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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.
| 201. |
When water is added to quick lime, the reaction isA. ExothermicB. EndothermicC. ExplosiveD. None of these |
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Answer» Correct Answer - A `CaO_((s))+H_(2)O_((l))rarrCa(OH)_(2)` is exothermic. |
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| 202. |
The heat change for the following reaction at 298K and at constant pressure is +7.3 kcal `A_(2)B(s) rarr 2A(s) + 1//2 B_(2)(g), DeltaH =+7.3 kcal` The heat change at constant volume would beA. 7.3 kcalB. More than 7.3C. ZeroD. None of these |
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Answer» Correct Answer - D `DeltaH=DeltaE+DeltanRT or DeltaE=DeltaH-DeltanRT` `thereforeDeltaE=+7.3-(1)/(2)xx0.002xx298=7.3-0.298=7` kcal. |
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| 203. |
The heat change for the following reaction at 298K and at constant pressure is +7.3 kcal `A_(2)B(s) rarr 2A(s) + 1//2 B_(2)(g), DeltaH =+7.3 kcal` The heat change at constant volume would be `C(s)+2S(s)rarrCS_(2)(l)` is known asA. Heat of vaporizationB. Heat of solutionC. Heat of fusionD. Heat of formation |
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Answer» Correct Answer - D According to the definition of heat of formation. |
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| 204. |
The standard heats of formation in kcal `mol^(-1)` of `NO_(2)(g)` and `N_(2)O_(4)(g)` are 8.0 and 2.0 respectively. The heat of dimerization of `NO_(2)` in kcal is `2NO_(2)(g) iff N_(2)O_(4)(g)`A. 10B. `- 6.0`C. `- 12.0`D. `- 14.0` |
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Answer» Correct Answer - D `DeltaH_("reaction")=DeltaH_(f)^(@)(N_(2)O_(4))-2DeltaH_(f)^(@)(NO_(2))` `=2-2(8)=-14 kcal`. |
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| 205. |
The surface of copper gets tarnished by the formation of copper oxide. `N_(2)` gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the `N_(2)` gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below `2Cu(s)+H_(2)O(g)rarrCu_(2)O(s)+H_(2)(g)` `P_(H_(2))` is the minimum partial pressure of `H_(2)` (in bar) needed to prevent the oxidation at 1250 K. The value of `ln(P_(H_(2)))` is_____ (Given : total pressure = 1 bar, R (universal gas constant)= 8 `JK^(-1) mol^(-1), ln(10)` = 2.3, Cu(s) and `Cu_(2)O(s)` are mutually immiscible. At `1250K :2Cu(s)+1//2O_(2)(g)rarrCu_(2)O(s),DeltaG^(theta)=-78,000J mol^(-1)` `H_(2)(g)+1//2O_(2)(g)rarrH_(2)O(g),DeltaG^(theta)=-1,78,000 J mol^(-1), G` the Gibbs energy |
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Answer» Correct Answer - A::D `2Cu(s)+(1)/(4)O_(2)(g)rarr1Cu_(2)O(s)" "DeltaG^(@)=-78 kJ` `[H_(2)(g)+(1)/(2)O_(2)rarrH_(2)O(g)" "DeltaG^(@)=-178 kJ]xx(-1)` Hence, `2Cu(s)+H_(2)O(g)rarrCu_(2)O+H_(2)(g)DeltaG^(@)=+100 kJ` `DeltaG=DeltaG^(@)+RT ln Q` `0=+100+(8)/(1000)xx1250ln.(P_(H_(2)))/(P_(H_(2)O))` `-(100xx1000)/(8)=1250ln.P_(H_(2))/(((1)/(100)xx1))` `ln P_(H_(2))=-14.6` |
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| 206. |
1 mole of conc. HCl requires X moles of dilute NaOH for neutralisation and 1 mole of concentrate `H_(2)SO_(4)` requires Y moles of small dilute NaOH then which of the following reaction is trueA. `Y=(1)/(2)X`B. `X=(1)/(2)Y`C. `X=2Y`D. None of these |
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Answer» Correct Answer - B `X=(1)/(2)Y`. |
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| 207. |
In which of the following neutralisation reactions, the heat of neutralisation will be highestA. `NH_(4)OH and CH_(3)COOH`B. `NH_(4)OH and HCl`C. `NaOH and CH_(3)COOH`D. `NaOH and HCl` |
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Answer» Correct Answer - D Heat of neutralization between strong acid and a strong base is about -13.7 kcal. |
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| 208. |
