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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1251. |
Calculate Q and w for the isothermal reversible expansion of one mole an ideal gas from an initial pressure of 1.0 bar to a final pressure of 0.1 bar at a constant temperature of 273 K respectively.A. 5.22kJ, -5.22 kJB. `-5.22 kJ, 5.22 kJ`C. 27.3 kJ, -27.3 kJD. `-27.3 kJ, 27.3 kJ` |
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Answer» Correct Answer - A In isothermal process change in internal energy is 0, so `q = -w` |
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| 1252. |
If the door of a refrigerator is kept open, then which of the following is trueA. gets heatedB. gets cooledC. neither get cooled nor gets heatedD. gets cooled or heated depending on the initial temperature ture of the room |
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Answer» Correct Answer - A If the door of a refrigerator is kept open, then the compressor has to run for a longer time releasing more heat to the surrounding. |
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| 1253. |
Which of the following is an exothermic reaction?A. Conversion of graphite to diamondB. Dehydrogenation of ethane to ethyleneC. Combustion of methaneD. Decomposition of water |
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Answer» Correct Answer - C Any process that gives off heat (that is, transfer thermal energy to surrounding) is called an exothermic process. Exo comes from Greek word meaning outside. Combustion is one of the many chemical reaction that release considerable quanitties of energy. In constrast, in an endothermic process, heat has to be supplied to the system by the surrounding. Endo comes from the Greek workk meaning inside. |
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| 1254. |
The following is(are) endothermic reaction (s):A. Combustion of methaneB. decomposition of waterC. Dehydrogenation of ethane to enthyleneD. Conversion of grpahite to diamond |
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Answer» Correct Answer - B::C::D All combustion reactions are accompained by evolution of heat i.e. they are exothermic. Hence (B, C, D) is correct option |
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| 1255. |
Diborane is a potential rocket fuel which undergoes combustion according to the reaction, `B_(2)H_(6) (g) + 3O_(2) (g) rarr B_(2) O_(3) (s) + 3H_(2) O (g)` From the following data, calculate the enthalpy change for the combustion of diborane: (i) `2 B (s) + ((3)/(2)) O_(2) (g) rarr B_(2) O_(3) (s) , Delta H = 0 1273 k J//mol` (ii) `H_(2) (g) + ((1)/(2)) O_(2) (g) rarr H_(2) O (l), Delta H = -286 kJ//mol` (iii) `H_(2) O (l) rarr H_(2)O (g) Delta H = 44 kJ//mol` (iv) `2B (s) + 3H_(2) (G) rarr B_(2) H_(6) (g), Delta H = 36 kJ//mol` |
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Answer» Correct Answer - `-2035 kj//`mole The concerned chemical reaction is `B_(2) H_(6 (g)) + 3O_(2 (g)) rarr B_(2)O_((s)) + 3H_(2) O_((g)), Delta H = ?` The enthalpy change can be calculated in the followingway. `Delta H = [Delta H_(B_(2) O_(3) (s)) + 3 Delta H_(H_(2) O (g))] - Delta H_(B_(2)H_(6)(s)), ( :. Delta H_(f)^(0) " of " O_(2) = 0) Delta H_(B_(2)O (g))` can be obtained by adding `Delta H_(H_(2)O(l))` and `Delta H_(H_(2) O(g))`, i.e. `-286 + 44 = - 242 kJ mol^(-1)` `Delta H = [-1273 + 3 -242] - 36 kJ mol^(-1) = - 1273 - 726 - 36 = - 2035 kJ mol^(-1)` |
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| 1256. |
When 1 mole of gas is heated at constant volume. Temperature is raised from `298` to `308K` . Heat supplied to the gas is `500J` . Then which stamenet is correct?A. `q = W = 500 J, Delta U = 0`B. `1 = Delta U = 500 J, W = 0`C. `q = W = 500 J, Delta U = 500`D. `Delta U = 0, q = W = - 500 J` |
