287 + Interview Questions in THERMODYNAMICS-AND-THERMOCHEMISTRY IN GENERAL AWARENESS Page 6 InterviewSolution

251.	In the complete combustion of butanol `C_(4)H_(9)OH(l)`, if `DeltaH` is enthalpy of combustion and `DeltaE` is the heat of combustion at constant volume, thenA. `DeltaH lt DeltaE`B. `DeltaH = DeltaE`C. `DeltaH gt DeltaE`D. `DeltaH, DeltaE` relation cannot be predicted
Answer» Correct Answer - C In the complete combustion of butanol `DeltaHgtDeltaE`.

Discussion

252.	The heat of formation of CO(g) and `CO_(2)(g)` are -26.4 kcal and -94.0 kcal respectively. The heat of of combustion of carbon monoxide will beA. `+26.4` kcalB. `-67.6` kcalC. `-120.6` kcalD. `+52.8 kcal
Answer» Correct Answer - B Aim: `CO+(1)/(2)O_(2)rarrCO_(2)` `DeltaH=DeltaH_(f)^(@)(CO_(2))-[DeltaH_(f)^(@)(CO)+(1)/(2)DeltaH_(f)^(@)(O_(2))]` `=-94.0-(-26.4)=-67.6 kcal`.

Discussion

253.	If `C+O_(2)rarrCO_(2)+94.2 kcal` `H_(2)+(1)/(2)O_(2)rarrH_(2)O+68.3 kcal` `CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O+210.8 kcal`
Answer» Correct Answer - B `C+O_(2)rarrCO_(2)+94.2 " kcal ...(i)"` `H_(2)+(1)/(2)O_(2)rarrH_(2)O+68.3 " kcal ...(ii)"` On multiplication of eq. (ii) by 2 and than adding in eq. (i) `C+2H_(2)+2O_(2)rarrCO_(2)+2H_(2)O+230.8 " kcal ...(iii)"` On subtracting eq. (iii) by following eq. `CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O+210.8" kcal. we get,"` `C+2H_(2)rarrCH_(4)DeltaH=20 kcal`.

Discussion

254.	Heats of combustion `(DeltaH^(@))` for C(s), `H_(2)(g)` and `CH_(4)(g)` are `-94, -68` and -213 kcal/mol. Respectively. The value of `DeltaH^(@)` for the reaction, `C(s)+2H_(2)(g)rarrCH_(4)(g)` isA. `- 85 kcal`B. `- 111 kcal`C. `-17 kcal`D. `- 170 kcal`
Answer» Correct Answer - C `C_((s))+2H_(2(g))rarrCH_(4(g))" ....(i)"` `C_((s))+O_(2(g))rarrCO_(2(g))DeltaH=-94 " kcal mol"^(-1)" ....(ii)"` `H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l))DeltaH=-68 " kcal mol"^(-1)" ....(iii)"` `CH_(4)+3//2O_(2)rarrCO_(2)+2H_(2)O` `DeltaH=-213 " kcal mol"^(-1)" ....(iv)"`

Discussion

255.	On combustion, carbon forms two oxides CO and `CO_(2)`, Heat of formation of `CO_(2)` gas is 94.3 kcal and that of CO is 26.0 kcal. Heat of combustion of carbon isA. 26.0 kcalB. `- 94.3` kcalC. 68.3 kcalD. `- 120.3` kcal
Answer» Correct Answer - B `C+O_(2)rarrCO_(2),DeltaH=-94.3` This is also heat of formation of `CO_(2)` `C+(1)/(2)O_(2)rarrCO, DeltaH=-26.0`.

Discussion

256.	Heat of formation of `CO_(2)(g), H_(2)O(l) and CH_(4)(g)` are - 94.0, - 68.4 and - 17.9 kcal respectively. The heat of combustion of methane isA. `- 212.9` kcalB. `- 136.8` kcalC. `- 304.3` kcalD. `- 105.2` kcal
Answer» Correct Answer - A Find `DeltaH" for "CH_(4(g))+2O_(2(g))rarrCO_(2(g))+2H_(2)O_((l))`

Discussion

257.	The standard molar heat of formation of ethane, `CO_(2)` and water (l) are respectively - 21.1, - 94.1 and - 68.3 kcal. The standard molar heat of combustion of ethane will beA. `- 372` kcalB. 162 kcalC. `- 240` kcalD. 183.5 kcal
Answer» Correct Answer - A `C_(2)H_(6)+(7)/(2)O_(2)rarr2CO_(2)+3H_(2)O` substitute the values.

