94420 + Interview Questions in Chemistry in Class 12 Page 603 InterviewSolution

30101.	The heats of neutralisation of CH_(3)COOH, HCOOH, HCN and H_(2)S are -13.2, -13.4, -2.9 and -3.8 kcal per equivalent respectively. Arrange these acids in increasing order of strength.
Answer» Solution :Since we know that the greater the (negative) value of the HEAT of NEUTRALISATION, the more is the strength of the acid, the given acids may be arranged in increasing order of strength as `HCNltH_(2)SltCH_(3)COOHltHCOOH`

Discussion

30102.	The heats of neutralisation of CH_3 - COOH , H-COOH,HCN and H_2 S are -13.2,-13.4, -2.9,"and" -3.8 kcal/equivalent respectively. The CORRECT increasing order of acid strength is____________.
Answer» `HCOOHltCH_3 COOHlt H_2 SltHCN` `HCNltH_2SltCH_3-COOHltH-COOH` `HCOOHltCH_3COOHltHCNltH_2S` `CH_3 -COOHltH_2SltHCNltH-COOH` ANSWER :B

Discussion

30103.	The heats of formation of PCl_(3) and PH_(3) are -306 kJ"mole"^(-1) and +8 kJ"mole"^(-1) respectively and the heats of atomization of phosphorus, chlorine and hydrogen are given by P_(s) to P_(g), DeltaH = 314 kJ"mole"^(-1) Cl_(2)(g) to 2Cl_(g), DeltaH = 242 kJ "mole"^(-1) H_(2)(g) to 2H_(2), DeltaH= 433 KJ "mole"^(-1) Calculate E_(p-cl) and E_(P-H).
Answer» ANSWER :319.33 KJ

Discussion

30104.	The heats of formation of Na_(2)B_(4)O_(7) (s) and Na_(2)B_(4)O_(7).10H_(2)O (s) are -742 and -1460 kcal respectively. Calculate the heat of hydration of Na_(2)B_(4)O_(7) in forming decahydrate.
Answer» ANSWER :(-718 KCAL)

Discussion

30105.	The heats of formation of CO_(2)(g) and H_(2)O(l) are -94.05 kcal and -68.32 kcal respectively. The heat of combustion of methyl alcohol (l) is -173.65 kcal. Calculate the heat of formation of liquid methyl alcohol.
Answer» ANSWER :(-57.04 KCAL)

Discussion

30106.	The heats of combustion of rhombic and monoclinic sulphur are respectively 70960 and 71030 calories. What will be the heat of conversion of rhombic sulphur to monoclinic
Answer» 70960 CALORIES 71030 calories `- 70` calories `+ 70` calories Solution :S (rhombic) `+O_(2)rarrSO_(2), DeltaH=70960 " CAL....(i)"` `S ("monoclinic") +O_(2)rarrSO_(2), DeltaH=71030 " cal....(ii)"` AIM: S (rhombic) `rarr` S (monoclinic) eq. (i)- eq. (ii) gives the required RESULT.

Discussion

30107.	The heats of combustion of yellow phosphorus and red phosphorous are - 9.19 KJ and - 8.78 KJ respectively, then heat of transition of yellow phosphorus to red phosphorous is
Answer» Solution :(i) `P_(4) ("yellow")+5O_(2)(G) rarr P_(4)O_(10)+9.19 KJ` (ii) `P_(4) ("RED")+5O_(2)(g) rarr P_(4)O_(10)+8.78 kJ` SUBTRACTING, `P_(4) ("yellow")-P_(4) ("red")=1.13 KJ` `rArr P ("yellow")=P_(4) ("rad")+1.13 KJ` So, HEAT of transition of yellow to red phosphorus is `-1.13 KJ`

Discussion

30108.	The heats of combustion of graphite and dimond at 298 K are -393 and -395 kJ/ "mole" respectively . The specific heats of these substances are 720 and 505 J kg^(-1)K^(-1) respectively. Calculate the heat of transformation of graphite into diamond at 273 K.
Answer» Solution :`DELTAH={-393-(395)}+(0.720-0.505)XX(12)/(1000)xx(298-273)]`

