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.
| 30901. |
The DeltaH_(t)^(@) for CO_(2)(g), CO(g) and H_(2)O (g) are -393.5, -110.5 and -241.8 kJ mol^(-1) respectively. The standard enthalpy change (in kJ) for the reaction. CO_(2)(g) + H_(2)(g) to CO(g) + H_(2)O(g) is |
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Answer» 524.1 |
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| 30902. |
the DeltaH_(t)^(@)for CO_(2)(g) , CO(g) and H_(2)O(g) are -393.5 , -110.5 and -241.8 kJ mol ^(-1) respectively, the standed enthalpy change ( in kJ) for the reaction CO_(2)(g)+ H_(2)(g)to CO(g) + H_(2)O(g) is |
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Answer» 524. 1 `C=1/2O_(2)toCO(g),DeltaH_(f)^(@)=-110.5 kJ mol^(-1)` ` H_(2)+1/2O_(2)to H_(2)O(g), DeltaH_(f)^(@)=-241.8 kJ mol^(-1)` Eq.(ii) + Eq.(III) -Eq.(i) ` Co_(2)(g)+H_(2)Oto CO(g)+ H_(2)O(g), DeltaH_(r)=412.2 kJ` |
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| 30903. |
The DeltaH_(f)^(@) of O_(3),CO_(3),NH_(3) and HI are 142.2, -393.3, -46.2 and +25.9 kJ per mol, respectively. The order of their increasing stabilities will be |
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Answer» `O_(3),CO_(2),NH_(3),HI` Energy released `prop` stability of compound `142.2 gt 25.9 gt -46.2 gt -393.2` i.e. `O_(3) gt HI gt NH_(3) gt CO_(2)` |
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| 30904. |
The DeltaH_(f)^(@) for CO_(2)(g), CO(g) and H_(2)O(g) are - 393.5, - 110.5 and - 241.8 kJ mol^(-1) respectively. The standard enthalpy change (in kJ) for the reaction CO_(2)(g)+H_(2)(g)rarrCO(g)+H_(2)O(g) is |
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Answer» 524.1 `"(REACTANT)"=(-110.5-241.08)-(-393.5)=41.2 KJ mol^(-1)`. |
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| 30905. |
The DeltaH^(0), DeltaG^(0) and DeltaS^(0) values for the reaction 2H_(2)O_(2)(l) to 2H_(2)O(l)+O_(2)(g) at 25^(@)C are - 196.0 KJ//"mole"-233.6 kJ//"mole" and +125.6 J//"mole". K respectively . Is there any temperature at which H_(2)O_(2)(l) is stable at 1 atm ? Assume that DeltaH and DeltaS values do not change with temperature. |
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Answer» |
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| 30906. |
The DeltaH values for reactions of ,C(s) + 1/2 O_2 (g) to CO(g) DeltaH =-100 KJCO(g) + 1/2 O_2 (g) to CO_2 (g) DeltaH = 200 KJThe heat of reaction for C(s) + O_2 (g) to CO_2 (g)is |
| Answer» Answer :A | |
| 30907. |
The DeltaH value of the reaction H_(2) + Cl_(2) hArr 2 HCl is -44.12 kcal . If E_(1) is the activation energy of the products , then for the above reaction |
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Answer» `E_(1) gt E_(2)` |
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| 30908. |
The DeltaH^(@), of thermal decomposition of BeCO_(3), MgCO_(3), SrCO_(3) and BaCO_(3) are in the order: |
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Answer» `BeCO_(3) LT MgCO_(3) lt CaCO_(3) lt SrCO_(3) lt BaCO_(3)` |
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| 30909. |
The DeltaH and DeltaS for a reaction at one atmospheric pressure are +30.558 kJ and 0.066 kJk^(-1) respectively. The temperature at which the free energy change will be zero and below of this temperature the nature of reaction would be |
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Answer» 483 K, spontaneous `0=+30.558-Txx0.066` `or T=(30.558)/(0.066)=463 K` If `(dG)_(T,P)=0` sign '=' mean. It is REVERSIBLE PROCESS. |
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| 30910. |
The Delta_(f) H^(@) (N_(2)O_(5), g) in kJ/mol on the basic of the following data is: {:(2NO(g)+O_(2)(g) rarr 2NO_(2)(g),,Delta_(r)H^(@)= -114" kJ/mol"),(4NO_(2)(g)+O_(2)(g) rarr 2N_(2)O_(5)(g),,Delta_(r)H^(@)= -102.6" kJ/mol"),(Delta_(f)H^(@) (NO, g)=90.2" kJ/mol",,):} |
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Answer» `15.1` Addition of (1), (2) and (3) EQUATION `N_(2)(g)+5/2O_(2)(g) rarr N_(2)O(g)""Delta_(f)H^(@) (N_(2)O_(5), g)=15.1 kJ//mol` |
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| 30911. |
The Delta H^(theta) for the reaction 4S(s) + 6O_(2)(g) rarr 4 SO_(3)(g) is -1583.2kJ. Standard enthalpy of formation of sulphur trioxide is: |
