4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 50 InterviewSolution

2451.	Which of the following option is correct ?A. `[(del lnK_(p))/(delT)]=(DeltaH^(@))/(RT^(2))`B. `(dellnK)/(delT)=(E_(a))/(RT^(2))`C. `[(dellnK_(p))/(delT)]=(DeltaU)/(RT^(2))`D. All of these
Answer» Correct Answer - D `(del)/(delT)lnK_(p)=(DeltaH)/(RT^(2))` `(del)/(delT)ln K= (E_(a))/(RT^(2))`

Discussion

2452.	If the enthalpy of combustion of benzene (l), carbon (s) and hydrogen (g)are `Q_(1) Q_(2)and Q_(3)` respectively, what will be enthalpy of formation of Benzene?A. `Q_(1)+6Q_(2)+Q_(3)`B. `6Q_(2)+Q_(1)+3Q_(3)`C. `6Q_(2)+3Q_(3)+Q_(1)`D. `6Q_(2)+3Q_(3)-Q_(1)`
Answer» Correct Answer - d

Discussion

2453.	Assertion (A): Pressure, volume, and temperature are all extensive properties. Reason (R ) : Extensive properties depend upon the amount and nature of the substance.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false.
Answer» Correct Answer - D

Discussion

2454.	Which of the following statement is correct ?A. The presence of reacting species in a covered beaker is an example of open systemB. There is an exchange of energy as well as matter between the system and the surroundings in a closed systemC. The presence of reactants in a closed vessel made up of copper is an example of a closed systemD. The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system
Answer» Correct Answer - C For a closed vessel made of copper, no matter can exchange between the system and the surrounding but energy exchange can occur through its walls. Presence of reaction species in a covered beaker-closed system and exchange of matter as well as energy-open-system. Presence of reactant in a closed vessel closed system and presence of reactant in thermos flask-isolated system

Discussion

2455.	Assertion : The presence of reactants in a closed vessel made of conducting material is an example of a closed system. Reason : In a closed system, there is no exchange of matter but exchange of energy is possible between the system and the surroundings.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - A

Discussion

2456.	Assertion (A): For a particular reaction, heat of combustion at constant pressure `(q_(P))` is always greater than that at constant volume `(q_(V))`. Reason (R ) : Combustion reactions are invariably accomplished by increase in number of moles.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false.
Answer» Correct Answer - D

Discussion

2457.	Calculate `DeltaG^(@)` (kJ/mol) at `127^(@)C` for a reaction with `K_("equilibrium")=10^(5)` :A. `-38.294`B. `-16.628`C. `-9.16`D. None of these
Answer» Correct Answer - A `DeltaG^(@)=-RTlnK=-8.314xx10^(-3)xx400ln10^(5)` `DeltaG^(@)=-38.294 " kJ"`

Discussion

2458.	(Figure) Shows three isothermal curves at temp `T_(1), T_(2) and T_(3)` `T_(3)gt T_(2)gt T_(1)`. A system changes its state by four paths a, b, c, and d, .Identify the path in which changes in internal energy of the system is maximum.
Answer» The change in internal energy vaties directly as the difference in temperatures of the system. In the path (d), difference in temperatures involved is `(T_(3)-T_(1))`, which is maximum. Therefore, change in internal energy would be maximum when the system follows the path (d).

Discussion

2459.	When reaction is at standard state at equilibrium, thenA. `DeltaH^(@)=0`B. `DeltaS^(@)=0`C. equilibrium constant K=0D. equilibrium constant K=1
Answer» Correct Answer - D At standard state on equilibrium `DeltaG^(@)=0` `DeltaG^(@)=0=-RTlnKimpliesK=1`

Discussion

2460.	Assertion :- At constant temp `0^(@)C` and 1 atm, the change `H_(2)O(s)rarr H_(2)O(l)Delta H` and `Delta E` both are zero. Reason :- During isothermal process H and E both remains constant.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false.
Answer» Correct Answer - D