Discussion

30109.	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 is
Answer» 676.5 `-676.5` `-110.5` 110.5 Solution :`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`

Discussion

30110.	The heats of combustion of C and CO are -393.5 and -283.5 kJ mol^(-1) respectively . The heat of formation (in kJ) of CO per mode is
Answer» `676.5` `-676.5` `-110.5` `110.5` ANSWER :C

Discussion

30111.	The heats of combustion of ammonia and hydrogen are 9.06 and 68.9 kcal respectively. Calculate the heat of formation of ammonia.
Answer» Solution :Given that, (i) `{:(NH_(3)(g)+(3)/(4)O_(2)(g) to (1)/(2)N_(2)(g)+(3)/(2)H_(2)O,,DeltaH=-9.06 "kcal"):}` (ii) `{:(H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),,DeltaH=-68.9 "kcal"):}` ( The negative signs are taken as the combustion process is exothermic) we have to calculate `DeltaH` of the FOLLOWING reaction : (iii) `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) to NH_(3)(g),DeltaH=?` Since `H_(2)` in Equation (iii) and in Equation (ii) is on ethe same side and `NH_(3)` in Equation (iii) and in Equation (i) is on OPPOSITE sides, we FIRST multiply Equation (ii) by (3)/(2) to equate number of MOLES of `H_(2)` in equations (ii) and (iii) . Then SUBTRACTING Equation (i) from equation (ii) `{i.e.,(3)/(2)xx"Equation"(ii)-"Equation"(i)},"we get",` `(3)/(2)H_(2)(g)+(3)/(4)O_(2)(g)-NH_(3)(g)-(3)/(4)O_(2)(g) to` `(3)/(2)H_(2)O(g)-(1)/(2)N_(2)(g)-(3)/(2)H_(2)O(g),` `DeltaH=(3)/(2)xx(-68.9)-(-9.06)` or `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) to NH_(3),DeltaH=-94.29 "kcal"`

Discussion

30112.	The heats of atomisation of PH_(3)(g) and P_(2)H_(4)(g) are 954 kJ mol^(-1) and 1485 kJ mol^(-1) respectively. The P-P bond energy in kJ mol^(-1) is
Answer» 213 426 318 1272 Solution :`PH_(3) (g) RARR P (g) + 3H (g), DELTA H^(@) = 954 kJ mol^(-1)` The bond energy is average energy for 3 P - H bonds as : Average bond energy of P - H bond `= (954)/(3)` `= 318 kJ mol^(-1)` `P_(2)H_(4)` has four P - H bonds and one P - P bond `B. E. (P - P) + 4 B. E. (P - H) = 1485 kJ mol^(-1)` `B. E. (P - P) + 4 xx 318 = 1485 kJ mol^(-1)` `B. E. (P - P) = 1485 - 4 xx 318` `= 1485 - 1272 = 213 kJ mol^(-1)`

Discussion

30113.	The heats of adsorption in physisorption lie in the range (in kJ/mol)
Answer» `40-400` `40-100` `10-40` `1-10` SOLUTION :For PHYSIOSORPTION the `DELTAH` VALUE is 10-40kH/mol.

Discussion

30114.	The heats evolved and absorbed when 100 g each of CuSO_(4)andCuSO_(4)*5H_(2)O are dissolved in water are 41.7 and 4.7 kJ/mole respectively. The heat of hydration of CuSo_(4) is (in kJ/mole)
Answer» `-66.59` `+11.72` `-78.2` `+78.2` ANSWER :C

Discussion

30115.	Railway wagon axles are made by heating rods of iron embedded in charcoal powerder. The process is known as
Answer» CASE HARDENING Sheradizing ANNEALING TEMPERING ANSWER :A