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Answer» `-3166.4kJ` |
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| 30912. |
The dehydromination of 2-bromobutane gives CH_(3)CH=CHCH_(3). The product is |
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Answer» HOFMANN product |
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| 30913. |
The dehydro halogenation of neopentyle bromide with alco.KOH mainly given: |
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Answer» 2-methy1-1butane |
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| 30914. |
The dehydrohalogenation of neopentyl bromide with alcoholic KOH mainly gives |
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Answer» 2-METHYL-1-butene In this reaction `1^@` carbonium ion is FORMED which rearranges to form `3^@` carbonium ion from which base obstruct proton Hence 2-methyl-2-butene is formed as a main product. `underset(1^@ "carbonium less soluble")(CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-overset+(CH_2) )overset"Methyl SHIFT"to CH_3-underset+oversetoverset(CH_3)(|)C-CH_2-CH_3 overset("Elimination of proton from" beta "carbon which is less hydrogenated")to underset"2-Methyl-2-Butene"(CH_3-oversetoverset(CH_3)(|)C=CH-CH_3)` |
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| 30915. |
The dehydration yield of cyclohexanol of cyclohexene is 75%. The yeild obtained when 100 g of cyclohexanol is dehydrated will be |
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Answer» 82.35 g 102 g of cyclohexanol GIVES cyclohexene = 84 g `THEREFORE " 100 g of cyclohexanol will give cyclohexene"` `=(84)/(102)xx100g=82.35g` This is the theoretical yield. As ACTUAL yield is 75%`, therefore, actual yield `=(75)/(100)xx82.35g=61.76g` |
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| 30916. |
The dehydration reaction is represented by : |
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Answer» <P>(II)(III)(S) |
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| 30917. |
The dehydration of neo=pentanol gives mainly: |
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Answer»
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| 30918. |
The dehydration of neopentanol gives mainly: |
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Answer»
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| 30919. |
The dehydration of 2-methyl butanol with conc. H_(2)SO_(4) gives |
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Answer» 2-METHYL butene as major product |
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| 30920. |
The dehalogenation of vicinal dihalides with zinc dust gives : |
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Answer» ALKENES |
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| 30921. |
The dehydration of alcohol is an example of …………………….. . |
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Answer» Bimolecular ELIMINATION REACTION |
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| 30922. |
The dehydration of alcohols to form ethers in presence of concentrated acid follows: |
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Answer» UNIMOLECULAR NUCLEOPHILIC SUBSTITUTION REACTION. |
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| 30923. |
The degreen of crystallinity of which of the following is highest |
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Answer» Atactic polyvinylchloride `(-CH_(2)-UNDERSET(Cl)underset(|)(Cl)-CH_(2)-overset(Cl)overset(|)(C)H-CH_(2)-underset(Cl)underset(|)(C)H-CH_2-overset(Cl)overset(|)(C)H-CH_(2)-underset(Cl)underset(|)(CH)-)_(n)` In this arrangementthe chlorine ATOMS are alternately arranged. the polymer is stereoregular and has high crystallinity. |
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| 30924. |
The degree of polarity of a covalent bond is given by the dipole moment (mu). Identify the incorrect order of dipole moment in the following. |
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Answer» `NH_(3)gtNF_(3)`
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| 30925. |
The degree of unsaturation of following compound C_(8)H_(12)O, C_(3)H_(5)N, C_(4)H_(8)O are respectively : |