Discussion

258.	The values of `DeltaH` for the combustion of ethene and ethyne are `-341.1` and `-310.0` kcal respectively. Which of the following is a better fuelA. `C_(2)H_(2)`B. `C_(2)H_(4)`C. Both of theseD. None of these
Answer» Correct Answer - B `DeltaH" for "C_(2)H_(4)=-341.1 K cal` its calorific value = `(-341.1)/(28)=-12.1 kcal//g`. `DeltaH` for `C_(2)H_(2)` =-310.0 kcal its Calorific value = `(-310.0)/(26)=-11.92`, hence `C_(2)H_(2)` is a better fuel.

Discussion

259.	In the reaction `CO_(2)(g)+H_(2)(g)rarrCO(g)+H_(2)O(g),DeltaH=80 kJ` `DeltaH` is known asA. Heat of formationB. Heat of combustionC. Heat of neutralizationD. Heat of reaction
Answer» Correct Answer - D `DeltaH` for this reaction is Heat of reaction.

Discussion

260.	Thermochemical reactions `C("graphite")+(1)/(2)O_(2)(g)rarrCO(g), DeltaH =- 110.5 kJ` `CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g),DeltaH=-283.2 kJ` From the above reaction, the heat of reaction for C(graphite) `+O_(2)(g)rarrCO_(2)(g)` will beA. `- 393.7` kJB. `+ 393.7` kJC. `- 172.7` kJD. ` + 172.7` kJ
Answer» Correct Answer - A eq. (i) + eq. (ii) gives the required result.

Discussion

261.	When a liquid boils, there isA. An increase in entropyB. A decrease in entropyC. An increase in heat of vaporizationD. An increase in freee pressure
Answer» Correct Answer - A Liquid `rarr` Vapour, entropy increases.

Discussion

262.	The spontaneous flow of heat is alwaysA. From low to high pressureB. From high to high pressureC. Unidirectional from lower temperature to higher temperatureD. Unidirectional from the higher to lower temperature
Answer» Correct Answer - D Heat is always flow from the highest to lower temperature.

Discussion

263.	An irreversible process occuring isothermally in an isolated system leads toA. Zero entropyB. An increase in the total entropy of the systemC. A decrease in the total entropy of the systemD. None of these
Answer» Correct Answer - B Entropy of the system increases as the process occur irreversibly and isothermally in an isolated system.

Discussion

264.	What is the free energy change `DeltaG` when 1.0mole of water at `100^(@)C` and 1 atm pressure is converted into steam at `100^(@)C` and 1 atm pressureA. 540 calB. `-9800` calC. 9800 calD. 0 cal
Answer» Correct Answer - D At equilibrium `DeltaG=0`.

Discussion

265.	The `DeltaS` for the vaporisation of 1 mol of water is 88.3 J/mole K. The value of `DeltaS` for the condensation of 1 mol of vapour will beA. 88.3 J/mol KB. `(88.3)^(2)` J/mol KC. `-88.3` J/mol KD. `(1)/(88.3)` J/mol K
Answer» Correct Answer - C For reverse reaction sign will be change.

Discussion

266.	One mole of water at `100^(@)C` is converted into steam at a constant pressure of 1 atm. The change in entropy is [heat of vaporisation of water at `100^(@)C = 540 cal//gm`]A. 8.74B. 18.76C. 24.06D. 26.06
Answer» Correct Answer - D The entropy change = `("heat of vaporization")//("temperature")` Here, heat of vaporisation = 540 cal/gm `=540xx18 cal mol^(-1)` Temperature of water = 100 + 273 = 373 K `therefore` entropy change = `(540xx18)/(373) = 26.06 cal mol^(-1) K^(-1)`.