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Answer» Correct Answer - B According to the first law of thermodynamics, `Delta U = q + W` Since `W = - P_("ext") Delta V)` we have `Delta U = q + (- P_("ext") Delta V)` At constant volume, `Delta V = 0`. Thus, `Delta U = q` Since heat is supplied to the system, `q` is positive. Thus, `Delta U = q = 500 J, W = 0` |
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| 1257. |
When 1 mole of gas is heated at constant volume. Temperature is raised from `298` to `308K` . Heat supplied to the gas is `500J` . Then which stamenet is correct?A. q=W=500J, `Delta U = 0`B. `q = Delta U = 500 J, W = 0`C. `q = W = 500J, Delta U = 0`D. `Delta U = 0, q = W = -500 J` |
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Answer» Correct Answer - B At constant volume `P Delta V = 0`, `:. Q = Delta U` |
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| 1258. |
Which of the following conditions reagarding a chemical process ensures its spontaneity at all temperature?A. `Delta H gt 0, DeltaS lt 0`B. `Delta H lt 0, DeltaS gt 0`C. `Delta H lt 0, DeltaS lt 0`D. `Delta H gt 0, DeltaS lt 0` |
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Answer» Correct Answer - B `DeltaG=DeltaH-TDeltaS,DeltaH=-ve, " "DeltaS=+ve` |
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| 1259. |
Which of the following conditions regarding a chemical process ensures its spontaneity at all temperatureA. `DeltaH gt 0, DeltaS lt O`B. `DeltaH lt 0, DeltaS gt O`C. `DeltaH lt 0, DeltaS lt O`D. `DeltaH gt 0, DeltaS lt O` |
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Answer» Correct Answer - B For spontaneity `Delta lt 0` , which is possible when `DeltaH lt 0, DeltaS gt 0.` |
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| 1260. |
When reaction is at standard state at equilibrium , then:A. `DeltaH^(@) =0`B. `DeltaS^(@) =0`C. equilibrium constant `K=0`D. equilibrium constant `K=1` |
| Answer» Correct Answer - D | |
| 1261. |
For the reaction : `C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(g)` if `Delta U^(@)= -1373 kJ mol^(-1)` at `298 K`. Calculate `Delta H^(@)` |
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Answer» Correct Answer - `-1368 kJ mol^(-1)` `DeltaH^(@)=DeltaU^(@)+Deltan_(g)RT` `= -1373+2xx(8.314xx298)/(1000)= -1368 kJ mol^(-1)` |
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| 1262. |
The difference between the heats of reaction at constant pressure and a constant volume for the reaction `2C_(6)H_(6)(l) + 15O_(2)(g) rarr 12CO_(2)(g) + 6H_(2)O(l)` at `25^(@)C` in `kJ` is |
| Answer» Correct Answer - `-7.432 kJ` | |
| 1263. |
In a thermodynamic process helium gas obeys the law `TP^(2//5)` = constant,If temperature of `2` moles of the gas is raised from `T` to `3T`, thenA. heat given to the gas is `9RT`B. heat given to the is zeroC. increase in internal energy is `6RT`D. work done by the gas is `-6RT`. |
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Answer» Correct Answer - B::C::D In adiabatic process `DeltaQ=0` `DeltaU=nc_(v)DeltaT=2((3)/(2)R)(3T-T)=6RT` ltBRgt `DeltaW=-DeltaU=-6RT , DeltaW=-DeltaU=-6RT` |
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| 1264. |
A fundamental goal of thermodynamics is the ……(a) prediction of spontaneity of the reaction. (b) determination of rate of the chemical reaction. (c) evaluation of the microscopic properties. (d) both (b) and (c) |
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Answer» (a) prediction of spontaneity of the reaction. |
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| 1265. |
Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. `DeltaG_(P.T) lt 0. DeltaG_(P.T) =0` implies the equilibrium condition and `DeltaG_(P.T) gt 0` corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : `" "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1)` The magnitude of `DeltaH` does not change much with the change in temperature but the entropy factor `TDeltaS` change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both `DeltaH` and `DeltaS` are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor `TDeltaS` will be small and may be less than `DeltaH, DeltaG` will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor `TDeltaS` Increases appreciably and when it exceeds `DeltaH, DeltaG` would become negative and the process would be spontaneous . For an expthermic process, both `DeltaH` and `DeltaS` would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when `T DeltaS gt DeltaH, DeltaG` will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor ,`TDeltaS` decreases rapidly and when `TDeltaS lt DeltaH, DeltaG` becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. The enthalpy change for a certain rection at 300 K is `-15.0` K cal `mol^(-1)` . The entropy change under these conditions is `-7.2` cal `K^(-1)mol^(-1)` . The free energy change for the reaction and its spontaneous/ non-spontaneous character will beA. `-12.84` Kcal`"mol"^(-1)`, spontaneousB. `-12.84` Kcal`"mol"^(-1)`,non-spontaneousC. `-17.16` Kcal`"mol"^(-1)`, spontaneousD. None of these |
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Answer» Correct Answer - A Use `DeltaG = DeltaH - TDeltaS` `DeltaG =- 15 -(33xx(-7.2))/(1000) =-12.84 "Kcal mol"^(-1)` |
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| 1266. |
Entropy is a state function and its value depends on two or three variable temperature (T), Pressure(P) and volume (V). Entropy change for an ideal gas having number of moles(n) can be determined by the following equation. `DeltaS=2.303 "nC"_(v)"log"((T_(2))/(T_(1))) + 2.303 "nR log" ((V_(2))/(V_(1)))` `DeltaS=2.303 "nC"_(P) "log"((T_(2))/(T_(1))) + 2.303 "nR log" ((P_(1))/(P_(2)))` Since free energy change for a process or a chemical equation is a deciding factor of spontaneity , which can be obtained by using entropy change `(DeltaS)` according to the expression, `DeltaG = DeltaH -TDeltaS` at a temperature `T` What would be the entropy change involved in thermodynamic expansion of `2` moles of a gas from a volume of `5` L to a volume of `50` L at `25^(@)` C [Given `R=8.3 J//"mole"-K]`A. `38.23` J/K molB. `26.76` J/KC. `20` J/kD. `28.23` J/K |
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Answer» Correct Answer - A `DeltaS=2.303 xx 2 xx8.3 "log"((50)/(5)) = 38.23 J//K` |
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| 1267. |
Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. `DeltaG_(P.T) lt 0. DeltaG_(P.T) =0` implies the equilibrium condition and `DeltaG_(P.T) gt 0` corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : `" "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1)` The magnitude of `DeltaH` does not change much with the change in temperature but the entropy factor `TDeltaS` change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both `DeltaH` and `DeltaS` are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor `TDeltaS` will be small and may be less than `DeltaH, DeltaG` will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor `TDeltaS` Increases appreciably and when it exceeds `DeltaH, DeltaG` would become negative and the process would be spontaneous . For an expthermic process, both `DeltaH` and `DeltaS` would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when `T DeltaS gt DeltaH, DeltaG` will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor ,`TDeltaS` decreases rapidly and when `TDeltaS lt DeltaH, DeltaG` becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. For the reaction at `298 K ,2A + B rarr C` `DeltaH =100 kcal` and `DeltaS=0.050 kcal K^(-1)`. If `DeltaH` and `DeltaS` are assumed to be constant over the temperature range, above what temperature will the reaction become spontaneous?A. 1000 KB. 1500 KC. 2000 KD. 2500 K |
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Answer» Correct Answer - C Reaction to be spontaneous `DeltaG lt 0 implies DeltaH-TDeltaS lt0 implies DeltaT gt (DeltaH)/(DeltaS) implies T gt (100xx10^(3))/(05xx10^(3)) implies T gt 2000` K. |
