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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.
| 2101. |
Enthalpy of neutralzation is defined as the enthalpy change when 1 mole of acid`/ /`base is completely neutralized by base `//`acid in dilute solution . For Strong acid and strong base neutralization net chemical change is`H^(+) (aq)+OH^(-)(aq)to H_(2)O(l)` `Delta_(r)H^(@)=-55.84KJ//mol``DeltaH_("ionization")^(@)` of aqueous solution of strong acid and strong base is zero . when a dilute solution of weak acid or base is neutralized, the enthalpy of neutralization is somewhat less because of the absorption of heat in the ionzation of the because of the absorotion of heat in the ionization of the weak acid or base ,for weak acid /base `DeltaH_("neutrlzation")^(@)=DeltaH_("ionization")^(@)+ Delta _(r)H^(@)(H^(+)+OH^(-)to H_(2)O)` If enthalpy of neutralization of `CH_(3)COOH` by NaOH is -49.86KJ`//`mol then enthalpy of ionization of `CH_(3)COOH` is:A. 5.98 kJ/molB. `-5.98` kJ/molC. 105.7 kJ/molD. None of these |
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Answer» Correct Answer - A `-49.86=(Delta_("Iron")H)_(CH_(3)COOH)-55.84` `(Delta_("Iron")H)_(CH_(3)COOH)=5.98 " kJ mol"^(-1)` |
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| 2102. |
Statement-1 : All the exothermic reactions are spontaneous. And Statement-2 : For a spontaneous reaction, `Delta`G must be negative.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is Not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - D | |
| 2103. |
A gas expands against a variable pressure given by `P = (20)/(V)` (where P in atm and V in L). During expansion from volume of 1 litre to 10 litre, the gas undergoes a change in internal energy of 400 J. How much heat is absorbed by the gas during expansion?A. 46 JB. 4660 JC. 5065.8 JD. 4260 J |
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Answer» Correct Answer - C `intdw=int-P.dV` `"implies "w=-int20.(dV)/(V)=-20ln.(V_(2))/(V_(1))` `w=-46.06 " L-atm"=-4665.8 J` `DeltaU=q+wimplies400=q-4665.8` `q = 5065.8 J` |
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| 2104. |
If the boundary of system moves by an infinitesimal amount, the work involved is given by `dw=-P_("ext")dV` for irreversible process `w=-P_("ext")DeltaV " "( "where "DeltaV=V_(f)-V_(i))` for reversible process `P_("ext")=P_("int")pmdP~=P_("int")` so for reversible isothermal process `w = -nRTln.(V_(f))/(V_(i))` 2mole of an ideal gas undergoes isothermal compression along three different plaths : (i) reversible compression from `P_(i)=2` bar and `V_(i) = 8L` to `P_(f) = 20` bar (ii) a single stage compression against a constant external pressure of 20 bar, and (iii) a two stage compression consisting initially of compression against a constant external pressure of 10 bar until `P_("gas")=P_("ext")`, followed by compression against a constant pressure of 20 bar until `P_("gas") = P_("ext")` Order of magnitude of work is :A. `w_(1) gt w_(2) gt w_(3)`B. `w_(3) gt w_(2) gt w_(1)`C. `w_(2) gt w_(3) gt w_(1)`D. `w_(1)=w_(2)=w_(3)` |
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Answer» Correct Answer - C In isothermal compression is `w_("One-stage")gtw_("Two-stage")gtw_("Reversible")` |
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| 2105. |
If the boundary of system moves by an infinitesimal amount, the work involved is given by `dw=-P_("ext")dV` for irreversible process `w=-P_("ext")DeltaV " "( "where "DeltaV=V_(f)-V_(i))` for reversible process `P_("ext")=P_("int")pmdP~=P_("int")` so for reversible isothermal process `w = -nRTln.(V_(f))/(V_(i))` 2mole of an ideal gas undergoes isothermal compression along three different plaths : (i) reversible compression from `P_(i)=2` bar and `V_(i) = 8L` to `P_(f) = 20` bar (ii) a single stage compression against a constant external pressure of 20 bar, and (iii) a two stage compression consisting initially of compression against a constant external pressure of 10 bar until `P_("gas")=P_("ext")`, followed by compression against a constant pressure of 20 bar until `P_("gas") = P_("ext")` Work done on the gas in single stage compression is :A. 36B. 72C. 144D. None of these |
