation en er gies E_{b} and E_{f} for the backward and forward reactions, respectively. In general [2005]
Enthalpy of reaction for an endothermic reaction ΔH = +ve hence for ΔH to be negative
Consider the reaction : N_{2 }+ 3H_{2} → 2 NH_{3} carried out at constant temperature and pressure. If ΔH and ΔU are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ? [2005]
ΔH = ΔU + Δ_{n}RT for N_{2} + 3H_{2 }—→ 2 NH_{3}
Δ_{n}g = 2 – 4 = – 2
∴ΔH =ΔU2RT or ΔU = ΔH+2RT ∴ΔU >ΔH
If the bond dissociation energies of XY, X_{ 2} and Y_{2} (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 and ΔH_{f} for the for mation of XY is – 200 kJ mole ^{1} . The bond dissociation energy of X_{2} will be [2005]
X_{2}+Y_{2}→2XY
ΔH=(BE)X−X+(BE)Y−Y–2(BE)X−Y
If (BE) of X−Y = a then (BE) of (X−X)=a
and (BE) of (Y−Y)=a/2
So, ΔHf(X−Y)=−200kJ
So, −400 (for 2 moles XY)=a+a/2–2a
⇒−400=−a/2
⇒a=800kJ
The bond dissociation energy of X2=800kJ/mol
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If T_{i} is the initial temperature and T_{f} is the final temperature, which of the following statements is correct? [2006]
NOTE : In a reversible process the work done is greater than in irreversible process. Hence the heat absorbed in reversible process would be greater than in the latter case. So
T_{f}(rev.) < T_{f }(irr.)
The standard enthalpy of formation (D_{f}Hº) at 298 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 [2006]
The standard enthalpy of formation of CH_{4} is given by the equation : C(s) + 2H_{2}(g) —→ CH_{4}(g)
Hence, dissociation energy of hydrogen and enthalpy of sublimation of carbon is required.
The enthalpy changes for the following processes are listed below : [2006]
Cl_{2}(g) → 2Cl(g), 242.3 kJ mol^{–1}
I_{2}(g) → 2I(g), 151.0 kJ mol^{–1}
ICl(g) → I(g) + Cl(g), 211.3 kJ mol^{–1}I_{2}(s) → I_{2}(g), 62.76 kJ mol^{–1}
Given that the standard states for iodine and chlorine are I_{2}(s) and Cl_{2}(g), the standard enthalpy of formation for ICl(g) is : [2006]
I_{2} (s) + Cl_{2}(g) —→ 2ICl(g)
ΔA = [ΔI_{2}(s) →I_{2}(g) + ΔHI–I + ΔHCl–Cl] – 2[ΔHI – Cl]
= 151.0 + 242.3 + 62.76 –2 × 211.3 = 33.46
= 16.73 kJ / mol
(ΔH – ΔU) for the formation of carbon monoxide (CO) from its elements at 298 K is [2006] (R = 8.314 J K^{–1} mol^{–1})
For the reaction,
ΔH = ΔU + D_{n}RT or ΔH – ΔU = D_{n}RT
= 1238.78 J mol^{–1}
In conversion of limestone to lime, CaCO_{3} (s) → CaO(s)+ CO_{2}(g) the values of ΔH° and ΔS° are + 179.1 kJ mol^{1} and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ΔH° and ΔS° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is [2007]
ΔG° = ΔH° – TΔS°
For a spontaneous reaction ΔG° < 0
Assuming that water vapour is an ideal gas, the internal energy change (ΔU) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (given : molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol^{–1} and R = 8.3 J mol^{–1} K^{–1}) will be [2007]
Given ΔH = 41 kJ mol^{–1} = 41000 J mol^{–1 }
T = 100°C = 273 + 100 = 373 K, n = 1
ΔU = ΔH – D_{n}RT = 41000 – (2 × 8.314 × 373)
Identify the correct statement regarding a spontaneous process: [2007]
Spontaneity of reaction depen ds on tendency to acquire minimum energy state and maximum randomness. For a spontaneous process in an isolated system the change in entropy is positive.