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| 1268. |
Entropy is a state function and its value depends on two or three variable temperature (T), Pressure(P) and volume (V). Entropy change for an ideal gas having number of moles(n) can be determined by the following equation. `DeltaS=2.303 "nC"_(v)"log"((T_(2))/(T_(1))) + 2.303 "nR log" ((V_(2))/(V_(1)))` `DeltaS=2.303 "nC"_(P) "log"((T_(2))/(T_(1))) + 2.303 "nR log" ((P_(1))/(P_(2)))` Since free energy change for a process or a chemical equation is a deciding factor of spontaneity , which can be obtained by using entropy change `(DeltaS)` according to the expression, `DeltaG = DeltaH -TDeltaS` at a temperature `T` An isobaric process having one mole of ideal gas has entropy change `23.03` J/K for the temperature range `27^(@)` C to `327^(@)` C . What would be the molar specific heat capacity `(C_(v))`?A. `(10)/("log"2) J//K mol`B. `(10)/("log"2) - 8.3 J//K` molC. `10 xx "log"2 J//K mol`D. `10 log2 + 8.3 J//k` mol |
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Answer» Correct Answer - B `DeltaS= 2.303 xx 1 xx C_(p)"log"((600)/(300)) = 23.03 implies " "C_(p) =(10)/("log"2)` `C_(v) = C_(P)-R (10)/("log"2) -8.3` |
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| 1269. |
Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. `DeltaG_(P.T) lt 0. DeltaG_(P.T) =0` implies the equilibrium condition and `DeltaG_(P.T) gt 0` corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : `" "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1)` The magnitude of `DeltaH` does not change much with the change in temperature but the entropy factor `TDeltaS` change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both `DeltaH` and `DeltaS` are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor `TDeltaS` will be small and may be less than `DeltaH, DeltaG` will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor `TDeltaS` Increases appreciably and when it exceeds `DeltaH, DeltaG` would become negative and the process would be spontaneous . For an expthermic process, both `DeltaH` and `DeltaS` would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when `T DeltaS gt DeltaH, DeltaG` will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor ,`TDeltaS` decreases rapidly and when `TDeltaS lt DeltaH, DeltaG` becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. For the reaction at `25^(@), X_(2)O_(4)(l) rarr 2XO_(2)(g)` `DeltaH=2.1 Kcal` and `DeltaS = 20 cal K^(-1)`. The reaction would beA. SpontaneousB. non-spontaneousC. at equilibriumD. Unpredictable |
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Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS=2.1 xx 10^(3) - 20 xx 298 lt 0` |
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| 1270. |
Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. `DeltaG_(P.T) lt 0. DeltaG_(P.T) =0` implies the equilibrium condition and `DeltaG_(P.T) gt 0` corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : `" "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1)` The magnitude of `DeltaH` does not change much with the change in temperature but the entropy factor `TDeltaS` change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both `DeltaH` and `DeltaS` are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor `TDeltaS` will be small and may be less than `DeltaH, DeltaG` will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor `TDeltaS` Increases appreciably and when it exceeds `DeltaH, DeltaG` would become negative and the process would be spontaneous . For an expthermic process, both `DeltaH` and `DeltaS` would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when `T DeltaS gt DeltaH, DeltaG` will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor ,`TDeltaS` decreases rapidly and when `TDeltaS lt DeltaH, DeltaG` becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. A reaction has a value of `DeltaH =-40 Kcal` at 400 k cal `mol^(-1)`. The reaction is spontaneous, below this temperature , it is not . The values fo `DeltaG` and `DeltaS` at 400 k are respectivelyA. `0,-0.1 cal K^(-1)`B. `0,100 cal K^(-1)`C. `-10` Kcal, `-100"cal"K^(-1)`D. `0,-100 cal K^(-1)` |