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Answer» Correct Answer - C `w_("irr")=-P_("ext")(V_(2)-V_(1))` `=-20((nRT)/(P_(2))-(nRT)/(P_(1)))` = 144 bar-L |
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| 2106. |
Statement-1 : `Delta H and Delta U` are the same for the reaction `H_(2)(g) + I_(2)(g) hArr 2 HI(g)` And Statement-2 : All reactants and products are in gaseous form.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is Not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - B | |
| 2107. |
The normal boiling point of a liquid X is 400 K. Which of the following statement is true about the process `X(l) to X(g)` ?A. at `400` K and `1` atm pressure `DeltaG=0`B. at `400` K and `2` atm pressure `DeltaG =+ve `C. at `400` K and `0.1` atm pressure `DeltaG=-ve`D. at `410` K and `1` atm pressure `DeltaG =+ve` |
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Answer» Correct Answer - A::B::C Boiling of a liquid at normal boiling point is a equilibrium process and on decreasing the pressure equilibrium will go forward and `DeltaG` will be negative and vice versa |
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| 2108. |
Which of the following statement is/are correctA. Reversible adiabetic process is iso entropie processB. `DeltaS_("system")` for irreversible adiabelic compression is greater than zeroC. `DeltaS_("system")` for free expension in zeroD. `DeltaS_("surrounding")`for irreversible isothermal compression is greater than zero |
| Answer» Correct Answer - A::B::D | |
| 2109. |
For a real gas having `a=4.105 atm -L^(2)//"mole"` and `b=(1)/(5.4) L//"mole"` .If it is at an initial temperture of 300 k, then which of the following process can cause liquifaction of the gas ?A. Isothermally decrease of pressureB. Isothermally increase of pressureC. Adiatbatic decrease of pressureD. both (b) and (c ) |
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Answer» Correct Answer - C |
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| 2110. |
If the external pressure is more than internal pressure then what will be the sign convention for the work done on a cylinder fitted with a frictionless and weightless pistion containing one mole of an ideal gas? |
| Answer» Correct Answer - `+W` | |
| 2111. |
One mole of oxygen is allowed to expand isothermally and reversibly from 5 `m^(3)` to 10 `m^(3)` at 300 K. Calculate the work done in expansion of the gas. |
| Answer» Correct Answer - 1-1728.98 J | |
| 2112. |
Statement-1 : `Delta H_(f)^(@)` is zero for oxygen `(O_(2))`. And Statement-2 : `Delta H_(f)^(@)` for all the elements at S.T.P. is zero.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is Not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - C | |
| 2113. |
Statement-1 : Work done in isothermal reversible process is more than irreversible process. And Statement-2 : Irreversible process is an infinitesimally slow process.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is Not a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - C | |
| 2114. |
What is the change internal energy when a gas contracts from 377 mL to 177 mL under a constant pressure of 1520 torr, while at the same time being cooled by removing 124 J heat ? `[Take :( 1 L atm ) =100 J)]`A. `-24 J`B. `-84 J`C. `-164 J`D. `-248 J` |
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Answer» Correct Answer - B |
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| 2115. |
Staetement -1: The magniyude of the work involed in an isothermal expansion is greater than that involved in an adiabatic expansion. Statement -2: P-v cure (pon y -axas and V on X-axas) decrease more repidly for reversible asiabatic expansion compared to reversible isothermal expansion starting from same initial state.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-5B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-5C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