Discussion

267.	For which of the processes is `DeltaS` negativeA. `H_(2)(g) rarr 2H(g)`B. `N_(2)(g)1atm rarr N_(2)(g) 8 atm`C. `2SO_(3)(g) rarr 2SO_(2)(g) + O_(2)(g)`D. `C_(("diamond")) rarr C_(("graphite"))`
Answer» Correct Answer - B Processes (a) and (c ) take palce with the increase of no. of moles of gaseous species and hence the disorder or entropy increases. (b) on increasing pressure, disorder or randomness decreases and so also the entropy (d) is endothermic process and `DeltaS` is positive.

Discussion

268.	A reaction is spontaneous at low temperature but non-spontaneous at high temperature. Which of the following is true for the reactionA. `DeltaH gt 0, DeltaS gt 0`B. `DeltaH lt 0, DeltaS gt 0`C. `DeltaH gt 0, DeltaS = 0`D. `DeltaH lt 0, DeltaS lt 0`
Answer» Correct Answer - D We know that `DeltaG=DeltaH-TDeltaS` When `DeltaHlt0` and `DeltaSlt0` then `DeltaG` will be negative at low temperatures and the reaction will be spontaneous.

Discussion

269.	Choose the reaction with negative `DeltaS` valueA. `2NaHCO_(3(s))rarrNa_(2)CO_(3(s))+CO_(2(g))+H_(2)O_((g))`B. `Cl_(2(g))rarr2Cl_((g))`C. `2SO_(2(g))+O_(2(g))rarr2SO_(3(g))`D. `2KClO_(3(s))rarr2KCl_((s))+3O_(2(g))`
Answer» Correct Answer - C `DeltaS` has negative value if `Deltan_(g)=-ve` i.e. number of gaseous moles decreasing during a reaction. For (a), `Deltan_(g)=2-0=2` For (b), `Deltan_(g)=2-1=1` For (c ), `Deltan_(g)=2-3=-1` (`DeltaS` negative) For (d), `Deltan_(g)=3-0=3` For (e), `Deltan_(g)`=1-0=1

Discussion

270.	For a reaction at `25^(@)C` enthalpy change and entropy changes are `-11.7 xx 10^(3) J mol^(-1)` and `-105 J mol^(-1)K^(-1)` respectively. What is the Gibbs free energyA. 15.05 kJB. 19.59 kJC. 2.55 KjD. 22.55 kJ
Answer» Correct Answer - B `DeltaG=DeltaH-TDeltaS, T=25+273=298 K` `=-11.7xx10^(3)-298xx(-105)=19590 J=19.59 kJ`.

Discussion

271.	`underset(S^(@)(298K)K^(-1)mol^(-1))(H^(+)(aq))+underset(-10.7)(OH^(-)(aq))rarrunderset(+70)(H_(2)O(l))` Standard entropy change for the above reaction isA. `60.3 JK^(-1) mol^(-1)`B. `80.7 JK^(-1) mol^(-1)`C. `-70 JK^(-1) mol^(-1)`D. `+10.7 JK^(-1) mol^(-1)`
Answer» Correct Answer - B `underset(S^(@)(298K)K^(-)mol^(-))(H_((aq))^(+))+underset(-10.7)(OH_((aq))^(-))rarrunderset(+70)(H_(2)O_((l)))` `DeltaS^(@)(298 K)=DeltaS_(P)-DeltaS_(R)=70-(-10.7+0)` `80.7 JK^(-)mol^(-1)`

Discussion

272.	For the equilibrium `H_(2)O(l)subH_(2)O(g)` at 1 atm and 298 KA. Standard free energy change is equal to zero `(DeltaG^(@)=0)`B. Free energy change is less than zero `(DeltaG lt 0)`C. Standard free energy change is less than zero `(DeltaG^(@)lt0)`D. Standard free energy change is greater than zero `(DeltaG^(@)gt0)`
Answer» Correct Answer - B For reaction `H_(2)O(l)rarrH_(2)O(g)` `Deltan=1` means positive.