Discussion

30116.	The heating process for the extraction of elements are quite old but highly acceptable method for the extraction of elements. Become in this process the elements produced is in the highly pure state. Mostly As, Sb,Ni, Zr,B etc are prepared by this principle 4 mumber of metal sulphides which may be reated first in air to partially convert them to the aride, and then further roasted in the absence of air, wing self reduction. For which of the given sulphides auto reduction is applicable
Answer» CuS PbS HgS All Solution :. METALS like Cu, HG, Pb EXTRACTED by AUTO reduction

Discussion

30117.	The heating process for the extraction of elements are quite old but highly acceptable method for the extraction of elements. Become in this process the elements produced is in the highly pure state. Mostly As, Sb,Ni, Zr,B etc are prepared by this principle 4 mumber of metal sulphides which may be reated first in air to partially convert them to the aride, and then further roasted in the absence of air, wing self reduction. How does impure Ni is purified
Answer» impure `Ni+CO_(2) to NiCO_(3) overset(Delta) to NiO+CO_(2) overset(Delta) to Ni(s)` impure `Ni+CO overset(50^(@)C) to Ni(CO)_(3)(g) overset(Delta) to Nis+Co(g) overset("Condensation") to Ni(s)` pure impure `Ni+CO_(2) overset(50^(@)C) toNi(CO)_(4),(s) overset(Delta) toNi(s)+CO(g)` impure `Ni+CO overset(50^(@)C) to Ni(CO)_(4) (g) overset(30^(@)C) to Ni(c) +CO(g)` Solution :Monds process : Condensation `UNDERSET("1m pure")(Ni)+4CO to Ni(CO)_(4) overset(Delta) to underset(Ni("pure"))underset(dar "condensation")(Ni_(g))+4CO`

Discussion

30118.	The heating process for the extraction of elements are quite old but highly acceptable method for the extraction of elements. Become in this process the elements produced is in the highly pure state. Mostly As, Sb,Ni, Zr,B etc are prepared by this principle 4 mumber of metal sulphides which may be reated first in air to partially convert them to the aride, and then further roasted in the absence of air, wing self reduction. The H_(2)(g)is not widely used as a reducing agent because
Answer» `H_(2)` decompose to ATOMIC HYDROGEN at higher TEMPERATURE `H_(2)` isomerises to orthohydrogen at higher temperature many metal form hydrides at lower temperature there is also a risk of explosion from hydrogen and dioxygen in air Solution :`H_(2)` combines with most of metals to form metal hydrides

Discussion

30119.	The heating process for the extraction of elements are quite old but highly acceptable method for the extraction of elements. Become in this process the elements produced is in the highly pure state. Mostly As, Sb,Ni, Zr,B etc are prepared by this principle 4 mumber of metal sulphides which may be reated first in air to partially convert them to the aride, and then further roasted in the absence of air, wing self reduction. In the purification Zr and Ti, which of the following is/are true
Answer» `Zr+2I_(2) to ZrI_(4) OVERSET("passed over") underset("the white hot w") to` the pure Zr is deposited on W :`Ti+2I_(2) to TiI_(4) overset("passed over") underset("the white hot w") to` the pure Ti is deposited on W :`Zr+2I_(2) to ZrI_(4) overset("MIXED with W") underset("& then heated") to ZrI_(4)` is reudced to `ZrI_(4)` Both a & b are correct Solution :Van Arkel method is USED to purfiy `Ti, Zr, HF...` `underset(("1M pure"))(Ti//Zn)+I_(2) to underset(Ti//r("pure")) underset(darr)(Ti//Zr) ("iodide")TiI_(4)(g)//Znrl_(4)` Pure Ti// Zr are deposited on W surface.