| Answer» Correct Answer - A | |
| 2116. |
Staetement -1: The magniyude of the work involed in an isothermal expansion is greater than that involved in an adiabatic expansion. Statement -2: P-v cure (pon y -axas and V on X-axas) decrease more repidly for reversible asiabatic expansion compared to reversible isothermal expansion starting from same initial state.A. Statement-1 is True, Statement -2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement -1 is True ,Starement -2 is True ,Statement-2 is not a correct explanation for Statement-1C. Statement-1 is True ,Statement-2 is False.D. Statement-1 is False ,Statement-2 is True. |
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Answer» Correct Answer - a |
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| 2117. |
Which of the following statement (s) is /are true?A. `DeltaE=0` for combustion of `C_(2)H_(6)(g)` in a sealed rigid adiabatic containerB. `Delta_(r)H^(@)` (S, monoclinic) `ne` 0C. If dissociation energy of `Ch_(4)(g) ` is 1656 KJ/mol and `C_(2)H_(6)(g)` is 2812 KJ/mol, then value of C-C bond energy will be 328 KJ/molD. If `DeltaH_(r)(H_(2)O,g)` =-242 KJ/mol, `DeltaH_("vap")(H_(2)O,l)` =44 KJ/mol then `Delta_(e)H^(@)(OH^(-), aq)` will be -142 KJ/mol |
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Answer» Correct Answer - A::B::C (A) `DeltaE` will be -ve for combustion reactions `DeltaH_(f)^(@)` (`S_(1)` rhombic )=0 (C) `CH_(4)(g) rarr C(g) + 4H(g) " " DeltaH=1656 KJ//"mole" =4DeltaH_(C-H)` `C_(2)H_(6)(g) rarr 2C(g) + 6H(g) " " DeltaH=2812 KJ//"mole"=6 DeltaH_(C-H) + DeltaH_(C-C)` so `DeltaH_(C-C) = (2812 - (6)/(4) xx 1656) KJ//"mole" = 328KJ//"mole"` (D) Can not be concluded from given data |
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| 2118. |
Which of the following stament (s) is/are false ?A. All adiabatic processes are isoentropic (orisentrophic) processesB. When `(DeltaG_(system ))_(T,P)lt 0:` the reaction must be exothermic.C. dG=VdP-SdT is applicable for closed system both PV and non -PV workD. the heat of vaproisaton of water at `100^(@)` C is `40.6 kJ//mol`. When 9 gm of water vapour condeneses to liquid at `100^(@)`C and 1 atm, then `DeltaS_(system)=54.42 J//K` |
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Answer» Correct Answer - a,b,c,d |
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| 2119. |
Which of the following stament (s) is/are false ?A. `DeltaE=0` for combutins of `C_(2)H_(6)(g)` in a sealed rigid adiabatic containerB. `Delta_(f)H^(@) (S," monoclinic" )ne 0`C. If dissociation enegry `CH_(4)(g)` is `1656 KJ//"mole"` and `C_(2)H_(6)(g)` is `2812 kJ//"mole"`,then value of C-C bond energy will be `328 kJ//"mol"`.D. If `DeltaH_(f)(H_(2) O,g,) =- 242 kJ//mol `, `DeltaH_(vap)(H_(2)O,l)= 44kJ//mol` then `Delta_(f)H^(@)(OH^(-),aq)` will be be-142 kJ/mol |
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Answer» Correct Answer - a,b,c,d |
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| 2120. |
The value of `DeltaH_("transition")` of C (graphite) `rarr` C (diamond) is 1.9 kJ/mol at `25^(@)C`. Entropy of graphite is higher than entropy of diamond. This implies that :A. C (diamond) is more thermodynamically stable than C (graphite) at `25^(@)C`B. C (graphite) is more thermodynamically stable than C (diamond) at `25^(@)C`C. diamond will provide more heat on complete combution at `25^(@)C`D. `Delta G_("transition")` of C (diamond) `rarr` C(graphite) is -ve |
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Answer» Correct Answer - B::C `C("graphite") rarr C("diamond")` `Delta H +ve, Delta S = -ve`, So the transition is non spontaneous at all temperatures. |
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| 2121. |