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below:
(using the data,
will be [2008]
The energy involved in the conversion of
is given by
Substituting various values from given data, we get
(–349) (–381) kJmol^{–1 }
= (120 – 349 – 381) kJ mol^{–1} = – 610 kJ mol^{–1}
i.e., the correct answer is (b)
Standard entropy of X_{2}, Y_{2} and X Y_{3} are 60, 40 and 50 J K^{–1} mol^{–1}, respectively. For the reaction,
, to be at equilibrium, the temperature will be [2008]
For a r eaction to be at equilibrium ΔG = 0.
Since ΔG = ΔH – TΔS so at equilibrium
ΔH – TΔS=0 or ΔH = TΔS For the reaction
ΔH= –30kJ (given)
Calculating ΔS for the above reaction, we get
=50 – (30 + 60) JK^{–1} = – 40 JK^{–1}
At equilibrium, TΔS =ΔH [∵ ΔG= 0]
∴ T x (–40) = –30x 1000 [∵ 1kJ = 1000J]
On the basis of the following thermochemical data :
(Δ_{ f }G°H^{+}_{ (aq)}= 0) [2009]
= 57.32kJ
= –286.20kJ
The value of enthalpy of formation of OH^{–} ion at 25° C is:
Given, for reaction
(i)H_{2}O (l) —→ H+ (aq.) + OH^{–} (aq.); ΔH_{r}= 57.32 kJ
(ii)= –286.20 kJ
For reaction (i)

–286.20 = ΔH°_{f }(H_{2}O, ℓ)
On replacing this value in equ. (iii) we have
57.32 =ΔH°_{f} (OH– , aq) – (–286.20)
ΔH°_{f} = –286.20+ 57.32 = –228.88 kJ
The standard enthalpy of for mation 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 [2010]
N_{2} + 3H_{2} —→ 2 NH_{3} ΔH = 2 x46.0 kJ mol^{–1 }
Let x be the bond enthalpy of N – H bond then
[Note : Enthalpy of formation or bond formation enthalpy is given which is negative but the given reaction involves bond breaking hence values should be taken as positive.]
ΔH = S Bond energies of products – S
Bond energies of reactants 2 × – 46 = 712 + 3 × (436) – 6x; – 92 = 2020 – 6x
6x = 2020 + 92 ⇒ 6x = 2112 ⇒ x = + 352 kJ/mol
For a particular reversible reaction at temperature T, ΔH and ΔS were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when [2010]
At equilibrium ΔG = 0
Hence, ΔG = ΔH – T_{e}ΔS = 0
For a spontaneous reaction
ΔG must be negative which is possible only if ΔH < TΔS
The entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of 10 dm^{3} to a volume of 100 dm3 at 27°C is: [2011]
Entropy change for an isothermal reversible process is given by
= 2 × 8.314 × 2.303 log
= 38.3 J mol^{–1} K^{–1}
The incorrect expression among the following is : [2012]
ΔG° = ΔH° – TΔS°; – RT ℓnK = ΔH° – TΔS°
A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0ºC. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be: [JEE M 2013]
(R = 8.314 J/mol K) (ln 7.5 = 2.01)
For complete combustion of ethanol, C_{2}H_{5}OH (l) + 3O_{2}(g)—→ 2CO_{2} ( g)+ 3H_{2}O (l), the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol^{–1 }at 25ºC. Assuming ideality the enthalpy of combustion, Δ_{c}H, for the reaction will be: (R = 8.314 kJ mol^{–1}) [JEE M 2014]
C_{2} H_{5}OH(l) + 3O_{2}( g)→ 2CO_{2}( g ) + 3H_{2}O(l)
Bomb calorimeter gives ΔU of the reaction
Given, ΔU = –1364.47 kJ mol^{–1 }
Δn_{g} = – 1 ΔH = ΔU + Δn_{g}RT
= 1364.47 = – 1366.93 kJ mol^{–1}
The heats of combustion of carbon an d car bon monoxide are –393.5 and –283.5 kJ mol^{–1}, respectively. The heat of formation (in kJ) of carbon monoxide per mole is : [JEE M 2016]
Given C(s) + O_{2}(g) → CO_{2}(g); ΔH = –393.5 kJ mol^{–1 }…(i)
= –393.5 kJ mol^{–1}…(i)
–283.5 kJ mol^{–1} …(ii)
∴ Heat of formation of CO = eqn(i) – eqn(ii)
= –393.5 – (–283.5) = –110 kJ
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