Discussion

30120.	The heating of which of the following gives pure chlorine .
Answer» `MnO_(2)` Bleaching powder `+ HCl` `PtCl_(4)` `NaCl + MnO_(2) + H_(2) SO_(4)` SOLUTION :Pure CHLORIDE obtains by heating of DRY platonic chloride `(PtCl_(4))` `PtCl_(4) overset( 374^(@)C)(rarr) PtCl_(2) + Cl_(2) overset( 58^(@)C)(rarr) Pt + 2Cl_(2)`

Discussion

30121.	The heating of phenyl-methyl ethers with HI produces
Answer» Iodobenzene Phenol Benzene Ethyl chlorides Solution :

Discussion

30122.	The heating of phenoylmethyl ethers with HI produces-
Answer» ETHYL chlorides iodobenzene phenol benzene Answer :C

Discussion

30123.	The heat required to raise the temperature of a body by 1 K is called
Answer» Specific heat Thermal capacity Water EQUIVALENT None of these Solution :Heat REQUIRED to rise the temperature of a BODY by 1 K called thermal capacity of the body.

Discussion

30124.	The heating of phenyl-methyl ethers with HIS produces
Answer» Benzene Ethyl chlorides Iodobenzene Phenol Answer :D

Discussion

30125.	The heat of vaporisation : Delta H_("vap"), of C Cl_(4) at 27^(@)C is 42 kJ/mole C Cl_(4)(l) rarr C Cl_(4) (g) : Delta H_("vap") = 42.0 kJ//"mole" If 1 mole of liquid CCl_(4) at 27^(@)C has entropy of 214 J/K mole, what is the entropy (in J/K-mol) of 1 mole of vapour in equilibrium with liquid at this temperature.
Answer» 74 454 354 254 Solution :If liquid `C Cl_(4)` is in EQUILIBRIUM with it's vapour's them `(DeltaH_("VAP "27^(@)C))/(300)=DeltaS_("vap")=S_(C Cl_(4)(g))-S_(C Cl_(4)(l))` `rArr (42, 000)/(300)=S_(C Cl_(4)(g))-214" J/K mole"` `140+214=354" J/K mole"`

Discussion

30126.	The heat of vaporisation of C Cl_(4) at 298 K is 43.0 kJ/"mole". C Cl_(4)(l) to C Cl_(4)(g),DeltaH=43.0 kJ If 1 mole of liquid C Cl_(4) at 298 K has an entropy of 214 J/K , what is the entropy of 1 mole of the vapour in equilibrium with the liquid at this temperature ?
Answer» Solution :`DeltaS=(DeltaH_(0))/(T)=(43.0xx10^(3))/(298)=144.3J//molK` But `DeltaS=S_(V)-S_(L)` or `S_(V)=DeltaS+S_(L)=144.3+214=358.3J//molK`

Discussion

30127.	The heat released during neutralisation is constant for the reaction of aqueous solutions of:
Answer» STRONG ACID and strong BASE Strong acid and WEAK base Strong base and weak acid In all the cases Answer :A

Discussion

30128.	The heat released by one mole of sugar from a bomb calorimeter ecperiment is 5648 KJ//mol. Calculate the enthalpy of combution per mole of sugar.
Answer» ANSWER :`5648 KJ//mol`

Discussion

30129.	The heat of sublimation of iodine is 24 cal g^(-1)at 50°C. If specific heat of solid iodine and its vapour are 0.055 and 0.031 cal g^(-1)respectively Calculate the heat of sublimation of iodine at 100°C.
Answer» Solution :`DeltaH_(2) - DeltaH_(1) = DeltaC_(p)(T_(2)-T_(1))` `DeltaH_(2) -24 =(0.031 -0.055)(100-50)` `RARR DeltaH_(2)=22.8 CAL g^(-1)`

Discussion

30130.	The heat of sublimation of iodine is 24"cal"g^(-1) at 50^(@)C. If specific heat of solid iodine and its vapour are 0.055 and 0.031"cal"g^(-1) respectively, the heat of sublimation of iodine at 100^(@)C is
Answer» `22.8"CAL"g^(-1)` `25.2"cal"g^(-1)` `-22.8"cal"g^(-1)` `-25.2"cal",g^(-1)` Solution :`DletaH_(2)-DeltaH_(1)=DeltaC_(P)(T_(2)-T_(1))` `DeltaH_(2)-24=(0.031-0.055)(100-50)` `impliesDeltaH_(2)=22.8"CALG^(-1)`