Discussion

2461.	Which one of the following is correct when an ideal gas is expanded adaibatically and reversibly ?A. `q=0,W=nC_(v) dT, DeltaU= nC_(v)dT`B. `q=0,W=0,DeltaU = nC_(v)dT`C. `q=0,W= nC_(v)d,DeltaU =0`D. None of the above
Answer» Correct Answer - A For adiabatic expansion, `q=0` As` DeltaU = q+W :. DeltaU = W` Further , `C_(v)= ((delU)/(delT))_(V) ` or `dU = C_(v)dT` For n moles , `W = n C_(v)dT` and `dU = nC_(v) dT`

Discussion

2462.	Which of the following statement is correct ?A. The presence of reaction speciesin a covered beaker is an example of open system.B. There is an exchange of energy as well as matter between the system and the surroundings in a closed system.C. The presence of reactants in a closed vessel made up of copper is an example of a closed system.D. The presence of reactants in a thermos flask or any other closed insolated vessel isan example of a closed system.
Answer» Correct Answer - c For a closed vessel made of copper , no matter can exchange between the systemand the surroundings but energy exchange can occur through its walls.

Discussion

2463.	1.0g sample of subtance A at `100^(@)C` is added to 100 mL of`H_(2)O` at`25^(@)C`. Using separate 100 mL portions of `H_(2)O` , the procedure is repeated with substance B and then with substance C. How will the final temperatures of the water compare ? `{:("Substance","Specific heat"),(A,0.60 J g^(-1).^(@)C^(-1)),(B,0.40 J g^(-1).^(@)C^(-1)),(C,0.20 J g^(-1).^(@)C^(-1)):}`A. `T_(C) gt T_(B)gt T_(A`B. `T_(B) gt T_(A)gt T_(C)`C. `T_(A) gtT_(B)gtT_(C)`D. `T_(A) =T_(B)=T_(C)`
Answer» Correct Answer - C Specific heat is the amount of heat required to raise the temperature of 1 g of substance through `1^(@)C` or it is the heat released when temperature of 1 g of the substance falls through `1^(@)C`. Greater the specific heat of the substance, greater is th heat released which is absorbed by water and hence greater is the rise in temperature of water or greater is the temperature ofwater. As specific heats ofA, B and C are in the order `A gt B gt C`, therefore, `T_(A) gt T_(B) gt T_(C)`

Discussion

2464.	Which of the following statements is correct?A. The presence of reacting species in a covered beaker is an example of open systemB. There is an exchange of energy as well as matter between the system and the surroundings in a closed systemC. The presence of reactants in a closed vessel made up of copper is an example of closed systemD. The presence of reactants in a thermos flask or other closed insulated vessel is an example of a closed system
Answer» Correct Answer - C For a closed vessel made up of copper, no matter can be exchanged between the system and the surrounding but energy exchange can occur through its walls

Discussion

2465.	A cannot engine has efficiency `(1)/(6)`. If temperature of sink is decreased by `62^(@)C` then its efficiency becomes `(1)/(3)` then the temperature of source and sink:A. `37^(@)C`B. `62^(@)C`C. `99^(@)C`D. `12^(@)C`
Answer» Correct Answer - C Here, `eta_(1)= 1/6, eta_(2)= 2xx1/6` From `eta_(1)= 1-(T_(1))/(T_(1))` `1/6= 1-(T_(2))/(T_(1)), (T_(2))/(T_(1))= 5/6 or T_(2)= (5T_(1))/6` Again, `eta_(2)= 1-((T_(2)-62))/(T_(1))` `((T_(2)-62))/(T_(1))= 1-1/3= 2/3` `3T_(2)-186= 2T_(1)` `3xx5/6T_(1)-186= 2T_(1)` `0.5T_(1)= 186` `T_(1)=372K= 372-273= 99^(@)C`

Discussion

2466.	From the data given below at 298 K for the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Calculate the enthalpy of formation of CH4(g) at 298 K. Enthalpy of reaction is = -89.5 kJ Enthalpy of formation of CO2(g) = -393kJ mol-1 Enthalpy of formation of H2O(l) = -286.0 kJ mol-1
Answer» ΔH = AH_fCO₂(g) + 2ΔH_fH₂O(l) – ΔH_fCH₄(g) – ΔH_f;O₂(g) -890.5 kJ = -393.5 kJ +2 × -286 kJ - Δ H_fCH₄(g) - 0 ΔH_fCH₄ - 75.0 kJ

2466.