Four grams of graphite is burnt in a bomb calorimeter of heat capacity `30 kJ K^(-1)` in excess of oxygen at 1 atmospheric pressure. The temperature rises from 300 to 304 K. What is the enthalpy of combustion of graphite (in kJ `mol^(-1)`)A. 360B. 1440C. `- 360`D. `- 1440` |
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Answer» Correct Answer - C `DeltaE=CxxDeltatxx(M)/(m)=30xx4xx(12)/(4)=360` `DeltaE=-360 kJ mol^(-1)` |
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| 209. |
The heat liberated when 1.89 g of benzoic acid burnt in a bomb calorimeter at `25^(@)C` increases the temperature of 18.94 kg of water by `0.632^(@)C`. If the specific heat of water at `25^(@)C` is 0.998 cal/g-deg, the value of the heat combustion of benzoic acid isA. 771.1 kcalB. 871.2 kcalC. 881.1 kcalD. 981.1 kcal |
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Answer» Correct Answer - A `Q=(18.94xx0.632xx0.998xx122)/(1.89)=771.1 kcal`. |
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| 210. |
Equal volumes of methanoic acid and sodium hydroxide are mixed. If x is the heat of formation of water, then heat evolved on neutralisation isA. More than xB. Equal to xC. Twice of xD. Less than x |
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Answer» Correct Answer - D As methanoic acid is weak acid, heat of neutralization `lt x`. |
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| 211. |
Enthalpy change for the reaction, `4H_((g))rarr2H_(2(g))` is - 869.6 kJ. The dissociation energy of H - H bond isA. `+ 217.4` kJB. `- 434.8` kJC. `- 869.6` kJD. `+ 434.8` kJ |
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Answer» Correct Answer - D The dissociation energy of H-H bond is `=(869.6)/(2)=434.8 kJ`. |
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| 212. |
The amount of heat measured for a reaction in a bomb calorimeter isA. `DeltaG`B. `DeltaH`C. `DeltaE`D. `PDeltaV` |
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Answer» Correct Answer - C The bomb calorimeter is used to measure `DeltaE`. |
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| 213. |
Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats `(C_(p)//C_(v))` of the mixture will beA. 0.83B. 1.5C. 3.3D. 1.67 |
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Answer» Correct Answer - D `(C_(P))/(C_(V))=(5//2R)/(3//2R)=(5)/(3)=1.67`. |
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| 214. |
For complete combustion of ethanol, `C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)`, the amount of heta produced as measured in bomb calorimeter, is 1364.47 kJ `mol^(-1)` at `25^(@)`. Assuming ideality the Enthalpy of combustion. `Delta_(C)H`, for the reaction will be `(R=8.314 kJ mol^(-1))`A. `-1366. 95 kJ mol^(-1)`B. `-1361.95 kJ mol^(-1)`C. `-1460.50 kJ mol^(-1)`D. `-1350.50 kJ mol^(-1)` |
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Answer» Correct Answer - A `C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)` `DeltaU=-1364.47 KJ//mol` `DeltaH=DeltaU+Deltan_(g)RT` `Deltan_(g)=-1` `DeltaH=-1364.47+[(-1xx8.314xx298)/(1000)]` `=-1364.47-2.4776` `=-1366.94 KJ//mol`. |
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| 215. |
The enthalpy change for the following processes are listed below `{:(Cl_(2)(g)=2 Cl(g),242.3 kJ mol^(-1)),(I_(2)(g)=2I(g),151.0 kJ mol^(-1)),(ICl(g)=I(g)+Cl(g),211.3 kJ mol^(-1)),(I_(2)(s)=I_(2)(g),62.76 kJ mol^(-1)):}` Given that the standard states for iodine and chlorine are `I_(2)(g)` and `Cl_(2)(g)`, the standard enthalpy of formation for ICl(g) isA. `-14.6 kJ mol^(-1)`B. `-16.8 kJ mol^(-1)`C. `+16.8 kJ mol^(-1)`D. `+244.8 kJ mol^(-1)` |
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Answer» Correct Answer - C `1//2I_(2)(s)+1//2Cl_(2)(g)rarrICl(g), DeltaH_(f)=?` `{:(Cl_(2)(g)rarr2Cl(g),,DeltaH=242.3 kJmol^(-1)),(I_(2)(s)rarrI_(2)(g),,DeltaH=62.76 kJ mol^(-1)),(I_(2)(g)rarr2I(g),,DeltaH=151.0 kJ mol^(-1)):}` `2Cl(g)+2I(g)rarr2ICl(g), DeltaH=-422.6 kJ mol^(-1)` `I_(2)(s)+Cl_(2)(g)rarr2ICl(g),` `thereforeH=-422.6 +151+62.76 +242.3=33.46 kJ` So heat of formation of `ICl=(33.46)/(2)=16.73 kJ mol^(-1)`. |