Discussion

30131.	The heat of reaction of N_(2)+3H_(2) to 2NH_(3) is -20 kcal. If the bond energies of H-H and N-H bonds are 104 and 93 kcal/ mole respectively, calculate the bond energy of N-=N bond.
Answer» ANSWER :(226 KCAL)

Discussion

30132.	The heat of reaction for C_(10)H_(8)(s)+12O_(2)(g)to 10CO_(2)(g)+4H_(2)O(l) at constant volume is -1228.2 kcal at 25^(@)C. The heat of reaction at constant pressure and same temperature is
Answer» `-1228.2` kcal `-1229.3` kcal `-1232.9` kcal `-1242.6` kcal Solution :`C_(10)H_(8)(s)+12O_(2)(g) to 10CO_(2)(g)+4H_(2)O(l)` We KNOW that, `DeltaH=DeltaE+Deltan_(g)RT` `therefore DeltaH=-1228.2xx10^(3)+(-2)(2)(298)` `=-1229392cal=-1229.392kcal`

Discussion

30133.	The heat of reaction (1)/(2) H_(2) (g) + (1)/(2) Cl_(2) (g) to HCl (g) at 27^(@) C is -22.1 k cal . Calculate the heat of reaction at 77^(@) C . The molar heat capacities at constant pressure at 27^(@) for hydrogen , chlorine & HCl are 6.82 , 7.70 & 6.80 cal mol^(-1) respectively .
Answer» SOLUTION :`-22.123` K CAL

Discussion

30134.	The heat of reaction at constant pressure is given by
Answer» `E_(P)-E_(R)` `E_(R)-E_(P)` `H_(P)-H_(R)` `H_(R)-H_(P)` SOLUTION :`DeltaH(H_(P)-H_(R))=q_(p)`.

Discussion

30135.	The heat of physisorption lie in the range of
Answer» 1 – 10 `KJ" mol"^(-1)` 20 to 40 `kJ" mol"^(-1)` 40 to 200 `kJ" mol"^(-1)` 200 to 400 `kJ" mol"^(-1)` ANSWER :B

Discussion

30136.	The extent of physical adsorption:
Answer» ZERO `20-50 LA mol-(1)` `200 - 500 kJ mol-(1)` very high. Answer :B

Discussion

30137.	The heat of neutralization of (i) CHCl_2 – COOH by NaOH is 12830 cal, (ii) HCl by NaOH is 13680 cal and (iii) NH_4OH by HCl is 12270 cal. What is the heat of neutralization of dichloro acetic acid by NH_4OH? Calculate the heats of ionization of dichloro acetic acid and NH_4OH.
Answer» Solution :(i) `CHCl_2COOH + OH^(-) to CHCl_(2)COO^(-) + H_2O, ""DeltaH_(1)= 12830 cal` (ii) `H^(+) + OH^(-) to H_2O, ""DeltaH_(2) = 13680 cal` (iii) `NH_4OH + H^(+) to NH_(4)^(+) + H_2O, ""DeltaH_3 = -12270 cal` Consider (iv) `CHCl_(2)COOH + NH_4OH to CHCl_2COO^(-) + NH_(4)^(+) + H_2O` (v) Will be obtained by APPLYING `[(i) + (iii) - (ii)]` `Delta H (iv) = Delta H_(1) + Delta H_(3) - DeltaH_(2) = -12830 - 12270 + 13680 = -11420 cal` Now, (vi) `CHCl_(2)COOH to CHCl_(2)COO^(-) + H^(+)` (VII) Will be obtained by applying (i) - (ii) `:. Delta H(v) = DeltaH_(1) - DeltaH_2 = 13680 - 12830 = 850 cal//mol` Similarly `DeltaH_("ionization")` for `NH_(4)OH = 13680 - 12270 = 1410 cal// mol`.