From the data given below at 298 K for the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Calculate the enthalpy of formation of CH4(g) at 298 K. Enthalpy of reaction is = -89.5 kJ Enthalpy of formation of CO2(g) = -393kJ mol-1 Enthalpy of formation of H2O(l) = -286.0 kJ mol-1

Answer»

ΔH = AH_fCO₂(g) + 2ΔH_fH₂O(l) – ΔH_fCH₄(g) – ΔH_f;O₂(g)

-890.5 kJ = -393.5 kJ +2 × -286 kJ - Δ H_fCH₄(g) - 0

ΔH_fCH₄ - 75.0 kJ

Discussion

2467.	Calculate the free energy change when 1 mole of NaCl is dissolved in water at 298 K. (Given: Lattice energy of NaCl = - 777.8 kJ mol-1), Hydration energy = -774.1 kJ mol-1 and ΔS = 0.043 kJK-1 mol-1 at 298 K.)
Answer» ΔH = Hydration energy – Lattice energy ΔH = -774.1 kJ mol^-1 (-777.8 kJ mol^-1) = 3.7 kJ mol^-1 ΔG = ΔH - TΔS = +3.7 kJ- 298 × 0.043 kJ = +3.7 kJ- 12.81 kJ ΔG = -9.11 kJ moH

2467.

Calculate the free energy change when 1 mole of NaCl is dissolved in water at 298 K. (Given: Lattice energy of NaCl = - 777.8 kJ mol-1), Hydration energy = -774.1 kJ mol-1 and ΔS = 0.043 kJK-1 mol-1 at 298 K.)

Answer»

ΔH = Hydration energy – Lattice energy

ΔH = -774.1 kJ mol^-1 (-777.8 kJ mol^-1) = 3.7 kJ mol^-1

ΔG = ΔH - TΔS = +3.7 kJ- 298 × 0.043 kJ = +3.7 kJ- 12.81 kJ

ΔG = -9.11 kJ moH

Discussion

2468.	Calculate `Delta G` (in joule) for the reaction `2A(g) to B(g)+C(g)` when mixture contains 1 mole of A, 2 moles of B and 1 mole of C at total pressure of 10 atm and 300 K. [Given : `G_(m)^(@),A(g)=40 "kJ mol"^(-1)`, `G_(m)^(@),B(g)=60 "kJ mol"^(-1)`, `G_(m)^(@),C(g)=20 "kJ mol"^(-1)`, `R=8.3 JK^(-1)mol^(-1) and "In " 2=0.7`]
Answer» Correct Answer - 1743

Discussion

2469.	4 mole `H_(2)O(g),2.5`mole `H_(2)`, 2.5 mole CO(g) and 1 mole inert gas He are kept at a total pressure of 10 bar in a flask containing C (graphite) at temperature T. Calculate `DeltaG` (inkJ) of the following reaction at the abovecondition :`C_("graphite")+H_(2)O(g)to H_(2)(g)+CO(g)` [Given : `Delta G_("formation",H_(2)O(g))^(@)=-230 " kJmol"^(-1)`, `Delta G_("formation",CO(g))^(@)=-130 " kJmol"^(-1) " and " 2.303 RT =10 kJ, "log"(5)/(4)=0.1`]
Answer» Correct Answer - 102