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| 216. |
Consider the following processes `{:(,DeltaH(kJ//mol)),(1//2ArarrB,+150),(3Brarr2C+D,-125),(E+Ararr2D,+350):}` For `B+DrarrE+2C,DeltaH` will beA. `-325 kJ//mol`B. `325 kJ//mol`C. `525 kJ//mol`D. `-175 kJ//mol` |
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Answer» Correct Answer - D `({:(,DeltaH,),((1)/(2)ArarB,+150,...(i)),(3Brarr2C+D,-125,...(ii)),(E+Ararr2D,+350,...(iii)),(,,):})/(B+DrarrE+2C" ")` `2xx` eq.(i) + eq. (ii) -eq. (iii) `DeltaH=300-125-350=-175` |
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| 217. |
For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, `DeltaU` and w correspond to `PCl_(5)(g) rarr PCl_(3) (g) + Cl_(2)(g)`A. `DeltaH = DeltaE`B. `DeltaH gt DeltaE`C. `DeltaH lt DeltaE`D. None of these |
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Answer» Correct Answer - B `PCl_(5)(g)rarrPCl_(3)(g)+Cl_(2)(g)` For this reaction `Deltan_(g)=2-1=1` `Deltan_(g)` is positive, i.e., there is an increase in the number of gaseous moles then `DeltaHgtDeltaE` |
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| 218. |
Which of the following units represents the largest amount of energyA. Electron voltB. ErgC. JouleD. Calorie |
| Answer» Correct Answer - D | |
| 219. |
Which of the following would be expected to have the largest antropy per moleA. `SO_(2)Cl_(2)(s)`B. `SO_(2)Cl_(2)(g)`C. `SO_(2)Cl_(2)(l)`D. `SO_(2)(g)` |
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Answer» Correct Answer - D Suppose the no. of gm, 1 gm for each case so the max. no. of mole expected for `SO_(2)(g)` which has to expected maximum entropy. |
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| 220. |
For melting of 3 moles of water at `0^(@)C` the `DeltaG^(@)` isA. ZeroB. `+ ve`C. `- ve`D. Unpredictable |
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Answer» Correct Answer - A For equilibrium `DeltaG^(@)=0` . |
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| 221. |
For conversion C(graphite) `rarr` C(diamond) the `DeltaS` isA. ZeroB. PositiveC. NegativeD. Unknown |
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Answer» Correct Answer - C Conversion of graphite into diamond is an endothermic reaction. So, heat of diamond is higher than that of graphite. But `DeltaS` would be negative for the conversion of graphite into diamond. |
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| 222. |
Entropy will not change for the reactionA. Crystallization of sucrose from the solutionB. Corrosion of ironC. Conversion of ice into waterD. Vaporisation of Camphor |
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Answer» Correct Answer - B The change in entropy in going from one state to another is independent of the path. |
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| 223. |
The positive value of `DeltaS` indicates thatA. The system becomes less disorderedB. The system becomes more disorderedC. The system is in equilibrium positionD. The system tends to reach at equilibrium position |
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Answer» Correct Answer - B `DeltaS=+ve` than the system is more disordered. |
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| 224. |
Entropy changes for the process, `H_(2)O_((l))rarrH_(2)O_((s))` at normal pressure and 274 K are given below `DeltaS_("system")=-22.13`, `DeltaS_("surr")=+22.05`, the process is non-spontaneous becauseA. `DeltaS_("system") " is " -ve`B. `DeltaS_("surr")" is +ve"`C. `DeltaS_(u)" is -ve"`D. `DeltaS_("system")ne DeltaS_("surr")` |
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Answer» Correct Answer - C `DeltaS_(u)=DeltaS_("system")+DeltaS_("surrounding")=-22.13+22.05 =-0.08` For a spontaneous process, `DeltaS_(u)` must be positive i.e., `DeltaS_(u)=DeltaS_("system")+DeltaS_("surrounding")gt0`. |
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| 225. |