Discussion

30138.	The heat of neutralization will be highest in
Answer» `NH_(4)OH and CH_(3)COOH` `NH_(4)OH and HCL` `KOH and CH_(3)COOH` `KOH and HCl` Solution :STRONG base (KOH) and strongacid (HCl).

Discussion

30139.	The heat of neutralization of NaOH with HCl is 57.3 KJ and with HCN is 12.1 KJ. The heat of ionization of HCN is
Answer» `+69.4 ` KJ `+45.2` KJ `-45.2` KJ `-69.4` KJ Answer :B

Discussion

30140.	The heat of neutralization of strong base and strong acid is 57.0 kJ. The heat released when 0.7 mole of HNO_3 solution is added to 0.50 mole of NaOH solution is :
Answer» 57.0kJ 28.5 kJ 11.40 kJ 34.9 kJ Answer :B

Discussion

30141.	The enthalpy of neutralisation of HCl by NaOH IS -55.9 kJand that of HCN by NaOH is -12.1 kJ mol^(-1). The enthalpy of ionisation of HCN is
Answer» `-68.0 KJ MOL^(-1)` `-43.8 kJ mol^(-1)` 68.0 kJ `mol^(-1)` 43.8 kJ `mol^(-1)` ANSWER :D

Discussion

30142.	The heat of neutralization of a strong acid by a strong base is a constant because
Answer» `57.0 kJ "mol"^(-1)` `13.7 KCAL "mol"^(-1)` `5.7 XX 10^(4) J "mol"^(-1)` `13.7 xx 10^(3) "cal" "mol"^(-1)` Answer :A::B::C::D

Discussion

30143.	The heat of neutralization of a strong acid and a strong alkali is 57.0 kJ mol^(-1). The heat released when 0.5 mole of HNO_(3) solution is mixed with 0.2 mole of KOH is
Answer» 57.0 KJ 11.4 kJ 28.5 kJ 34.9 kJ Solution :0.2 MOLE will neutralize 0.2 mole of `HNO_(3)` heat EVOLVED `= 57xx0.2=11.4 kJ`.

Discussion

30144.	The heat of neutralisation will be highest for which of the following neutralisation reactions?
Answer» `NH_4OH and HCL` `NH_4OH and CH_3COOH` `NAOH and HCl` `NaOH and CH_3COOH` ANSWER :C

Discussion

30145.	The heat of neutralisation of HCl by NaOH is -55.9 kJ//"mole". If the heat of neutralisation of HCN by NaOH is -12.1 kJ//"mole", the energy of dissociation of HCN is
Answer» `-43.8 KJ` `43.8 kJ` `68 kJ` `-68 kJ` ANSWER :B

Discussion

30146.	the heat of neutralisation of acetic acid and sodium hydroxide is -50.6 kJ eq^(-1). Find the heat of dissociation of CH_(3)COOH if the heat of neutralisation of a strong acid and a strong base is -55.9 Kj eq^(-1).
Answer» Solution :We have, `DeltaH` (neutralisation) `=DeltaH ( "ionisation of" CH_(3)COOH)+DeltaH(H^(+)+OH^(-))` `:.DeltaH("ionisation of" CH_(3)COOH)=-50.6-(-55.9)=5.3 "kJ mol"^(-1)` [DeltaH (ionisation of NaOH)=0 as NaOH is a strong BASE]

Discussion

30147.	The heat of neutralisation of HCl by NaOH is-55.9 kJ/mol. If the heat of neutralisation of HCN by NaOH is -12.1 kJ/mol. The energy of disso-ciation of HCN is:
Answer» `-43.8 KJ` 43.8kJ 68 kJ `-68 kJ` ANSWER :B

Discussion

30148.	The heat of neutralisation of a strong dibasic acid in dilute solution by NaOH is nearly
Answer» `-27.4 KCAL// EQ` `13.7 kcal//"MOLE"` `-13.7 kcal//eq` `-13.7 kcal/"mole"` ANSWER :C