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Answer» Correct Answer - D Clearly at `400` K reaction is in equilibrium `implies DeltaG=0 implies DeltaS =(DeltaH)/(T) =- (40000)/(400) =- 100 cal K^(-1)` |
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| 1271. |
Entropy is a state function and its value depends on two or three variable temperature (T), Pressure(P) and volume (V). Entropy change for an ideal gas having number of moles(n) can be determined by the following equation. `DeltaS=2.303 "nC"_(v)"log"((T_(2))/(T_(1))) + 2.303 "nR log" ((V_(2))/(V_(1)))` `DeltaS=2.303 "nC"_(P) "log"((T_(2))/(T_(1))) + 2.303 "nR log" ((P_(1))/(P_(2)))` Since free energy change for a process or a chemical equation is a deciding factor of spontaneity , which can be obtained by using entropy change `(DeltaS)` according to the expression, `DeltaG = DeltaH -TDeltaS` at a temperature `T` For a reaction `M_(2)O(s) rarr 2M(s) + (1)/(2) O_(2)(g) , DeltaH = 30 KJ//mol` and `DeltaS=0.07 KJ/mole "at" 1` atm. Calculate upto which temperature the reaction would not be spontaneous.A. `T gt 428.6` KB. `T gt 300.8` KC. `T lt 300.8` KD. `T lt 428.6` K |
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Answer» Correct Answer - D For reaction to be spontaneous, `DeltaG lt 0= DeltaH-TDeltaS lt 0` `implies T gt (DeltaH)/(DeltaS) implies T gt 426.6 K.` if `T lt 426.6 K` so, reaction is non spontaneous. |
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| 1272. |
Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. `DeltaG_(P.T) lt 0. DeltaG_(P.T) =0` implies the equilibrium condition and `DeltaG_(P.T) gt 0` corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : `" "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1)` The magnitude of `DeltaH` does not change much with the change in temperature but the entropy factor `TDeltaS` change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both `DeltaH` and `DeltaS` are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor `TDeltaS` will be small and may be less than `DeltaH, DeltaG` will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor `TDeltaS` Increases appreciably and when it exceeds `DeltaH, DeltaG` would become negative and the process would be spontaneous . For an expthermic process, both `DeltaH` and `DeltaS` would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when `T DeltaS gt DeltaH, DeltaG` will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor ,`TDeltaS` decreases rapidly and when `TDeltaS lt DeltaH, DeltaG` becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. When `CaCO_(3)` is heated to a high temperature , it undergoes decomposition into CaO and `CO_(2)` whereas it is quite stable at room temperature . The most likely explanation of it, isA. The enthalpy of reaction `(DeltaH)` overweighs the term `TDeltaS` at high temperature.B. The term `TDeltaS` overweighs the enthalpy of reaction at high temperature.C. At high temperature , both enthalpy of reaction and entropy change become negative.D. None of these |
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Answer» Correct Answer - B `DeltaG=DeltaH-TDeltaS` If `TDeltaS` dominates `DeltaH` at high temperature then `DeltaG lt 0` hence `CaCO_(3)` decomposes at high temperature. |
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| 1273. |
Which kind of process is superheating of steam. |
| Answer» Superheating of steam is isbaric process because temperature of steam increase in this process at constant pressure. | |
| 1274. |
The reaction `SO_(2)(g) +1//2O_(2)(g) rarr SO_(3)(g)` should beA. EndothermicB. ExothermicC. `DeltaH = 0`D. Unperdictable |