Discussion

2470.	Identify the four digit number abcd if it is equal to sum of code numbers of all those processes in which `Delta S gt 0`. [For example if only process P1 satisfies the condition then fill 0035 if both P1 and P6 satisfy then fill 0046.]
Answer» Correct Answer - 163

Discussion

2471.	Calculate the entropy change when `2 mol` of an idela gas expand isothermally and reversibly from an initial volume of `2 dm^(3)` to `20 dm^(3)` at `300 K`.
Answer» `DeltaS = 2.303 nR log_(10) ((V_(2))/(V_(1)))` `= 2.303 xx2xx 8.314 xx log_(10) ((10)/(i)) = 38.24 J K^(-1)`

Discussion

2472.	1 kg of water is heated from `40^(@) C "to" 70^(@)C`,If its volume remains constatn, then the change in internal energy is (specific heat of water = 4148 J `kg^(-1 K^(-1))`A. 2.44x`10^(5)J`B. 1.62x`10^(5)J`C. 1.24xx`10^(5)J`D. 2.62x`10^(5)J`
Answer» Correct Answer - C Since volume of water remains constant , then work done `DeltaW=PDV=0` According to first pair of thermodynamics `dQ=dU+dW,dU=dQ=msDeltaT` `=1xx4148xx(70-40)=4148xx30` `=124440 J=1.244xx10^(5)J`

Discussion

2473.	Specific heat of constatn pressure of a diatomic gas having molar mass `M` is approximately equal toA. `(gamma R)/(M(gamma-1))`B. `(gamma)/(RM)`C. `(M)/(R(gamma-1))`D. `(gamma RM)/(gamma +1)`
Answer» We know that: `C_(P) - C_(V) = R` `C_(P) - C_(V) = (R )/(M) ..(i)` `(C_(P))/(C_(V)) = gamma …(ii)` Substituting equation (ii) in (i), we get `C_(P) - (C_(P))/(gamma) = (R )/(M)` `C_(P) ((gamma-1))/(gamma) = (R )/(M)` `C_(P) = (R gamma)/((gamma - 1)M)`

Discussion

2474.	Rubber bands comprise of loosely packed chains of atoms.When stretched, the chain of atoms get neatly lined up in rows and hens the entrops of system decreases whereas when contacted the chains get tangled up in a mess increasing entropy of system. Also on strectching rubber band, it is observed that any substance brought in its contact gets heated up. Based on this information, answer the following questions. What should happen when a rubber band is heated ?A. It should expand and get strectched.B. It should contract.C. Bonds/Attractions between the molecules of rubber band will keep on weakening.D. The chains of moleculles will get more tangled.
Answer» Correct Answer - b,c,d

Discussion

2475.	Rubber bands comprise of loosely packed chains of atoms.When stretched, the chain of atoms get neatly lined up in rows and hens the entrops of system decreases whereas when contacted the chains get tangled up in a mess increasing entropy of system. Also on strectching rubber band, it is observed that any substance brought in its contact gets heated up. Based on this information, answer the following questions. Which of the following statement (s) is/are true regarding the change `"Rubber band"_("streched") rarr "Rubberband"_("unstreached")`?A. The change is spontaneous.B. The process is endothrmic in nature.C. Entropy of the rubber band is increasing in the process.D. Entropy of surroundings is decreacing in the process.
Answer» Correct Answer - a,b,c,d

Discussion

2476.	The efficiency of a heat engine cannot be 100%. Explain why?
Answer» The efficiency of heat engine ƞ =1 - T₂/T₁ The efficiency will be 100% or 1, if T₂ = 0 K. Since the temperature of 0 K cannot be reached, a heat engine cannot have 100% efficiency.

2476.

The efficiency of a heat engine cannot be 100%. Explain why?

Answer»

The efficiency of heat engine ƞ =1 - T₂/T₁

The efficiency will be 100% or 1, if T₂ = 0 K.

Since the temperature of 0 K cannot be reached, a heat engine cannot have 100% efficiency.