A container has hydrogen and oxygen mixture in ratio of 4 : 1 by weight, thenA. Internal energy of the mixture decreasesB. Internal energy of the mixture increasesC. Entropy of the mixture increasesD. Entropy of the mixture decreases |
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Answer» Correct Answer - C Entropy of the mixture increases due to increase in impurity. |
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| 226. |
If 900 J/g of heat is exchanged at boiling point of water, then what is increase in entropyA. `43.4 J//mol`B. `87.2 J//"mole"`C. `900 J//"mole"`D. Zero |
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Answer» Correct Answer - A `DeltaS_("vap")=((900xx18))/(373)=43.4 JK^(-1) mol^(-1)`. |
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| 227. |
Which of the following are not state functions (I) q+w (II) q (III) w (IV) H-TSA. (I), (II) and (III)B. (II) and (III)C. (I) and (IV)D. (II), (III) and (IV) |
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Answer» Correct Answer - B |
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| 228. |
Choose the reaction in which `DeltaH` is not equal to `DeltaU`A. `C_((gr))+O_(2(g))rarrCO_(2(g))`B. `C_(2)H_(4(g))+H_(2(g))rarrC_(2)H_(6(g))`C. `2C_((gr))+H_(2(g))rarrC_(2)H_(2(g))`D. `H_(2(g))+I_(2(g))rarr2HI_((g))` |
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Answer» Correct Answer - B `DeltaH=DeltaU+Deltan_(g)RT` `"When "Deltan_(g)=0, DeltaH=DeltaU` For (a), `Deltan_(g)=0` `(b) Deltan_(g)=1-2=-1` (c) `Deltan_(g)=0` (d)` Deltan_(g)=2-2=0` `(e) Deltan_(g)=2-2=0` |
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| 229. |
The enthalpy of solution of sodium chloride is 4 kJ `mol^(-1)` and its enthalpy of hydration of ions is `-784kJ mol^(-1)`. Then the lattice enthalpy of NaCl `("in kJ mol"^(-1))` isA. `+788`B. `+4`C. `+398`D. `+780` |
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Answer» Correct Answer - A `DeltaH_("sol")=DeltaH_("lattice")+DeltaH_("hydration")` |
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| 230. |
The work done by the gas liberated when 50 g of iron (molar mass 55.85 g `mol^(-1)`) reacts with hydrochloric acid in an open beaker at `25^(@)C`A. ZeroB. `-2.2 kJ`C. 2.2 kJD. 0.22 kJ |
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Answer» Correct Answer - B `w=-PV=-nRT` `w=-(50)/(55.85)xx8.314xx298` `w=-2218.05 =-2.218 kJ` |
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| 231. |
Enthalpy for the reaction `C+O_(2)rarrCO_(2)` isA. PositiveB. NegativeC. ZeroD. None |
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Answer» Correct Answer - B It is a combustion reaction, `DeltaH=-ve`. |
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| 232. |
Which expression is correct for the work done in adiabatic reversible expansion of an ideal gasA. `W=nRTln.(V_(2))/(V_(1))`B. `W=n_(e)ln.(T_(2))/(T_(1))`C. `W=PDeltaV`D. `W=-int_(1)^(2)PdV` |
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Answer» Correct Answer - C `W=PDeltaV` |
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| 233. |
Enthalpy change for reaction, `(1)/(2)H_(2)+(1)/(2)Cl_(2)rarrHCl`, is calledA. Enthalpy of combinationB. Enthalpy of reactionC. Enthalpy of formationD. Enthalpy of fusion |
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Answer» Correct Answer - C Enthalpy of formation of HCl. |
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| 234. |
The enthalpy of neutralization is about 57.3 kJ for the pairA. `HCl and NH_(4)OH`B. `NH_(4)OH and HNO_(3)`C. HCl and NaOHD. `CH_(3)COOH and NaOH` |
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Answer» Correct Answer - C Heat of neutralization between strong acid and a strong base is about -13.7 kcal i.e., 57.3 kJ. |
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| 235. |
The enthalpy of neutralization of HCN by NaOH is `- 12.13 kJ mol^(-1)`. The enthalpy of ionisation of HCN will beA. 4.519 kJB. 54.10 kJC. 451.9 kJD. 45.19 kJ |
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Answer» Correct Answer - D `-57.4+x=-12.13` `x=45.2`. |
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| 236. |