Discussion

273.	For a spontaneous change, free energy change `DeltaG` isA. PositiveB. NegativeC. ZeroD. Can be positive or negative
Answer» Correct Answer - B For spontaneous change `DeltaG=-ve`.

Discussion

274.	In a chemical reaction `DeltaH` is 150 k J and `DeltaS` is `100 J K^(-1)` at 300 K, then `DeltaG` isA. 200 k JB. 333 k JC. 300 k JD. 120 k J
Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS` `=150-300(100xx10^(-3))=150-30=120 kJ`.

Discussion

275.	For a chemical reaction, `DeltaG` will always be negative ifA. `DeltaH` and `TDeltaS` both are positiveB. `DeltaH` and `TDeltaS` both are negativeC. `DeltaH` is negative and `TDeltaS` is positiveD. `DeltaH` is positive and `TDeltaS` is negative
Answer» Correct Answer - C Gibbs-Helmholtz equation is as follows : `DeltaG=DeltaH-TDeltaS` From Gibbs-Helmholtz equation, it is clear that `DeltaG` will always be negative, if `DeltaH` is negative and `TDeltaS` is positive.

Discussion

276.	Assertion : Heat absorbed in a reaction at constant temperature and constant volume is -`DeltaG`. Reason : `DeltaG` should be negative for the reaction to be spontaneous.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 the assertionC. If assertion is true but reason is falseD. If assertion is false but reason is true.
Answer» Correct Answer - D Heat absorbed in a reaction at constant temperature and constant volume (Qv)=E.

Discussion

277.	If `DeltaG` is negative, the reaction will beA. At equilibriumB. Not possibleC. Both (a) and (b)D. Possible
Answer» Correct Answer - D For a feasible reaction `DeltaG` should be negative.

Discussion

278.	In an irreversible process taking place at constant T and P and in which only pressure-volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteriaA. `(dS)_(V,E)lt0,(dG)_(T,P)lt0`B. `(dS)_(V,E)gt0,(dG)_(T,P)lt0`C. `(dS)_(V,E)=0,(dG)_(T,P)=0`D. `(dS)_(V,E)=0,(dG)_(T,P)gt0`
Answer» Correct Answer - B `(dS)_(V,E)gt0, (dG)_(T, P) lt0`.

Discussion

279.	For a particular reversible reaction at temperature `T, DeltaH` and `DeltaS` were found to be both +ve. If is the temperature `T_(e)` at equilibrium, the reaction would be spontaneous whenA. `T=T_(e)`B. `T_(e) gt T`C. `T gt T_(e)`D. `T_(e)` is 5 time T
Answer» Correct Answer - C `DeltaG=DeltaH-TDeltaS" "[DeltaH=+ve, DeltaS=+ve]` `DeltaG=+ve-T_(e)(+ve)DeltaG=+ve-T_(e)(+ve)` if `T gt T_(e)` then `DeltaG=-ve` (spontaneous).

Discussion

280.	The incorrect expression among the following isA. `(DeltaG_("system"))/(DeltaS_("total"))=-T`B. In isothermal process, `W_("reversible")=-nRTln.(V_(f))/(V_(i))`C. `ln K=(DeltaH^(@)-TDeltaS^(@))/(RT)`D. `K=e^(-DeltaG^(@//RT))`
Answer» Correct Answer - C `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `-RT ln K=DeltaH^(@)-TDeltaS^(@)` `ln K=(DeltaH^(@)-TDeltaS^(@))/(RT)`

Discussion

281.	A Beckmann thermometer is used to measureA. High temperatureB. Low temperatureC. Normal temperatureD. All temperature
Answer» Correct Answer - B Beckmann thermometer is used to measure low temperture.

Discussion

282.	The equilibrium constant of a reaction at 298 K is `5xx10^(-3)` and at 1000 K is `2xx10^(-5)`. What is the sign of `DeltaH` for the reactionA. `DeltaH=0`B. `DeltaH` is negativeC. `DeltaH` is positiveD. None of these
Answer» Correct Answer - B Equilibrium constant decreases with temperature and `DeltaH` also decreases so it is -ve.