Discussion

2477.	Air pressure in a car increases during driving. Explain Why?
Answer» During driving as a result of the friction between the tyre and road ,the temperature of The tyre and the air inside it increases. Since volume of the tyre does not change, due to increase in temperature ,pressure of the increases (due to pressure law ).

Discussion

2478.	Can the whole of work be converted into heat?
Answer» Yes ,Through friction.

Discussion

2479.	In a Carnot engine, temperature of the sink is increased. What will happen to its efficiency?
Answer» We know ƞ = 1 – T₂/T₁ On increasing the temperature of the sink (T₂), the efficiency of the Carnot engine will decrease

2479.

In a Carnot engine, temperature of the sink is increased. What will happen to its efficiency?

Answer»

We know ƞ = 1 – T₂/T₁

On increasing the temperature of the sink (T₂), the efficiency of the Carnot engine will decrease

Discussion

2480.	What happen to the internal energy of a gas during (i) isothermal expansion (ii) adiabatic Expansion?
Answer» (i) In isothermal expansion ,temperature remains constant.Therefore internal energy which is a function of temperature will remain constant. (ii)for adiabatic change dQ = 0 and hence first law of thermodynamics becomes 0 = dU + dW dW = - dU During expansion, work is done by the gas i.e. dW is positive. Hence ,dU must be negative. Thus ,in an adiabatic expansion , the internal energy of the system will decrease.

2480.

What happen to the internal energy of a gas during (i) isothermal expansion (ii) adiabatic Expansion?

Answer»

(i) In isothermal expansion ,temperature remains constant.Therefore internal energy which is a function of temperature will remain constant.

(ii)for adiabatic change dQ = 0 and hence first law of thermodynamics becomes

0 = dU + dW

dW = - dU

During expansion, work is done by the gas i.e. dW is positive. Hence ,dU must be negative.

Thus ,in an adiabatic expansion , the internal energy of the system will decrease.

Discussion

2481.	What is the coefficient of performance (β) or a Carnot refrigerator working 30°C and 0°C?
Answer» Given: T₂ = 0°C = 273 K T₁ = 30°C = 273 + 30 = 303 K β = ? From the relation, β = \(\frac{T_2}{T_1-T_2},\) or, β = \(\frac{273}{303-273}\) = \(\frac{273}{30}\) = 9.1

2481.

What is the coefficient of performance (β) or a Carnot refrigerator working 30°C and 0°C?

Answer»

Given:

T₂ = 0°C = 273 K

T₁ = 30°C = 273 + 30 = 303 K

β = ?

From the relation,

β = \(\frac{T_2}{T_1-T_2},\)

or, β = \(\frac{273}{303-273}\)

= \(\frac{273}{30}\)

= 9.1

Discussion

2482.	At `0^(@)C` and normal atmospheric pressure, the volume of 1 gram of water increases from `1 c.c "to" 1.091 c.c` on freezing. What will be the change in its internal energy? Normal atmospheric pressure is `1.013xx10^(5)N//m^(2)` and latent heat of melting of ice `= 80 cal//gram`.
Answer» Correct Answer - `-80.0022 cal` Here, `dV= 1.091-1= 0.091 c c` `=0.091xx10^(6)m^(3)` `P=1 atm = 1.013xx10^(5)N//m^(2) : L = 80 cal//g` Heat given out by `1 g` of water on frezzing, `dQ= -mL = -1xx80 cal` External work done by water in freezing `dW= pdV= 1.013xx10^(5)xx0.091xx10^(-6)` `=0.0092= (0.0092)/(4.18)cal. = 0.0022 cal`. `dU= dQ-dW= -80-0.0022 cal`. `dU= dQ-dW= -80-0.0022= -80.0022 cal`.