Which of the following fuels will have the highest calorific value `(kJ//kg)`A. CharcoalB. KeroseneC. WoodD. Dung |
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Answer» Correct Answer - B Out of given substances, kerosene oil has maximum calorific value. |
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| 237. |
The calorific value of fat isA. Less than that of carbohydrate and proteinB. Less than that of protein but more than carbohydrateC. Less than that of carbohydrate but more than that of proteinD. More than that of carbohydrate and protein |
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Answer» Correct Answer - D Fat has highest calorific value among the protein, carbohydrate and fat. |
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| 238. |
One mole of methanol when burnt in `O_(2)`, gives out 723 kJ `mol^(-1)` heat. If one mole of `O_(2)` is used, what will be the amount of heat evolvedA. 723 kJB. 924 kJC. 482 kJD. 241 kJ |
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Answer» Correct Answer - C `CH_(3)OH(l)+(3)/(2)O_(2)(g)rarrCO_(2)(g)+2H_(2)O(l) DeltaH=-723 kJ` By burning with `(3)/(2)O_(2)(g)` heat evolved =-723 kJ `therefore` By burning with 1 mole `O_(2)(g)=(-723xx2)/(3)=-482 kJ`. |
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| 239. |
The energy evolved is highest for which of the following reactionsA. `F+e^(-)rarrF^(-1)`B. `Cl+e^(-)rarrCl^(-)`C. `S+2e^(-)rarrS^(2-)`D. `O+2e^(-)rarrO^(2-)` |
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Answer» Correct Answer - B Due to high electron affinity of Cl the highest energy is evolved. |
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| 240. |
For an exothermic reactionA. H of the products is less than H of the reactantsB. H of the products is more than H of the reactantsC. H of the products is equal to H of the reactantsD. `DeltaH` is always positive |
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Answer» Correct Answer - A `H_("product")ltH_("reactant")` for exothermic reaction. |
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| 241. |
Reaction, `H_(2)(g)+I_(2)(g)rarr2HI(g)DeltaH=-12.40 kcal`. According to this, the heat of formation of HI will beA. 12.4 kcalB. `-12.4` kcalC. `-6.20` kcalD. 6.20 kcal |
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Answer» Correct Answer - C Heat of formation is for 1 mole. Hence `DeltaH_(f)^(@)(HI)=-12.40//2=-6.20 kcal`. |
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| 242. |
In the reaction for the transition of carbon in the diamond form to carbon in the graphite form, `DeltaH` is `-453.5` cal. This points out thatA. Graphite is chemically different from diamondB. Graphite is as stable as diamondC. Graphite is more stable than diamondD. Diamond is more stable than graphite |
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Answer» Correct Answer - C `C_(D)rarrC_(G),DeltaH=-453.5 cal`. i.e. energy of `C_(G)` is less and thus more stable. |
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| 243. |
How much energy is released when 6 mole of octane is burnt in air? Given `DeltaH_(f)^(@) " for " CO_(2)(g), H_(2)O(g) and C_(8)H_(18)(l)` respectively are - 490, -240 and + 160 kJ/molA. `- 6.2 kJ`B. `- 37.4 kJ`C. `- 35.5 kJ`D. ` - 20.0 kJ` |
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Answer» Correct Answer - B `C+O_(2)rarrCO_(2) , DeltaH_(F)=-490 kJ//mol - (i)` `H_(2)+1//2O_(2)rarrH_(2)O, DeltaH_(F)=-240 kJ//mol - (ii)` `8C+9H_(2)rarrC_(8)H_(18), DeltaH_(F)=+160 kJ//mol - (iii)` applying (i)`xx8+(ii)xx9-(iii)` `C_(8)H_(18)+(25)/(2)O_(2)rarr8CO_(2)+9H_(2)O` `DeltaH^(@)=-3920-2160-160=-6240 kJ//mol` `=37440 kJ//mol =-37.4 kJ`. |
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| 244. |
The rusting of iron takes place as follows `2H^(+)+2e^(-)+1//2O_(2)rarrH_(2)O(l), E^(@)=+1.23 V` `Fe^(2+)+2e^(-)rarrFe(s), E^(@)=-0.44V` Calculate `DeltaG^(@)` for the net processA. `-322 kJ mol^(-1)`B. `-161 kJ mol^(-1)`C. `-152 kJ mol^(-1)`D. `-76 kJ mol^(-1)` |