Which of the following statement(s) is/are false?A. All adiabatic processes are isoentropic (or isenthermic) processesB. When `(DeltaG_("system"))_(TP) lt 0`, the reaction must be exothermicC. `dG=VdP-SdT ` is applicable for closed system, both PV and non-PV workD. the heat of vaporisation of water at `100^(@)C` is 40.6 KJ/mol . When 9 gm of water vapour condenses to liquid at `100^(@)C` of 1 atm , then `DeltaS_("system")=54.42 J//K` |
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Answer» Correct Answer - A::B::C::D (A) Reversible adiabatic process is isoentropic (B) Reaction is spontaneous , need not be exothermic. (C) Only when PV work is innvoled (D) `DeltaS_("system") =(DeltaH_("System"))/(T_(b)) =-(40600)/(373xx2) =-54.42 J//K` |
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| 2122. |
Consider the reactions: (P) `S ("rhombic") +(3)/(2)O_(2)(g) to SO_(3)(g), " "DeltaH_(1)` (Q) S`("Monoclinic")0 +(3)/(2)O_(2)(g) to SO_(2)" "DeltaH_(2)` (R) `S("rhiombic")+O_(3)(g) to SO_(3)(g)" "DeltaH_(3)` (S) `S("monoclince")+O_(3)(g) to SO_(3)(g)" "DeltaH_(4)`A. `DeltaH_(1)ltDeltaH_(2)lt DeltaH_(4)` (magnitude only)B. `DeltaH_(1)ltDeltaH_(3)lt DeltaH_(4)`(magnitude only)C. `DeltaH_(1)ltDeltaH_(2)=DeltaH_(3)lt DeltaH_(4)`(magnitude only)D. `DeltaH_(1) + DeltaH_(4)=DeltaH_(2)lt DeltaH_(3)` |
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Answer» Correct Answer - a,b,d |
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| 2123. |
The normal boiling point of a liquid X is 400 K. Which of the following statement I dtrue about the process `X(l) to X(g)` ?A. At 400 k and 1 atm pressure `DeltaG=0`B. At 400 k and 2 atm pressure `DeltaG=+ve`C. At 400 k and 0.1 atm pressure `DeltaG=-ve`D. At 410 k and 1 atm pressure `DeltaG=+ve` |
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Answer» Correct Answer - a,b,c,d |
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| 2124. |
Bond dissociation enthalpy and bond enthalpy are not the same forA. `O_(2) (g)`B. `N_(2) (g)`C. `CH_(4) (g)`D. `F_(2) (g)` |
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Answer» Correct Answer - C They are not the same for polyatomic molecules such as methane because in every dissociation step, bonds are broken is different species. Note that both bond dissociation enthalpy and bond enthalpy have the same symbols. |
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| 2125. |
Which of the following statement(s) is/are false?A. All adiabatic process are isoentropic (or isentropic) processesB. When `(DeltaG_("system"))_(T,P)lt0,` the reaction must be exothermicC. dG=VdP-SdT is applicable for closed system, both PV and non-PV workD. The heat of vaporisation of water at `100^(@)C` is 40.6 kJ/mol. When 9 gm of water vapour condenses to liquid at `100^(@)C` and 1 atm, then `DeltaS_("system")` = 54.42 J/K |
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Answer» Correct Answer - A::B::C::D (a) Reversible adiabatic process is iso-entropic process (b) When `(DeltaG_("system"))_(T,P)lt0` then process may be endothermic / exothermic (c )`dG=VdP -SdT`, only PV work consider (d) `DeltaS=-(9)/(18)xx(40.6 xx 1000)/(373)=-54.42` |
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| 2126. |
The following is(are) endothermic reaction (s):A. Combustion of methaneB. decomposition of waterC. dehydrogenation of etheane to ethyleneD. conversion pf graphite to diamond. |
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Answer» Correct Answer - b,c,d |
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| 2127. |
Which of the indicated relationship is correct for the following exothermic reaction carried out at constant pressure? `CO(g) +3H_(2)(g) rarrCH_(4)(g)+H_(2)O(g)`A. `DeltaU = DeltaH`B. `DeltaU gt DeltaH`C. `w lt 0`D. `q gt 0` |
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Answer» Correct Answer - B `DeltaH=DeltaU+Deltan_(g)RT, " "Deltan_(g)=-2, " "DeltaH lt DeltaU` `w=-Deltan_(g)RT, " "wgt 0 , " "q=-ve` |