Discussion

2483.	In a refrigerator, one removes heat from a lower temperature and deposites to the surrounding at a higher temperature. In this process, machanical work has to be done, which is provided by an elecrtic motor. If the motor is of `1KV` power, and heat is transferred from `-3^(@)C "to" 27^(@)C`, find the heat taken out of the refrigerator per second assuming its efficiency is `50%` of a perfect engine.
Answer» Here, power of motor, `W= 1kW` `T_(1)= 27^(@)C= (27+273)K= 300 K, T_(2)= -3^(@)C= (-3+273)K= 270K` `eta = 1-(T_(2))/(T_(1))= 1-(270)/(300)= 1/(10)`. Efficiency of refrigerator `= 0.5 eta= 1/(20)` `COP= beta=(Q_(2))/W= (1-eta)/(eta)= (1-1/20)/(1//20)=19` `Q_(2)= 19W = 19W :.` Heat taken out of refrigerator/sec `= Q_(2)= 19kW`

Discussion

2484.	Assertion : The internal energy of an isothermal process does not change. Reason : The internal energy of a system depends only on pressure of the system.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of t he assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false
Answer» Correct Answer - C

Discussion

2485.	Statement-1 : In an adiabatic process, change in internal energy of a gas is equal to work done on/by the gas in the process. Statement-2 : This is because temp.of gas remains constant in an adiabatic process.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of t he assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false
Answer» Correct Answer - C

Discussion

2486.	Consider a process shown in the figure. During this process the work done by the system A. Continuously increasesB. Continuously decreasesC. First increases, then decreasesD. First decreases, then increases
Answer» Correct Answer - A

Discussion

2487.	In the figure given two processes `A and B` are shown by which a thermodynamic system goes from initial to final state F. if `DeltaQ_(A)` and `DeltaQ_(B)` are respectively the heats supplied to the systems then A. `DeltaQ_(A) = DeltaQ_(B)`B. `DeltaQ_(A) ge DeltaQ_(B)`C. `DeltaQ_(A) lt DeltaQ_(B)`D. `DeltaQ_(A) gt Delta_(B)`
Answer» Correct Answer - b

Discussion

2488.	The `K_(P)` for reaction `A + B iff C + D ` is 1.34 at `60^(@)C` and 6.64 at ` 100 ^(@)C` . Determine the standard free energy change of this reaction at each temperature and `DeltaH^(@)` for the reaction over this range of temperature ?
Answer» Correct Answer - `- 810 J//mol ; - 5872 J//ml and 41.3 kJ //mol`. `DeltaG=- 2.303 RT log K` `DeltaG_(1) = - 810J//mol` `DeltaG_(2) = - 5872J//mol` `log .(K_(2))/(K_(1))= (DeltaH)/(2.303 R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `DeltaH = 41.3 kJ mol^(-1)`

Discussion

2489.	In the reaction `CS_(2)(l)+3O_(2)(g)rarrCO_(2)(g)+2SO_(2)(g)DeltaH= -265 kcal` The enthalpies of formation of `CO_(2)` and `SO_(2)` are both negative and are in the ratio `4:3`. The enthalpy of formation of `CS_(2)` is `+26 kcal//mol`. Calculate the enthalpy of formation of `SO_(2)`.A. `-90 kcal//mol`B. `-52 kcal//mol`C. `-78 kcal//mol`D. `-71.7 kcal//mol`
Answer» Correct Answer - D

Discussion

2490.	An ideal gas expands from `1xx10^(-3)m^(3)` to `1xx10^(-2)m^(3)` at `300K` againts a constant pressure of `1xx10^(5)Nm^(-2)` . The work done is :A. `900 k J`B. `-900 k J`C. `270 k J`D. `-900 k J`
Answer» Correct Answer - d Work done during expansion (or by the system) `=-PxxDeltaV=-1xx10^(5)[10^(-2)-10^(-3)]` `=-1xx10^(5)xx9xx10^(-3)=-900J`