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Answer» Correct Answer - A `Fe(s)rarrFe^(2+)+2e^(-), " "E^(@)=0.44 V` `(2H^(+)+2e^(-)+1//2O_(2)rarrH_(2)O(l), " "E^(@)=+1.23 V)/(Fe(s) +2H^(+)+1//2O_(2)rarrFe^(2+)+H_(2)O,)` `E_("cell")^(@)=0.44+1.23=1.67 V` `thereforeDeltaG^(@)="-nF " E_("cell")^(@)=-2xx96500xx1.67=-322 kJ`. |
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| 245. |
`DeltaG` in `Ag_(2)Orarr2Ag+1//2O_(2)` at a certain temperature is -10kJ/mole. Pick the correct statementA. `Ag_(2)O` decomposes to Ag and `O_(2)`B. Ag and `O_(2)` combines to form `Ag_(2)O`C. Reaction is in equilibriumD. Reaction does not take place |
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Answer» Correct Answer - A `DeltaG-ve` means the process is spontaneous. |
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| 246. |
Given, `NH_(3(g))+3Cl_(2(g))iffNCl_(3(g))+3HCl_((g)),-DeltaH_(1)` `N_(2(g))+3H_(2(g))iff2NH_(3(g)),-DeltaH_(2)` `H_(2(g))+Cl_(2(g))iff2HCl_((g)),DeltaH_(3)` The heat of formation of `NCl_(3(g))` in the terms of `DeltaH_(1), DeltaH_(2)` and `DeltaH_(3)` isA. `DeltaH_(f)=-DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`B. `DeltaH_(f)=DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`C. `DeltaH_(f)=-DeltaH_(1)-(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`D. None of these |
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Answer» Correct Answer - C Find `(1)/(2)N_(2)+(3)/(2)Cl_(2)rarrNCl_(3),DeltaH_(f)` Multiply Eq. (ii) by 1/2 and add to Eq. (i), Eq. (iii) by 3/2 and subtract from Eq. (i), we get `DeltaH_(f)=-DeltaH_(l)-(DeltaH_(2))/(2)-[+(3)/(2)DeltaH_(3)]` `=-DeltaH_(l)-(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)` |
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| 247. |
Consider the Born- Haber cycle for the formation of an ionic compound given below and identify the compound (Z) formed. A. MXB. `M^(+) X_((g))^(-)`C. `M^(+)X_((s))^(-)`D. `M^(-)X_((s))^(-)` |
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Answer» Correct Answer - C The energy released when the requisite number of positive and negative ions are condensed into crystal to form one mole of the compound is called lattice energy `M_((g))^(+)+X_((g))^(-)overset(DeltaH_(5))rarrMunderset((Z))(X_((s))),DeltaH_(5)rarr` Lattice energy. |
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| 248. |
The entropy values (in `JK^(-1) mol^(-1)`) of `H_(2(g))` = 130.6, `Cl_(2(g))=223.0` and `HCl_((g))=186.7` at 298 K and 1 atm pressure. Then entropy change for the reaction `H_(2(g))+Cl_(2(g))rarr2HCl_((g))` isA. `+540.3`B. `+727.3`C. `-166.9`D. `+19.8` |
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Answer» Correct Answer - D `DeltaS^(@)=2S_(HCl)^(@)-(S_(H_(2))^(@)+S_(Cl_(2))^(@))` `=2xx186.7-(130.6+223.0)=19.8 JK^(-1) mol^(-1)`. |
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| 249. |
The heats of combustion of carbon and carbon monoxide are - 393.5 and -283.5 KJ `mol^(-1)`, respectively. The heat of formation (in kJ) of carbon monoxide per mole isA. 676.5B. `-676.5`C. `-110.5`D. 110.5 |
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Answer» Correct Answer - C `C(s)+O_(2)(g)rarrCO_(2)(g), DeltaH=-393.5 kJ//mol` `CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g) , DeltaH=-283.5 kJ//mol` `C(s)+(1)/(2)O_(2)(g)rarrCO(g) , DeltaH=-393.5+283.5 kJ//mol=-110 kJ//mol` |
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| 250. |
In which of the following reactions does the heat change represent the heat of formation of waterA. `2H_(2)(g)+O_(2)(g)rarr2H_(2)O(l),DeltaH=-116 kcal`B. `H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l),DeltaH=-58 kcal`C. `H^(+)(aq.)+OH^(-)(aq.)rarr2H_(2)O(l), DeltaH=-13.7 kcal`D. `C_(2)H_(2)(g)+2(1)/(2)O_(2)(g)rarr2CO_(2)(g)+H_(2)O(l), DeltaH=-310 kcal` |
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Answer» Correct Answer - B Heat of formation is the formation of one mole of the substance from its elements. |
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