Discussion

283.	In an isobaric process, the ratio of heat supplied to the system (dQ) and work done by the system (dW) for diatomic gas isA. `1 : 1`B. `7 : 2`C. `7 : 5`D. `5 : 7`
Answer» Correct Answer - B `-(dQ)/(dW)=(dQ)/(dQ-dE), dE=dW+dQ, -dW=dQ-dE` `=(nc_(p)dT)/(nc_(p)dT-nc_(v)dT)=(c_(p))/((c_(p)-c_(v)))` `=(7R)/(2R){"for diatomic of gas "c_(p)=(7R)/(2)}`.

Discussion

284.	The following two reactions are known `Fe_(2)O_(3)(s)+3CO(g)rarr2Fe(s)+3CO_(2)(g), DeltaH=-26.8 kJ` `FeO(s)+CO(g)rarrFe(s)+CO_(2)(g), DeltaH=-16.5 kJ` The value of `DeltaH` for the following reaction `Fe_(2)O_(3)(s)+CO(g)rarr2FeO(s)+CO_(2)(g)` isA. `+ 10.3` kJB. `- 43.3` kJC. `- 10.3` kJD. `+ 6.2 kJ`
Answer» Correct Answer - D (a) - 2(b) i.e. - 26.8 - 2(-16.5)= 6.2 kJ.

Discussion

285.	Using the Gibbs energy change, `DeltaG^(@) =+63.3 kJ`, for the following reaction, `Ag_(2)CO_(3)(s) " "2Ag^(+)(aq)+CO_(3)^(2-)(aq)` the `K_(sp)" of "Ag_(2)CO_(3)(s)` in water at `25^(@)C` is `(R=8.314 J K^(-1)mol^(-1))`A. `2.9 xx 10^(-3)`B. `7.9 xx 10^(-2)`C. `3.2 xx 10^(-26)`D. `8.0 xx 10^(-12)`
Answer» Correct Answer - D `DeltaG^(@)=-2.303 " RT log Ksp"` `63300=-2.303xx8.314xx298 log K_(sp)` `K_(sp)=8xx10^(-12)`

Discussion

286.	Standard enthalpy of vapourisation `Delta_("vap")H^(@)` for water at `100^(@)C` si 40.66 kJ `mol^(-1)`. The internal energy of vaporization of water at `100^(@)C` (in kJ `mol^(-1)`) is (Assume water vapour to behave like an ideal gas)A. `+37.56`B. `-43.76`C. `+43.76`D. `+40.66`
Answer» Correct Answer - A `DeltaH=DeltaE+Deltan(g)RT` `40.66xx1000=DeltaE+(1)xx8.314xx373` `DeltaE=37.56 kJ mol^(-1)`

Discussion

287.	Assertion : Zeroth law can also be termed as law of thermal equilibrium. Reason : Two objects in thermal equilibrium with the third one, are in thermal equilibrium with each other.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 the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false.
Answer» Correct Answer - A Zeroth (law of temperature) can also be summarized as two objects at different temperature in thermal contact with each other tend to move towards the same temperature.

Discussion

Explore topic-wise InterviewSolutions in .

In the complete combustion of butanol `C_(4)H_(9)OH(l)`, if `DeltaH` is enthalpy of combustion and `DeltaE` is the heat of combustion at constant volume, thenA. `DeltaH lt DeltaE`B. `DeltaH = DeltaE`C. `DeltaH gt DeltaE`D. `DeltaH, DeltaE` relation cannot be predicted

The heat of formation of CO(g) and `CO_(2)(g)` are -26.4 kcal and -94.0 kcal respectively. The heat of of combustion of carbon monoxide will beA. `+26.4` kcalB. `-67.6` kcalC. `-120.6` kcalD. `+52.8 kcal

If `C+O_(2)rarrCO_(2)+94.2 kcal` `H_(2)+(1)/(2)O_(2)rarrH_(2)O+68.3 kcal` `CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O+210.8 kcal`