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Answer» `4,3,2` |
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| 30926. |
The degree of ionization of HF in 1.00 m aqueous solution is (freezing = 1.86^(@)C) is : |
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Answer» 0.06 `i=(DeltaT_(f))/(K_(f)xxm)=(0.197)/(1.86xx0.1)=1.059` For dissociation `ALPHA=((i-1))/((n-1))=((1.059-1))/((2-1))-0.059` `=5.9%~~6%` |
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| 30927. |
The degree of ionization of 0.2 m weak acid HX is 0.3. If the K_(f) for water = 1.85. What will be the freezing point of solution ? |
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Answer» `-0.360^(@)C` Total mole `= (1-0.3)+0.3+0.3=1.3` `i=(1.3)/(1)=1.3` `Delta T_(f)=ixxmxxK_(f)` `= 1.3xx1.85xx0.2` `= 0.480^(@)C` `THEREFORE` Freezing point `= 0^(@)C-0.480^(@)C` `= -0.480^(@)C` |
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| 30928. |
The degree of ionization increases |
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Answer» with increase in concentration of the SOLUTION |
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| 30929. |
The degree of hydrolysis of a salt of weak acid and weak base in its 0.1 M solutionis 0.2 M, the precentage hydrolysis of the salt should be |
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Answer» `50%` |
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| 30930. |
The degree of hydrolysis of a salt of weak acid and strong base is = 0.5. The equation to be used to calculate the accurate value of the degree of hydrolysis (h) is |
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Answer» `h=sqrt(K_(W)/(K_(a)*K_(b)))` |
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| 30931. |
The degree of hydrolysis in hydrolic equilibrium A^(-) + H_(2)O hArr HA + OH^(-) at salt concentration of 0.001 M is (K_(1) = 1 xx 10^(-5)) |
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Answer» `1 xx 10^(-3)` `K_(h) = alpha^(2)C, alpha = sqrt((K_(h))/(C)) = sqrt((1 xx 10^(-9))/(0.001)) = 1 xx 10^(-3)`. |
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| 30932. |
The degree of hardness of water is usually expressedin terms of |
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Answer» ppm by WEIGHT of `MgSO_4` |
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| 30933. |
The degreeof hardness of water is usually expressed in terms of |
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Answer» ppm by weightof `mgSO_(4)` |
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| 30935. |
The degree of dissociationof water is 1.8 xx10^(-9) at 298 K . Calculatetheionization constant and Ionic product of water at 298 K . |
| Answer» SOLUTION :`1XX10^(-14)` | |
| 30936. |
The degree of dissociation ofPCl_5 (alpha) obeying the equiliberium, PCl_5 ⇌ PCl_3 + Cl_2is approximately realted to the pressure at equilibrium by : |
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Answer» <P>``alpha PROP P` |
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| 30937. |
The degree of dissociation of water at 25^(@)C is 1.9 xx 10^(-7)% and density of 1.0 gm^(-3). The ionic constant for water is |
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Answer» `1.0 xx 10^(-14)` `K = 1.9 xx 10^(-9) xx 1.9 xx10^(-9) xx 1000/18` `= 2.0 xx 10^(-16)` |
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| 30938. |
The degree of dissociation of water in a 0.1 M aqueous solution of HCl ata certain temperature t^(@)C is 3.6 xx 10^(-15). The temperature t must be : [density of water at t^(@)C = 1 gm//mL.] |
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Answer» `lt 25^(@)C` Hence temperature MUST be `gt 25^(@)C`. |
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| 30939. |
The degree of dissociation of Ca(NO_(3))_(2) in a dilute aqueous solution containing 7g of the salt per 100g of water at 100^(@)C is 70 per cent. If the vapour pressure of water at 100^(@)C is 760mm, calculate the vapour pressure of the solution. |