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| 2128. |
Among the following , the state function (s) is(are):A. internal energyB. irrevcersible expansion workC. reversible expansion workD. molar enthalpy |
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Answer» Correct Answer - a,d |
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| 2129. |
Place the following systems in order of increasing randomnes: a. `1mol` of a gas `X` b. `1mol` of a solid `X` c. `1 mol` of a liquid `X` |
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Answer» Order of increasing randomnes: `1 mol` of solid `X lt 1mol` of liquid `X lt 1 mol` of gas `X`. |
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| 2130. |
When hydrogen gas is burnt in chlorine 2000 cals of heat is librated during the formation of 3.65 g of `HCl, Delta H` of formation of `HCl` isA. 2 K.calB. `-20` K.calC. `+20` K.calD. `-2` K.cal |
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Answer» Correct Answer - B `1/2 H_(2)+1/2 Cl_(2) rarr HCl` |
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| 2131. |
Which of following equations does not correspond to the standard enthalpy of atomization?A. `H_(2) (g) rarr 2 H (g)`B. `CH_(4) (g) rarr C (g) + 4 H (g)`C. `Na (s) rarr Na (g)`D. `Br_(2) (g) rarr 2Br (g)` |
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Answer» Correct Answer - D `Delta_(a) H^(@)` is the enthalpy change when one mole of the substance in its stantard state dissociates into isolated atoms in the gas phase. The standard state of bromine is `Br_(2)` (liquid). In case of `H_(2) (g)`,k the enthalpy of atomization is the same as bond dissociation enthalpy. In `CH_(4) (g)`, it is just the enthalpy of atomization. |
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| 2132. |
A monoatomic ideal gas undergoes a process in which the ratio of `P` to `V` at any istant is constant and equal to unity. The molar heat capacity of the gas isA. `(4R)/(2)`B. `(3R)/(2)`C. `(5R)/(2)`D. `0` |
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Answer» A monoatomic ideal gas undergoes a process in which the ratio of `P` to `V` at any instant is constant and equals `1`. `(P)/(V) =` constant `PV^(-1) =` constant `PV^(gamma) =` constant `gamma =-1` Monatomic gas: `C_(V) = (3)/(2)R` Molar heat capacity `= C_(V) +(R )/(1- gamma)` `= (3)/(2)R +(R )/(1-(-1))` `=(3)/(2)R +(R )/(2) = (4R)/(2)` |
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| 2133. |
For the process `H_(2)O(l) (1 "bar", 373 K) rarr H_(2)O(g) (1"bar", 373 K)` the correct set of thermodynamic parameters isA. `DeltaG = 0, DeltaS =+ve`B. `DeltaG =0, DeltaS =- ve`C. `DeltaG =+ve, DeltaS =0`D. `DeltaG =- ve, DeltaS =+ve` |
| Answer» The conversion of a liquid into a gas occurs with an increases in randomness or entropy, therefore the entropy change is positive. At the boiling point, liquid and gas are in equilibrium with each other, therefore, free energy change is equal to zero. | |
| 2134. |
The combustion of `1 mol` of benzene takes place at `298 K` and `1 atm`. After combustion, `CO_(2)(g)` and `H_(2)O(l)` are produced and `3267.0 kJ` of heat is librated. Calculate the standard entalpy of formation, `Delta_(f)H^(Θ)` of benzene Given: `Delta_(f)H^(Θ)CO_(2)(g) = -393.5 kJ mol^(-1)` `Delta_(f)H^(Θ)H_(2)O(l) = -285.83 kJ mol^(-1)`.A. `-48.51` kJ/MoleB. `48.51` kJ/MoleC. `-24.5 kJ`D. `24.5 kJ` |
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Answer» Correct Answer - B `DeltaH=H_(p)-H_(R)` `6C+3H_(2) rarr C_(6)H_(6)" "DeltaH= ?` |
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| 2135. |
Which of the following statement is .are correct ?A. Internal energy can be written as U=f(P,T) for a substance (no physical or chemical change )B. Absolute value of entropy can be determinedC. The heat absored during the isothernal expansion of an ideal gas against vacuum is zeroD. During an adiabatic reversible expansion of an ideal gas, temperature of the system increases. |