Discussion

2491.	For a spontaneous reaction, `DeltaG` , equilibrium constant `(K)` and `E_(cell)^(0)` will be respectively:A. `-ve, gt1,+ve`B. `+ve,gt1,-ve`C. `-ve,lt1,-ve`D. `-ve,gt1,-ve`
Answer» Correct Answer - a For a spontaneous reaction `DeltaG=-ve`. Aslo `Kgt1` and `E_("cell")^(@)=+ve`

Discussion

2492.	Assertion (A): The heat of neutralisation of perchloric acid, `HCIO_(4)`, with `NaOH` is same as that of `HCI` with `NaOH`. Reason (R ) : Both `HCI` and `HCIO_(4)` are strong acids.A. If both (A) and (R ) are correct, and (R ) is the correct explanation for (A).B. If the both (A) and (R ) are correct, but(R ) is not a correct explanation for (A).C. If (A) is correct, but (R ) is incorrect.D. If (A) is incorrect, but (R ) is correct.
Answer» Heat of neutralisation for strong acid and storng base is equal to `-57.3 kJ mol^(-1). HCIO_(4)` and `HCI` both are strong acid.

Discussion

2493.	Calculate the equilibrium constant Kp for the reaction given below if ΔG° =-10.632 kJ at 300 K. CO2(g) + H2(g) ⇌ CO(g) + H2O(g)
Answer» ΔG° =-2.303 RT log Kp or -10.632 x 10³ =-2.303 x 8.314 x 300 log K_p K_p = 70.95

2493.

Calculate the equilibrium constant Kp for the reaction given below if ΔG° =-10.632 kJ at 300 K. CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

Answer»

ΔG° =-2.303 RT log Kp

or -10.632 x 10³ =-2.303 x 8.314 x 300 log K_p

K_p = 70.95

Discussion

2494.	The value of `DeltaH_(O-H)` is `109 kcal mol^(-1)`. Then formation of one mole of water in gaseous state from `H(g)` and `O(g)` is accompanied byA. Release of `218 kcal` of enegryB. Release of `109 kcal` of enegryC. Absorption of `218 kcal` of enegryD. Unpredicatable
Answer» `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g)` `DeltaH =- (2 xx Delta_(O-H)H^(Theta))` `DeltaH =- 2 xx 109 =- 218 kcal`

Discussion

2495.	A certain reaction is at equilbrium at `82^(@)C and Delta H` for this reaction is 21.3 kJ. What would be the value of `Delta S` (in `JK^(-1)mol^(1)`) for the reaction?
Answer» `60.0 JK^(-1) mol^(-1)`

Discussion

2496.	If `Delta G gt 0`, then the nature of the process will be (spontaneous/non-spontaneous)?
Answer» Correct Answer - Non-spontaneous.

Discussion

2497.	For a process `Delta S_("total") = 0` and `Delta G = 0` was obtained. What does it mean?
Answer» The process is in equilibrium

Discussion

2498.	The entropy change in the conversion of water to ice at 274 K for the system is -22.13 `JK^(-1)mol^(-1)` and for surrounding is +22.05 `JK^(-1)mol^(-1)`. State whether the process is spontaneous or non-spontaneous.
Answer» Correct Answer - Non-spontaneous.

Discussion

2499.	The entropy change in the conversion of water to ice at 273 K for the system is -21.99 `JK^(-1) mol^(-1)` and that of surrounding is +21.99 `JK^(-1)mol^(-1)`. State whether the process is spontaneous or non-spontaneous.
Answer» At this temperature, water and ice will be at equilibrium

Discussion

2500.	Integral enthalpy of solution of KCI, when 1 mole of it is dissolved in 20 mole water is +15.90 kJ. When 1 mole of it is dissolved in 200 mole water. `Delta H` is 15.58 kJ. Calculate enthalpy of hydration or dilution.
Answer» Correct Answer - 2.68 kJ

Discussion

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Integral enthalpy of solution of KCI, when 1 mole of it is dissolved in 20 mole water is +15.90 kJ. When 1 mole of it is dissolved in 200 mole water. `Delta H` is 15.58 kJ. Calculate enthalpy of hydration or dilution.