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Answer» Solution :`{:(1,,,,,,,,),(Ca(NO_(3))_(2),,to,,Ca^(2+),,+,,2NO_(3)^(-)),(1-x,,,,x,,,,2x):}` van.t Hoff factor `i=(1-x+x+2x)/(1)=(1+2x)/(1)=1+2xx0.7=2.4` Now, `i=("observed lowering of VAPOUR PRESSURE")/("normal lowering of vapour pressure")` `:.` obs. lowering of vap. PRESS. `=ixx` nor.lowering of vap.press. `p^(0)-p_(obs.)=ixx{(n)/(n+N).p^(0)}` `760-p_(obs)=2.4xx(7//164)/((7)/(164)+(100)/(18))xx760` `{{:(Ca(NO_(3))_(2)=164),(H_(2)O=18):}}` `=13.90` `:.p_(obs.)=760-13.90=746.10mm` |
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| 30940. |
The degree of dissociation of N_2O_4 according to the equation N_2O_4 iff 2NO_2 at 70^@C and 1 atm is 65.6%. Calculate the apparent mol. wt. of N_2O_4under the given conditions. |
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Answer» |
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| 30941. |
The degree of dissociation of dinitrogen tetroxide N_(2)O_(4) (g) to 2NO_(2)(g) atature T and total pressure P is alpha. Which one of the following is the correct expression for the equilibrium constant (K_(p)) at this temperature ? |
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Answer» <P>`(2 ALPHA)/((1-alpha^(2)))` Total moles at equ `=1-alpha+2alpha=1+alpha` `K_(e )=(P_(NO_(2))^(2))/(P_(N_(2)O_(4)))=(((2alpha)/(1+alpha)P)^(2))/((1-alpha)/(1+alpha)P)=(4alpha^(2)P)/(1-alpha^(2))` |
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| 30942. |
The degree of dissociation of dinitrogen tetroxide N_(2)O_(4)(g)rarr 2NO_(2)(g)at temperature T and total pressure P is alpha. Which one of the following expressions is correct for the equilibrium constant at this temperature ? |
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Answer» <P>`(2ALPHA)/(1-ALPHA^(2))` |
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| 30943. |
The degree of dissociation of acetic acid in 0.1 M solution is 0.04 . Calculate K_a for acetic acid. Where alpha is the degree of dissociation,C is the concentrationof the acid in moles/ lit. |
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Answer» SOLUTION :`K_a=(alpha^2 C)/(1-alpha)=(02xx.02xx0.1)/(1-0.02)` `K_a=(0.4xx10^(-4))/(0.98) = 4.08xx10^(-5)` |
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| 30944. |
The degree of dissociation of an electrolyte does not depend on |
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Answer» NATURE of electrolyte |
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| 30945. |
The degree of dissociation of an electrolyte in aqueous solution depends on A) Temperature B) Concentration of the electrolyte C) Nature of the electrolyt |
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| 30946. |
The degree of dissociation of 100 mL of pure water at 25^(@)C is |
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Answer» `1.8xx10^(-16)` |
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| 30947. |
The degree of dissociation of 0.1 M weak acid HA is 0.5%. If 2 mL of 1.0MHA solution is diluted to 32 mL the degree of dissociation of acid and H_(3)O^(+)ion concentration in the resulting solution will be respectively |
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Answer» `0.02and3.125xx10^(-4)` MOLARITY of diluted solution `2xx1=32xxM,M=1//16(C_(2))` `alpha_(2)=sqrt((K_(a))/(C_(2)))=0.005sqrt(16)=0.02` `H_(3)O^(+)=C_(2)alpha_(2)=(0.02)/(16)=1.25xx10^(-3)M` |
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| 30948. |
The degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be |
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Answer» `10^(-3)` `K = C ALPHA^(2) = 0.1 xx (10^(-4))^(2) = 10^(-9)`. |
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| 30949. |
The degree of dissociation is 0.4 at 400 K and 1 atm for the gaseous reactionPCl_5 iff PCl_5 + Cl_2Assuming ideal behaviour of all the gases, calculate the density of equilibrium mixture at 400 K and 1 atm. |
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Answer» <P> SOLUTION :APPLY `p = (DRT)/(M_(MIX)`4.535g/L |
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| 30950. |
The degree of dissociation in a weak electrolytic increases |
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Answer» On INCREASING |
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