Heats of combustion `(DeltaH^(@))` for C(s), `H_(2)(g)` and `CH_(4)(g)` are `-94, -68` and -213 kcal/mol. Respectively. The value of `DeltaH^(@)` for the reaction, `C(s)+2H_(2)(g)rarrCH_(4)(g)` isA. `- 85 kcal`B. `- 111 kcal`C. `-17 kcal`D. `- 170 kcal`

On combustion, carbon forms two oxides CO and `CO_(2)`, Heat of formation of `CO_(2)` gas is 94.3 kcal and that of CO is 26.0 kcal. Heat of combustion of carbon isA. 26.0 kcalB. `- 94.3` kcalC. 68.3 kcalD. `- 120.3` kcal

Heat of formation of `CO_(2)(g), H_(2)O(l) and CH_(4)(g)` are - 94.0, - 68.4 and - 17.9 kcal respectively. The heat of combustion of methane isA. `- 212.9` kcalB. `- 136.8` kcalC. `- 304.3` kcalD. `- 105.2` kcal

The standard molar heat of formation of ethane, `CO_(2)` and water (l) are respectively - 21.1, - 94.1 and - 68.3 kcal. The standard molar heat of combustion of ethane will beA. `- 372` kcalB. 162 kcalC. `- 240` kcalD. 183.5 kcal

The values of `DeltaH` for the combustion of ethene and ethyne are `-341.1` and `-310.0` kcal respectively. Which of the following is a better fuelA. `C_(2)H_(2)`B. `C_(2)H_(4)`C. Both of theseD. None of these

In the reaction `CO_(2)(g)+H_(2)(g)rarrCO(g)+H_(2)O(g),DeltaH=80 kJ` `DeltaH` is known asA. Heat of formationB. Heat of combustionC. Heat of neutralizationD. Heat of reaction

Thermochemical reactions `C("graphite")+(1)/(2)O_(2)(g)rarrCO(g), DeltaH =- 110.5 kJ` `CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g),DeltaH=-283.2 kJ` From the above reaction, the heat of reaction for C(graphite) `+O_(2)(g)rarrCO_(2)(g)` will beA. `- 393.7` kJB. `+ 393.7` kJC. `- 172.7` kJD. ` + 172.7` kJ

When a liquid boils, there isA. An increase in entropyB. A decrease in entropyC. An increase in heat of vaporizationD. An increase in freee pressure

The spontaneous flow of heat is alwaysA. From low to high pressureB. From high to high pressureC. Unidirectional from lower temperature to higher temperatureD. Unidirectional from the higher to lower temperature

An irreversible process occuring isothermally in an isolated system leads toA. Zero entropyB. An increase in the total entropy of the systemC. A decrease in the total entropy of the systemD. None of these

What is the free energy change `DeltaG` when 1.0mole of water at `100^(@)C` and 1 atm pressure is converted into steam at `100^(@)C` and 1 atm pressureA. 540 calB. `-9800` calC. 9800 calD. 0 cal

The `DeltaS` for the vaporisation of 1 mol of water is 88.3 J/mole K. The value of `DeltaS` for the condensation of 1 mol of vapour will beA. 88.3 J/mol KB. `(88.3)^(2)` J/mol KC. `-88.3` J/mol KD. `(1)/(88.3)` J/mol K

One mole of water at `100^(@)C` is converted into steam at a constant pressure of 1 atm. The change in entropy is [heat of vaporisation of water at `100^(@)C = 540 cal//gm`]A. 8.74B. 18.76C. 24.06D. 26.06

For which of the processes is `DeltaS` negativeA. `H_(2)(g) rarr 2H(g)`B. `N_(2)(g)1atm rarr N_(2)(g) 8 atm`C. `2SO_(3)(g) rarr 2SO_(2)(g) + O_(2)(g)`D. `C_(("diamond")) rarr C_(("graphite"))`

A reaction is spontaneous at low temperature but non-spontaneous at high temperature. Which of the following is true for the reactionA. `DeltaH gt 0, DeltaS gt 0`B. `DeltaH lt 0, DeltaS gt 0`C. `DeltaH gt 0, DeltaS = 0`D. `DeltaH lt 0, DeltaS lt 0`