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Answer» `DeltaH = DeltaU + Deltan_(g)RT` or `DeltaH ~~Deltan_(g)RT` `Deltan_(g) = n_(P) - n_(R) =(1-(1+(1)/(2))) =- (1)/(2)` `:.DeltaH =- ve` or exothermic |
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| 1275. |
Consider the following reaction: `2NaN_(3)(s) rarr 2Na(s) +3N_(2)(g)` The enthalpy change for this reaction at `1atm` is equal to:A. `-2Delta_(f)H^(Theta) [NaN_(3)(s)]`B. `+2Delta_(f)H^(Theta)[NaN_(3)(s)]`C. `3Delta_(f)H^(Theta)[N_(2)(s)+2Delta_(f)H^(Theta)(Na(s)]`D. `-Delta_(f)H^(Theta)[NaN_(3)(s)]` |
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Answer» `Delta_(f)H = DeltaH_(P) - DeltaH_(R)` `[{:("Enthalpy of elements in their"),("stable state is assumed to be"),("zero".):}]` `=(2 xx H_(Na) +3 xx H_(N_(2))) -2H_(NaN_(3))` `=(2 xx 0+3 xx0) - 2H_(NaN_(3))` `=- 2H_(NaN_(3))` |
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| 1276. |
Calculate the enthalpy of the following reaction `:` `H_(2)C = CH_(2)(g) + H_(2)(g) rarr H_(3)C - CH_(3)(g)` The bond energies of `C-H,C-C,C=C` and `H-H` are 414,347,615 and 435 `kJ mol^(-1)` respectively |
| Answer» Correct Answer - `-125kJ mol^(-1)` | |
| 1277. |
The given enthalpy diagram reperesents which of the following reactions ? A. Enthalpy diagram for endothermic reactionB. Enthalpy diagram for exothermic reactionC. Enthalpy diagram for reversible reactionD. Enthalpy diagram for non-spontaneous reaction |
| Answer» Correct Answer - B | |
| 1278. |
In endothermic reactions,A. reactants have more energy than productsB. reactants have less energy than productsC. reactants and products have same energyD. reactants have lower temperature than products. |
| Answer» Correct Answer - B | |
| 1279. |
Bond energies of H - H bond is 80 kJ/mol, I - I bond is 100 kJ/mol and for H - I bond is 200 kJ/mol, the enthalpy of the reaction : `H_(2)(g) + I_(2)(g) rarr 2HI(g)` isA. `-120` kJB. `200` kJC. `+100` kJD. `+120` kJ |
| Answer» Correct Answer - B | |
| 1280. |
Calculate the enthalpy change for the reaction `H_(2)(g)+I_(2)(g) rarr 2HI(g)` Given that the bond energies of H-H,I-I and H-I are433, 151 and 299 `kJmol^(-1)` respectively. |
| Answer» Correct Answer - `-14 kJ` | |
| 1281. |
Which of the following expressions regarding entropy is not correct?A. `DeltaS_("system")=(q)/(T)`B. `DeltaS_("system")=DeltaS_("total")+DeltaS_("surrounding")`C. `DeltaS=DeltaS_("final")-S_("initial")`D. `DeltaS_("total")=DeltaS_("system")+DeltaS_("surrounding")` |
| Answer» Correct Answer - B | |
| 1282. |
Match the column I with column II and mark the appropriate choice. A. (A) `rarr` (iii), (B) `rarr` (i), (C) `rarr` (iv), (D) `rarr` (ii)B. (A) `rarr` (ii), (B) `rarr` (iv), (C) `rarr` (i), (D) `rarr` (ii)C. (A) `rarr` (ii), (B) `rarr` (iv), (C) `rarr` (iii), (D) `rarr` (i)D. (A) `rarr` (iii), (B) `rarr` (ii), (C) `rarr` (i), (D) `rarr` (iv) |
| Answer» Correct Answer - A | |
| 1283. |
Read the following statements regarding spontaneity of a process and mark the appropriate choice. (i) When enthalpy factor is absent then randomness factor decides spontaneity of a process. (ii) When randomness factor is absent then enthalpy factor decides spontaneity of a process. (iii) When both the factors take place simultaneously, the magnitude of both of factors decide spontaneity of a process.A. Statements (i) and (ii) are correct and (iii) is incorrect.B. Statement (iii) is correct, (i) and (ii) are incorrect.C. Statements (i), (ii) and (iii) are correct.D. Statements (i), (ii) and (iii) are incorrect. |
| Answer» Correct Answer - C | |
| 1284. |
In a spontaneous process the system undergoesA. No energy changeB. Lowering of free energyC. Lowering of entropy of universeD. Increase in internal energy always |
| Answer» Correct Answer - B | |
| 1285. |
A quantity of an ideal gas at `20^(@)C` reversibly expands against a constant pressure of 2.0 atm from 1.0 L to 2.0 L. Calculate the work doneA. `-101.3` JB. `-202.6` JC. `-844` JD. `-448`J |
| Answer» Correct Answer - B | |
| 1286. |
Calculate DeltaH (in kcal) for the following phase transformation : A (s,1 bar, `200`K) rarr A (g, `1` bar ,`200` K) Standard melting poing of A = `200` K Standard boioling poing of A = `300 K` Standard boiling ping of A = `300` K Latent heat of fusion of A at `200` K = `60 cal//g` Latent heat of vaporisation of A at `300 K = 410 cal//`g Molar mass of A = `50 g//`mole `C_(Vm)` of A (s)= 5cal `K^(-1)` `mol^(-1)` `C_(Vm)` of A (l) = `10 cal K^(-1)mol^(-1)` `C_(Vm)` of A (g)=`3cal K^(-1) mol^(-1)` `R=2 cal K^(-1) mol^(-1)` |
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Answer» Correct Answer - 24 |
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| 1287. |
Select the correct option with respect to reversible process:A. `DeltaG` will always be zero for isobaric process.B. `DeltaS_(system)` wil always be greater than zero for a closed system.C. `DeltaS_(universe)` will always be zero for a closed system.D. Both A and C option are correct |
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Answer» Correct Answer - C |
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| 1288. |
It is observed that an combustion of 5.6 g of but-1ene(g) 70 kcal of heat is liberated in a closed rigid vessel at 300 K. What could be a possible value of `abs(DeltaH_(combustion)^(@))` of gaseous cyclone propane [Take : R=2 cal`//`K mole]?A. 490 kcalB. 491.5 kcalC. 49 kcalD. 500 kcal |
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Answer» Correct Answer - C |
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| 1289. |
It is observed that on combustion of 4.2 g of gaseous propene in a closed rigid vessel 49 kcal of heat was liberated at 300 K. What could be a possible value of `abs(DeltaH^(@) "combustion")` of gaseous cyclone propane[Take : R=2 cal`//`K mole]?A. 490 kcalB. 491.5 kcalC. 49 kcalD. 500 kcal |
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Answer» Correct Answer - D |
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| 1290. |
A substance P has a standard boiling point of 450K. Which of the following options contain correct set of thermodynamic parameters for the following reactions? `P(l)(1atm,420K) toP(g)(1atm,420K)`A. `DeltaG lt0, DeltaHgt0,DeltaSgt0`B. `qgt0, Wgt0, DeltaUgt0`C. `DeltaH =0, DeltaU=0,qgt0`D. `DeltaG gt 0,qgt0,DeltaUgt0` |
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Answer» Correct Answer - D |
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| 1291. |
One gram sample of oxygen undergoes free expansion from 0.75L to 3.0 L at 298 K. Calculate `DeltaS`, q,w `DeltaH "and" DeltaE`A. `DeltaS=0.36 JK^(-1)`B. W=227.97 JC. q=-227.97 JD. `DeltaH=107.28 J` |
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Answer» Correct Answer - A Since . Expansion occurred at constant temperature , `DeltaS=nR "In" (V_(2))/(V_(1)) =(1)/(32) xx 8.314 "In" (3.0)/(0.75)=0.36 JK^(-1)` Since , this is case of free expansion , `P_("ext")=0 " " implies " "-W=P_("ext")DeltaV=0,q=0` Also , since , `DeltaT=0 implies DeltaH=DeltaE=0.` |
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| 1292. |
Given that: `DeltaG_(F)^(@)(CuO) =-30.4 "Kcal"//"mole"` `DeltaG_(f)^(@)(Cu_(2)O)=-34.98 Kcal//"mole" " "T=298K` Now on the basis of above data which of the following predications will be most appropriate under the standard conditons and reversible reaction.A. Finely divided form fof CuO Kept in excess `O_(2)` would be completely converted to `Cu_(2)O`B. Finely divided form of `Cu_(2)O` kept in excess `O_(2)` would be Completely converted to CuOC. Finely divided form of CuO kept in excess `O_(2)` would be converted to a mixture of CuO and `Cu_(2)O` (having more of CuO)D. Finely divided form of CuO kept in excess `O_(2)` would be converted to a mixture of CuO and `Cu_(2)O` (having more of `Cu_(2)O`) |