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Answer» Correct Answer - a,b,c,d |
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| 2136. |
For the transition `C` (diamond) `rarr C` (graphite), `Delta H = - 1.5 kJ` it follows thatA. diamond is exothermicB. graphite is endothermicC. graphite is more stable than diamondD. diamond is more stable than graphite |
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Answer» Correct Answer - C During the transformation of diamond to graphite, energy is released. This implies that graphite has lower energy content and, hence, is more stable than diamond. |
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| 2137. |
Ammonium chloride when dissolved in water leads to a cooling sensation. The dissolution of ammonium chloride at constant temperature is accompanied byA. Increase in entropyB. Decrease in entropyC. No change in entropyD. No change in enthalpy |
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Answer» `NH_(4)CI +aq rarr NH_(3) +HCI` On product side, the number of moles is greater than the reactant side. Therefore, the entropy increases. |
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| 2138. |
Work done by the system in isothermal reversible process is `w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1))`. Also in case of adiabatic reversible process work done by the system is given by: `w_(rev.) = (nR)/(gamma -1) [T_2 - T_1]`. During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy `DeltaS = q_(rev.)/T`. The entropy changes also occurs during transformation of one state to other end expressed as `DeltaS = DeltaH/T`. Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and `DeltaG=DeltaH - T DeltaS`. A chemical change will definitely be spontaneous if:A. `DeltaH=-ve, DeltaS=-ve` and low temperatureB. `DeltaH=+ve, DeltaS=-ve` and high temperatureC. `DeltaH=-ve, DeltaS=+ve` and any temperatureD. `DeltaH=+ve,DeltaS=+ve` and `TDeltaS lt DeltaH` |
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Answer» Correct Answer - c `DeltaG=DeltaH-TDeltaS` `=-ve-Txx(+ve)=-ve` |
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| 2139. |
A certain mass of gas expanded from ( 1L, 10 atm) to (4 L, 5 atm) against a constant external pressure of 1 atm. If initial temperature of gas is 300 K and the heat capacity of process is `50 J//^(@)C`. Then the enthalpy change during the process is : (`1"L" "atm" underline ~ 100 J`)A. `DeltaH=15 kJ`B. `DeltaH=15.7 kJ`C. `DeltaH=14.4 kJ`D. `DeltaH=14.7 kJ` |
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Answer» Correct Answer - D |
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| 2140. |
The heat of formation of `H_(2)O_((l))` is `-286.2` kJ. The heat of formation of `H_(2)O_((g))` is likely to beA. `-286.2 kJ`B. `-290.78 kJ`C. `-335.2 kJ`D. `-242.76 kJ` |
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Answer» Correct Answer - D `lt -286.2` as energy is absorbed for `l rarr g` |
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| 2141. |
Which of the following parameters correctly represent conditions for a spontaneous process with no non P-V work involved? (P) `(dH)_(S,P)lt0` (Q) `(dU)_(S,P)lt0` (R ) `(dG)_(T,P)lt0` (S) `(dH)_(P,T)gt0` (T) `(d,S)_(universe)gt0` (U) `(dS)_(U,V)gt0`A. Options (Q),(R ) and (T) are correctB. Options (P),(T) and (U) are correctC. Options (P),(R ) and (T) are incorrectD. Options (Q) and (R ) are incorrect |
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Answer» Correct Answer - B |
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| 2142. |
Which of the following is incorrect?A. Calorific value of fat is more than that of carbohydrate and protein.B. `Delta_(C ) H^(@)` is always negativeC. Calforific value of kerosene is less than that of coal.D. Butane is the main component of cooking gas. |