Choose the reaction with negative `DeltaS` valueA. `2NaHCO_(3(s))rarrNa_(2)CO_(3(s))+CO_(2(g))+H_(2)O_((g))`B. `Cl_(2(g))rarr2Cl_((g))`C. `2SO_(2(g))+O_(2(g))rarr2SO_(3(g))`D. `2KClO_(3(s))rarr2KCl_((s))+3O_(2(g))`

For a reaction at `25^(@)C` enthalpy change and entropy changes are `-11.7 xx 10^(3) J mol^(-1)` and `-105 J mol^(-1)K^(-1)` respectively. What is the Gibbs free energyA. 15.05 kJB. 19.59 kJC. 2.55 KjD. 22.55 kJ

`underset(S^(@)(298K)K^(-1)mol^(-1))(H^(+)(aq))+underset(-10.7)(OH^(-)(aq))rarrunderset(+70)(H_(2)O(l))` Standard entropy change for the above reaction isA. `60.3 JK^(-1) mol^(-1)`B. `80.7 JK^(-1) mol^(-1)`C. `-70 JK^(-1) mol^(-1)`D. `+10.7 JK^(-1) mol^(-1)`

For a spontaneous change, free energy change `DeltaG` isA. PositiveB. NegativeC. ZeroD. Can be positive or negative

In a chemical reaction `DeltaH` is 150 k J and `DeltaS` is `100 J K^(-1)` at 300 K, then `DeltaG` isA. 200 k JB. 333 k JC. 300 k JD. 120 k J

For a chemical reaction, `DeltaG` will always be negative ifA. `DeltaH` and `TDeltaS` both are positiveB. `DeltaH` and `TDeltaS` both are negativeC. `DeltaH` is negative and `TDeltaS` is positiveD. `DeltaH` is positive and `TDeltaS` is negative

If `DeltaG` is negative, the reaction will beA. At equilibriumB. Not possibleC. Both (a) and (b)D. Possible

For a particular reversible reaction at temperature `T, DeltaH` and `DeltaS` were found to be both +ve. If is the temperature `T_(e)` at equilibrium, the reaction would be spontaneous whenA. `T=T_(e)`B. `T_(e) gt T`C. `T gt T_(e)`D. `T_(e)` is 5 time T

The incorrect expression among the following isA. `(DeltaG_("system"))/(DeltaS_("total"))=-T`B. In isothermal process, `W_("reversible")=-nRTln.(V_(f))/(V_(i))`C. `ln K=(DeltaH^(@)-TDeltaS^(@))/(RT)`D. `K=e^(-DeltaG^(@//RT))`

A Beckmann thermometer is used to measureA. High temperatureB. Low temperatureC. Normal temperatureD. All temperature

The equilibrium constant of a reaction at 298 K is `5xx10^(-3)` and at 1000 K is `2xx10^(-5)`. What is the sign of `DeltaH` for the reactionA. `DeltaH=0`B. `DeltaH` is negativeC. `DeltaH` is positiveD. None of these

In an isobaric process, the ratio of heat supplied to the system (dQ) and work done by the system (dW) for diatomic gas isA. `1 : 1`B. `7 : 2`C. `7 : 5`D. `5 : 7`

Using the Gibbs energy change, `DeltaG^(@) =+63.3 kJ`, for the following reaction, `Ag_(2)CO_(3)(s) " "2Ag^(+)(aq)+CO_(3)^(2-)(aq)` the `K_(sp)" of "Ag_(2)CO_(3)(s)` in water at `25^(@)C` is `(R=8.314 J K^(-1)mol^(-1))`A. `2.9 xx 10^(-3)`B. `7.9 xx 10^(-2)`C. `3.2 xx 10^(-26)`D. `8.0 xx 10^(-12)`

Standard enthalpy of vapourisation `Delta_("vap")H^(@)` for water at `100^(@)C` si 40.66 kJ `mol^(-1)`. The internal energy of vaporization of water at `100^(@)C` (in kJ `mol^(-1)`) is (Assume water vapour to behave like an ideal gas)A. `+37.56`B. `-43.76`C. `+43.76`D. `+40.66`