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Answer» Correct Answer - B `(2) Cu_(2)O(s) +(1)/(2)O_(2)(g) hArr 2CuO(s)` `DeltaG_("reaction")^(@) =[2xx(-30.4)]-[-34.98]=-25.82 `Kcal and `-25.82 xx 10^(3) =2.303 xx 2xx 298 `logK `therefore K~~10^(19), ` a very high value, hence reaction will be almost complete with a trace of `Cu_(2)O`. |
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| 1293. |
For the reaction takes places at certain tempreature `NH_(4) HS(s) hArrNH_(3)(g) + H_(2)S(g)`, if eqilibrium pressure is X bar, then `Delta_(r) G^(@)` would beA. `-2RT ln X`B. `- RT ln (X- ln2)`C. `-2RT (ln X- ln2)`D. None of these |
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Answer» Correct Answer - C `DeltaG^(@) = - 2.303 RT log Keq` ` = - RT In K_(P)` `P_(NH^(3)) = P_(H_(2)S) = (x)/(2) "bar"` `= - RT ln (("x")/(2))^(2) = - 2RT (ln X - ln 2)` |
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| 1294. |
It is necessary to define the 'standard state'. Give reason. |
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Answer» State functions are very useful because change in then values depends only on the initial and final states of the system and not on how change is carried out. |
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| 1295. |
It is preferable to determine a change in enthalpy than change in internal energy. Give reason. |
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Answer» This is because of the reason that most of the processes are carried in open container i.e., at constant pressure. Since the volume can change therefore, the changes in heat energy are enthalpy changes (ΔH) and not internal energy changes (ΔE). Therefore, it is preferable to determine a change in enthalpy that change in internal energy. |
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| 1296. |
Define bond enthalpy. |
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Answer» Bond enthalpy:- The bond enthalpy is the amount of energy necessary to break bonds in mole of a gaseous covalent substance to form products in the gaseous state. The bond enthalpy for H2 is 435.0 kJ mol-1. |
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| 1297. |
Calculate the enthalpy of the following reaction: `H_(2)C =CH_(2)(g) +H_(2)(g) rarr CH_(3)-CH_(3)(g)` The bond enegries of `C-H, C-C,C=C`, and `H-H` are `99,83,147`,and `104kcal` respectively. |
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Answer» The reaction is: `underset(H)underset(|)overset(H)overset(|)(C)=underset(H)underset(|)overset(H)overset(|)C(g)+H-H(g)rarrH-underset(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(H)overset(|)(C)-H(g), DeltaH^(Theta) =` ? `DeltaH^(Theta) =` Sum of bond energies of reactants -Sum of the bond energies of products `[DeltaH_(C=C) +4 xx DeltaH_(C-H) + DeltaH^(H-H)] -[DeltaH_(C-C) +6 xx DeltaH_(C-H)]` `(147 +4 xx 99 +104)-(83+6 xx 99) =- 30 kcal` |
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| 1298. |
For spontenous processA. `Delta_("total") S= 0`B. `Delta_("total")S gt 0`C. `Delta_("total")S lt 0`D. None of these |
| Answer» `Delta_("total")S = Delta_(sys)S +Delta_(surr)S =+ve` for spontaneous process | |
| 1299. |
Calculate `DeltaH` of the reaction, `H-underset(CI)underset(|)overset(H)overset(|)C-CI(g)rarrC(s) +2H(g) +2CI(g)` Bond enegry for `C-H` bond and `C-CI` bond are `400 kJ` and `320kJ`, respectively. |
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Answer» `DeltaH^(Theta) =` Sum of bond energies of reactants -Sum of bond energies of products `=[2 xx (C -H) +2 xx (C -C1)] = 0` [All the prodcuts are free atoms] `= 2xx 400 + 2xx 320` `= 800 + 640 = 1440kJ` |
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| 1300. |
For the reversible process, the value of `DeltaS` is given by the expression:A. `(q_(rev))/(T)`B. `T - q_(rev)`C. `q_(rev) xx T`D. `q_(rev) - T` |
| Answer» `DeltaS = (q_(rev))/(T)` | |