Discussion

30149.	The heat of neutralisation of HCl and NaOH is 57.3 kJ "mol"^(-1). The amount of heat liberated when 0.25 mol of HCl reacts with 1 mol of NaOH is :
Answer» 57.3 kJ 14.325 kJ 28.65kJ 114.6 kJ SOLUTION :Heat liberated for NEUTRALISATION of 0.5 mol of `H^(+) = 57.3 xx 0.5 = 28.65 kJ`

Discussion

30150.	The heat of neutralisation of strong acid and strong base is 57.0 kJ. The heat released when 0.5 mol of HNO_3 is added to 0.2 mol of NaOH solution is:
Answer» 57.0 kJ 11.40 kJ 28.5 kJ 34.9 kJ Solution :0.2 MOLE of `HNO_(3)` reacts with 0.2 mole of NaOH HEAT RELEASED `= 57.0 XX 0.2 = 11.40 kJ`

Discussion

Explore topic-wise InterviewSolutions in .

The heats of neutralisation of CH_(3)COOH, HCOOH, HCN and H_(2)S are -13.2, -13.4, -2.9 and -3.8 kcal per equivalent respectively. Arrange these acids in increasing order of strength.

The heats of neutralisation of CH_3 - COOH , H-COOH,HCN and H_2 S are -13.2,-13.4, -2.9,"and" -3.8 kcal/equivalent respectively. The CORRECT increasing order of acid strength is____________.

The heats of formation of Na_(2)B_(4)O_(7) (s) and Na_(2)B_(4)O_(7).10H_(2)O (s) are -742 and -1460 kcal respectively. Calculate the heat of hydration of Na_(2)B_(4)O_(7) in forming decahydrate.

The heats of formation of CO_(2)(g) and H_(2)O(l) are -94.05 kcal and -68.32 kcal respectively. The heat of combustion of methyl alcohol (l) is -173.65 kcal. Calculate the heat of formation of liquid methyl alcohol.

The heats of combustion of rhombic and monoclinic sulphur are respectively 70960 and 71030 calories. What will be the heat of conversion of rhombic sulphur to monoclinic

The heats of combustion of yellow phosphorus and red phosphorous are - 9.19 KJ and - 8.78 KJ respectively, then heat of transition of yellow phosphorus to red phosphorous is

The heats of combustion of graphite and dimond at 298 K are -393 and -395 kJ/ "mole" respectively . The specific heats of these substances are 720 and 505 J kg^(-1)K^(-1) respectively. Calculate the heat of transformation of graphite into diamond at 273 K.

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 is

The heats of combustion of C and CO are -393.5 and -283.5 kJ mol^(-1) respectively . The heat of formation (in kJ) of CO per mode is

The heats of combustion of ammonia and hydrogen are 9.06 and 68.9 kcal respectively. Calculate the heat of formation of ammonia.

The heats of atomisation of PH_(3)(g) and P_(2)H_(4)(g) are 954 kJ mol^(-1) and 1485 kJ mol^(-1) respectively. The P-P bond energy in kJ mol^(-1) is

The heats of adsorption in physisorption lie in the range (in kJ/mol)

The heats evolved and absorbed when 100 g each of CuSO_(4)andCuSO_(4)*5H_(2)O are dissolved in water are 41.7 and 4.7 kJ/mole respectively. The heat of hydration of CuSo_(4) is (in kJ/mole)

Railway wagon axles are made by heating rods of iron embedded in charcoal powerder. The process is known as

The heating of which of the following gives pure chlorine .

The heating of phenyl-methyl ethers with HI produces

The heating of phenoylmethyl ethers with HI produces-

The heat required to raise the temperature of a body by 1 K is called

The heating of phenyl-methyl ethers with HIS produces

The heat of vaporisation of C Cl_(4) at 298 K is 43.0 kJ/"mole". C Cl_(4)(l) to C Cl_(4)(g),DeltaH=43.0 kJ If 1 mole of liquid C Cl_(4) at 298 K has an entropy of 214 J/K , what is the entropy of 1 mole of the vapour in equilibrium with the liquid at this temperature ?