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Answer» Correct Answer - C Fluid fossil fuels such as kerosene are popular because they release more heat per gram tha coal does. Kerosene is apporximately `C_(12) H_(26)`, the thermochemical equation is `C_(12) H_(26) (1) (37)/(2) O_(2) (g) rarr 12 CO_(2) (g) + 13 H_(2) O (g)` `Delta_(C ) H^(@) = - 7513 kJ` The average calorific value quoted for carbohydrates and fats are `4.0 cal g^(-1)` and `9.0 kcal g^(-1)`, respectively. Note that fats contain more than twice the fuel value per gram as do carbhydrates. Thus, by storing its fuel s fat, the body can store more fuel for a given mass of body tissue. |
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| 2143. |
Charge in enthalpy and change in internal energy are equal at room temperature forA. Combustion of glucoseB. Combustion of enthyleneC. Combustion of methaneD. Combustion of ethyl alcohol |
| Answer» Correct Answer - A | |
| 2144. |
`DeltaH` value for the manufacture of `NH_(3)` is `DeltaH=-91.8 kJ`. The correct thermo chemical equation for dissociation of `NH_(3)` isA. `N_(2)+3H_(@)-91.8 rarr 2NH_(3)`B. `2NH_(3) rarr N_(2) +3H_(2) DeltaH=+91.8 kJ`C. `2NH_(3) rarr N_(2)+3H_(2)DeltaH=-91.8 kJ`D. `1/2 N_(2)+3/2 H_(2) rarr NH_(3)DeltaH=+91.8 kJ` |
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Answer» Correct Answer - B `2NH_(3) rarr N_(2)+3H_(2), DeltaH=91.8 kJ` |
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| 2145. |
Which of the following has maximum standard enthalpy of combustion per gram?A. `C_(3) H_(8)`B. `C_(4) H_(10)`C. `C_(2) H_(4)`D. `C_(2) H_(2)` |
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Answer» Correct Answer - B During combustion. `C` is oxidized to `CO_(2)` and `H` is oxidized to `H_(2) O`. Thus, saturated hydrocarbons have higher caloriffic values than the corresponding unsaturated hydrocarbons. More the number of `C` and `H` atoms, greater is the calorific value. |
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| 2146. |
At `25^(@)C` the heat of formation of `H_(2)O_((l))` is `-285.9" kJ mole"^(-1)` and that for `H_(2)O_((g))` is `-242.8" kJ mole"^(-1)`. The heat of vaporization of water at the same temperature isA. `43.1" kJ mole"^(-1)`B. `242.8" kJ mole"^(-1)`C. `-43.1" kJ mole"^(-1)`D. `-242.8" kJ mole"^(-1)` |
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Answer» Correct Answer - A `DeltaH=H_(H_(2)O_((g)))-H_(H_(2)O_((l)))` |
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| 2147. |
Calculate `C-H` bond energy from the following data : `Delta_(f)H" "[C(g)]=716.68 kJ//"mole"` `Delta_(f)H" "[H(g)]=217.97 kJ//"mole"` `Delta_(f)H" "[CH_(4)(g)]=-74.81 kJ//"mole"`A. 1663.37 kJB. 415.84 kJC. 179.17 kJD. 74.81 kJ |
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Answer» Correct Answer - B |
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| 2148. |
Calculate bond energy of `X-X` bond from the following data. `DeltaH_(f) "of" X(g)=+300 kJ//"mole"` A. 400 kJB. 200 kJC. 100 kJD. 150 kJ |
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Answer» Correct Answer - A |
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| 2149. |
Enthapy of combustion `(Delta_(C ) H)` depens on (i) whether combustion is carried out in constant volume calorimeter of constant pressure calorimeter (ii) the physical state of the substances (iii) the temperature at which combustion is carried out (iv) the amount of xygen presentA. (i),(ii),(iii)B. (i),(ii),(iii),(iv)C. (iii),(iv)D. (i),(ii) |
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Answer» Correct Answer - A Enthalpy of combustion is independent of the amount of oxygen used because combustion, by definition, is always carried out in excess of oxygen. |
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| 2150. |
Can water be made to boil without heating? |
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Answer» Yes, by reducing pressure on water, boiling point of water can be brought down to room temperature. |
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