The heat released during neutralisation is constant for the reaction of aqueous solutions of:

The heat released by one mole of sugar from a bomb calorimeter ecperiment is 5648 KJ//mol. Calculate the enthalpy of combution per mole of sugar.

The heat of sublimation of iodine is 24 cal g^(-1)at 50°C. If specific heat of solid iodine and its vapour are 0.055 and 0.031 cal g^(-1)respectively Calculate the heat of sublimation of iodine at 100°C.

The heat of sublimation of iodine is 24"cal"g^(-1) at 50^(@)C. If specific heat of solid iodine and its vapour are 0.055 and 0.031"cal"g^(-1) respectively, the heat of sublimation of iodine at 100^(@)C is

The heat of reaction of N_(2)+3H_(2) to 2NH_(3) is -20 kcal. If the bond energies of H-H and N-H bonds are 104 and 93 kcal/ mole respectively, calculate the bond energy of N-=N bond.

The heat of reaction for C_(10)H_(8)(s)+12O_(2)(g)to 10CO_(2)(g)+4H_(2)O(l) at constant volume is -1228.2 kcal at 25^(@)C. The heat of reaction at constant pressure and same temperature is

The heat of reaction (1)/(2) H_(2) (g) + (1)/(2) Cl_(2) (g) to HCl (g) at 27^(@) C is -22.1 k cal . Calculate the heat of reaction at 77^(@) C . The molar heat capacities at constant pressure at 27^(@) for hydrogen , chlorine & HCl are 6.82 , 7.70 & 6.80 cal mol^(-1) respectively .

The heat of reaction at constant pressure is given by

The heat of physisorption lie in the range of

The extent of physical adsorption:

The heat of neutralization of (i) CHCl_2 – COOH by NaOH is 12830 cal, (ii) HCl by NaOH is 13680 cal and (iii) NH_4OH by HCl is 12270 cal. What is the heat of neutralization of dichloro acetic acid by NH_4OH? Calculate the heats of ionization of dichloro acetic acid and NH_4OH.

The heat of neutralization will be highest in

The heat of neutralization of NaOH with HCl is 57.3 KJ and with HCN is 12.1 KJ. The heat of ionization of HCN is

The heat of neutralization of strong base and strong acid is 57.0 kJ. The heat released when 0.7 mole of HNO_3 solution is added to 0.50 mole of NaOH solution is :

The enthalpy of neutralisation of HCl by NaOH IS -55.9 kJand that of HCN by NaOH is -12.1 kJ mol^(-1). The enthalpy of ionisation of HCN is

The heat of neutralization of a strong acid by a strong base is a constant because

The heat of neutralization of a strong acid and a strong alkali is 57.0 kJ mol^(-1). The heat released when 0.5 mole of HNO_(3) solution is mixed with 0.2 mole of KOH is

The heat of neutralisation will be highest for which of the following neutralisation reactions?

The heat of neutralisation of HCl by NaOH is -55.9 kJ//"mole". If the heat of neutralisation of HCN by NaOH is -12.1 kJ//"mole", the energy of dissociation of HCN is

the heat of neutralisation of acetic acid and sodium hydroxide is -50.6 kJ eq^(-1). Find the heat of dissociation of CH_(3)COOH if the heat of neutralisation of a strong acid and a strong base is -55.9 Kj eq^(-1).

The heat of neutralisation of HCl by NaOH is-55.9 kJ/mol. If the heat of neutralisation of HCN by NaOH is -12.1 kJ/mol. The energy of disso-ciation of HCN is:

The heat of neutralisation of a strong dibasic acid in dilute solution by NaOH is nearly

The heat of neutralisation of HCl and NaOH is 57.3 kJ "mol"^(-1). The amount of heat liberated when 0.25 mol of HCl reacts with 1 mol of NaOH is :

The heat of neutralisation of strong acid and strong base is 57.0 kJ. The heat released when 0.5 mol of HNO_3 is added to 0.2 mol of NaOH solution is: