52671 + Interview Questions in Chemistry in Class 11

1.	Calculate E^@ For the cell : Al\|Al^(3+)(1M)\|\|Cu^(2+)(1M)Cu Given E^@(Al^(3+)\|Al)=-1.66V and E^@(Cu^(2+) \|Cu)=0.34V
Answer» ANSWER :2.0 V

Discussion

2.	Calculate distance of closet approach by an alpha -particle of KE=2.5 MeV being scattered by gold nucleus (Z=79).
Answer» Solution :`KE_(i) PE_(F) = (K(79e)(2e))/(r )` Remember : `(1eV=1.6xx10^(-19)J)/(1MeV=1.6xx10^(-13)J)` `implies 2.5xx1.6xx10^(-13)J=(9xx10^(9)xx79xx2xx(1.6xx10^(-19))^(2))/(r )implies r = 9.1xx10^(-14)m`

Discussion

3.	Calculate E^(@)for the cell : AI\|AI^(3+)(1 M)\|\|Cu^(23+)(1M)\|Cu Given : E_(AI^(3+)//AI)^(@) and E_(Cu^(2+)//Cu)^(@) as -1.66 V and + 0.34 V respectively
Answer» SOLUTION :`E_(cell)^(@)=E_(CU^(2+),Cu)^(@)-E_(AI^(3+),AI)^(@)=0.34-(-1.66)=2.0V`

Discussion

4.	Calculate density of HCl gas having density 8kg//m^(3) at -40^(@)C temperature.
Answer» ANSWER :`424.58xx10^(3)PA`

Discussion

5.	Calculate Delta_(r)H^(@) for the reaction, H_(2)(g) + 1/2O_(2)(g) to H_(2)O(g) Given that bond enthalpies of H-H bond , O = O bond and O - H bond are 433 kJ mol^(-1), 492 kJ"mol"^(-1) and 464 kJ"mol"^(-1)
Answer» Solution :For the reaction : `H -H(G) + 1/2(O = O) (g) to 2O - H(g)` `DeltaH` = (Bond enthalpy of H - H) + `1/2`(Bond enthalpy of O = O) - 2(Bond enthalpy of O - H) ` = 433 + 1/2(492) - 2(464) = 433 + 246 - 928 = -249 kJ`.

Discussion

6.	Calculate DeltaS for ice at 275 K temperature.
Answer» SOLUTION :`T= t+273` `273+t=275 therefore t=2^(@) C` At this temperature ICE melts SPONTANEOUS. `therefore Delta S = +ve`

Discussion

7.	Calculate Delta_(r)G for thereaction at 27^(@)C H_(2)(g)+2Ag^(+)(aq)hArr2Ag(s)+2H^(+)(aq) Given : P_(H2)=0.5 bar,[Ag^(+)]=10^(-5)M, [H^(+)]=10^(-3)M,Delta_(r)G^(@)[Ag^(+)(aq)]=77.1kJ//mol
Answer» `-154.2kJ//mol` `-178.9kJ//mol` `-129.5kJ//mol` NONE of these ANSWER :C

Discussion

8.	Calculate Delta_(r)G^(0) for (NH_(4)Cl, s) at 310K. Given: Dleta _(f)H^(0) for (NH_(4)Cl, s) = - 314.5 KJ/mol, Delta_(r ) C_(P) = 0S_(N_(2)(g))^(0) = 192JK^(-1) mol^(-1), S_(H_(2) (g))^(0) = 130.5 JK^(-1) mol^(-1), S_(Cl_(2)(g))^(0) = 233JK^(-1) mol^(-1), S_(NH_(4)Cl(s))^(0) = 99.5 JK^(-1) mol^(-1). All given data at 300K
Answer» `-198.56kJ//mol` `-426.7 kJ//mol` `2-2.3 kJ//mol` none of these Solution :`(1)/(2) N_(2) + 2H_(2)+ (1)/(2) Cl_(2) rarr NH_(4) Cl` `Delta G_(gamma)^(0) = Delta H_(gamma)^(0) - T Delta S_(gamma)^(0)` `Delta H_(310)^(0) = Delta H_(300)^(0) = - 314.5` `Delta S_(gamma)^(0) = 99.5 - ((1)/(2) xx 192 + 2 xx 130.5 + (1)/(2) xx 233)` `= - 374` `Delta G^(0) = - 314.5 - (310 xx (-374))/(1000)` `= - 198.56` Kj/mol

Discussion

9.	Calculate Delta_(r) G^(@) for conversion of oxygen to ozone. 3 //2 O_(2)(g) rarrO_(3)(g) at298 K , if K_(p) for this conversion is 2. 47 xx 10^(-29).
Answer» Solution :`Delta_(r)G^(@) = -2.303 RT log K_(p) = - (2.303) ( 8.314 JK^(-1)mol^(-1)) ( 298 K) log ( 2.47 XX 10^(-29))` `=163229 J mol^(-1) = 163.2 kJ mol^(_1)` [For standard conditions ` T = 298 K`, and log `(2.47 xx 10^(-29)) = - 29 + 0.3927 = - 28.6073]`

Discussion

10.	Calculate Delta_(r )G^(@) for the reaction for which the value of K_(p) is 1.5 xx 10^(10) at 300 K.
Answer» <P> Solution :`Delta G^(@) = -2.303 RT log_(10)K_(p)` `= -2.303 XX 8.314 xx 300 xx log 1.5 xx 10^(10)` `= -2.303 xx 8.314 xx 300 xx (log 1.5 + log 10^(10))` `= -2.303 xx 8.314 xx 300 xx (0.1761 + 10)` `= -2.303 xx 8.314 xx 300 xx 10.1761 = 58452.97 J mol^(-1)`. `= -57270 J`

Discussion

11.	Calculate Delta_(r ) G^(@) for the conversion of oxygen to ozone, (3)/(2)O_(2) (g) rarr O_(3) at 298 K, given partial pressure equilibrium constant is 2.47 xx 10^(-29).
Answer» <P> Solution :`Delta_(R )G^(@) = -2.303 RT log_(10) K_(p)` `= -2.303 xx 8.314 xx 298 xx log 2.47 xx 10^(-29)` `= -163229 J mol^(-1) = -163.229kJ mol^(-1)`

Discussion

12.	Calculate Delta_r G^Ө for conversion of oxygen to ozone , 3//2 O_(2(g)) to O_(3(g)) at 298 K . If K_P for this conversion is 2.47xx10^(-29) .
Answer» SOLUTION :We know `Delta_rG^Ө=-2.303 RT LOG K_P` and `R=8.314 JK^(-1) "MOL"^(-1)` THEREFORE , `Delta_rG^Ө=-2.303 (8.314 J K^(-1) "mol"^(-1))xx(298 K) (log 2.47 xx 10^(-29))` `=163000 "J mol"^(-1) =163 "mol"^(-1)`

Discussion

13.	Calculate Delta_(h)H^(ɵ) for the addition of 1 mol of H_2 to (i) benzene (C_(6)H_(6)) and (ii) 1,3-cyclohexadiene (C_(6)H_(8)) from frigure. Strategy (i): Target is to find Delta_(h)H^(ɵ) of the following reaction benzene +H_(2)to1,3-Cyclohexadiene Thus, look for the data involving these two compound. strategy (ii). We need to find Delta_(h)H^(ɵ) of the following reaction: 1,3-C_6H_8+H_2toC_6H_(12) Thus, we look for the data involving these two compounds.
Answer» Solution :(a). `C_(6)H_(6)+3H_2toC_(6)H_(12)`, `Delta_(h)H^(ɵ)=-208Kgmol^(-1)` (b). `1.3-C_6H_6+2H_2toC_6H_(12)`, `Delta_(h)H^(ɵ)=-232kJmol^(-1)` Substracting second equation (b) from the first (a), we get `C_6H_6+H_2to1,3-C_6H_6`, `Delta_(h)H^(ɵ)24kJmol^(-1)` (ii). (a). `1,3-C_6H_6+2H_2toC_6H_(12)toC_6H_(12)`, `Delta_(h)H^(ɵ)-232mol^(-1)` (b). `C_6H_(10)+H_(2)toC_6H_(12)`, `Delta_(h)H^(ɵ)-120kJmol^(-1)` SUBTRACTING second equation (b) from the first (a), we get `1,3-C_6H_8+H_2toC_6H_(10)`,`Delta_(h)H^(ɵ)112kJmol^(-1)`

Discussion

14.	Calculate DeltaH_(f)^(0) for the reactions. CO_(2_((g)))+H_(2_((g)))rarrCO_(2_((g)))+H_(2)O_((g)) " given that "DeltaH_(f)^(0) " for " CO_(2_((g))),CO_((g))and H_(2)O_((g)) " are " -393.5,-111.31 and -242 " KJ.mol"^(-1)respectively .
Answer» Solution :`DeltaH_(f)^(0)(CO_(2))=-393.5 "KJ MOL "^(-1)` `DeltaH_(f)^(0)(CO) =-111.31 " kJ.mol"^(-1)` `DeltaH_(f)^(0)(H_(2)O)=-242"kJmol"^(-1)` `CO_(2_((g)))+H_(2_((g)))rarrCO_((g))+H_(2)O_((g))` `DeltaH_(r)^(0)=? ` `DeltaH_(r)^(0)=sum(DeltaH_(r)^(0))_("product")-sum(DeltaH_(r)^(0))_("REACTANT")` `DeltaH_(r)^(0)=[DeltaH_(r)^(0)(CO)+DeltaH_(r)^(0)(H_(2)O)]-[DeltaH_(r)^(0)(CO_(2))+DeltaH_(r)^(0)(H_(2))]` `=[-111.31+(-242)]-[-393.5+(0)]` `DeltaH_(r)^(0)=[-353.31]+393.5` `DeltaH_(r)^(0)=40.19` `DeltaH_(r)^(0)=+40.19 "kJ mol "^(-1)`

Discussion

15.	Calculate DeltaH (in kcal) for the following phase transformation : A (s,1 bar, 200K) rarr A (g, 1 bar ,200 K) Standard melting poing of A = 200 K Standard boioling poing of A = 300 K Standard boiling ping of A = 300 K Latent heat of fusion of A at 200 K = 60 cal//g Latent heat of vaporisation of A at 300 K = 410 cal//g Molar mass of A = 50 g//mole C_(Vm) of A (s)= 5cal K^(-1) mol^(-1) C_(Vm) of A (l) = 10 cal K^(-1)mol^(-1) C_(Vm) of A (g)=3cal K^(-1) mol^(-1) R=2 cal K^(-1) mol^(-1)
Answer» ANSWER :24

Discussion

16.	Calculate DeltaH of the reaction H - underset(Cl) underset(\|) overset(H) overset(\|) (C ) - Cl(g)rarr C(g) +2H(g) + 2Cl(g) Theaverage bond energies of C-H bond and C-Cl bond are 416kJ and 325 kJ mol^(-1) respectively.
Answer» Solution :In the given REACTION, TWO MOLES of C-H BOND and two molesof C-Cl bond are broken. Hence energy absorbed `= 2 xx 416 + 2 xx 325` `= 1485kJ`

Discussion

17.	Calculate DeltaG^( Theta ) for conversion of oxygen to ozone,(3)/(2) O_(2(g)) to O_(3(g)) at 298 K. If K_(p) for this conversion is 2.47 xx 10^(-29).
Answer» <P> Solution :We KNOW `Delta_(r) G^( Theta ) = -2.303 "RT LOG" K_(p) and R = 8.314 "JK"^(-1) "mol"^(-1)` THEREFORE, `Delta_( r) G^( Theta ) = -2.303 (8.314 "JK"^(-1) "mol"^(-1) ) (298 K) (log 2.47 xx 10^(-29) )` `= 163000 "J mol"^(-1)` `=163 "kJ mol"^(-1)`

Discussion

18.	Calculate DeltaG^0 for conversion of oxygen to ozone 3//2O_2 hArr O_(3(g)) at 298 K, if K_p for the conversion is 2.47 xx 10^(-29)in standard pressure units.
Answer» Solution :`DeltaG^0=-"2.303 RT LOG" K_p` where `R=8.314 "JK"^(-1) "MOL"^(-1)` `K_p=2.47xx10^(-29)` T=298 K `DeltaG^0=-2.303(9.314)(298) log (2.47xx10^(-29))` `DeltaG^0=16300 "J mol"^(-1)` `DeltaG^0=16.3 "KJ mol"^(-1)`

Discussion

19.	Calculate DeltaG^@ for the following cell reaction
Answer» `-305 kJ//"MOL"` `-301 kJ//"mol"` `305 kJ//mol` `301 kJ//mol` ANSWER :B

Discussion

20.	Calculate DeltaG^(@) for conversion of oxygen to ozone 3//20_(2)(g)toO_(3)(g) at 298K if Kp=2.47xx10^(-29)
Answer» Solution :`DELTAN^(@)=-2.303RT"log"K_(p)=2.303xx8.314xx298"log"(2.47xx10^(-29))` `=163000` JL mol`=163` KJ mol.

Discussion

21.	Calculate DeltaG and DeltaG^(@) for the reaction :A + B hArr C + Dat 27^(@)C .Equilibrium constnat(K) for this reaction = 10^(2)
Answer» Solution :`DeltaG = 0` because the reactin is in equilibrium. `DeltaG^(@) = - 2.303 RT logK = -2.303 XX 8.314 JK^(-1) MOL^(-1) xx 300 K log 10^(2)` `= - 1148 J mol^(-1) = - 11.488 kJ mol^(-1)`

Discussion

22.	Calculate delta H_(r)^(0) for the reaction CO_(2)(g) + H_(2)(g) rarr CO(g) + H_(2)O(g) given that Delta H_(r)^(0) for CO_(2) (g), CO (g) and H_(2)O (g) are - 393.5, -111.31 and - 242 KJ mol^(-1) respectively.
Answer» Solution :GIVEN: `Delta H_(F)^(0) CO_(2) = -393.5kJ mol^(-1)` `DeltaH_(f)^(0) CO = -111.31kJ mol^(-1)` `DeltaH_(f)^(0) (H_(2)O) = -242kJ mol^(-1)` `CO_(2)(g) + H_(2)(g) rarr CO(g) + H_(2)O(g)` `DeltaH_(r)^(0) =?` `DeltaH_(r)^(0) = Sigma(DeltaH_(f)^(0))_("products")- Sigma (DeltaH_(f)^(0))_("REACTANTS")` `DeltaH_(r)^(0) = [DeltaH_(f)^(0) (CO) + DeltaH_(f)^(0) (H_(2)O)]- [DeltaH_(f)^(0) (CO_(2)) + DeltaH_(f)^(0) (H_(2))]` `DeltaH_(r)^(0) = [-111.31 + (-242)] -[-393.5+ (0)]` `Delta H_(r)^(0) = [-353.31] + 393.5` `DeltaH_(r)^(0) = 40.19` `DeltaH_(r)^(0) = +40.19 kJ mol^(-1)`

Discussion

23.	Calculate DeltaH_f^@ for the reaction CO_2(g) +H_2(g) to CO(g) + H_2O(g) given that DeltaH_f^0 for CO_2(g),CO(g) and H_2O(g) are -393.5 , -111.31 and -242 "kJ mol"^(-1) respectively.
Answer» Solution :`DELTA H_(r )^(0)(CO_(2))=-393.5 "kJ mol"^(-1)` `Delta H_(r )^(0)(CO)=-111.31"kJ mol"^(-1)` `Delta H_(r )^(0)(H_(2)O)=-242 "kJ mol"^(-1)` `CO_(2(g))+H_(2(g))rarr CO_((g))+H_(2)O_((g))` `Delta H_(r )^(0)=?` `Delta H_(r )^(0)=sum (Delta H_(r )^(0))_(("product"))=sum (Delta H_(r )^(0))_(("REACTANT"))` `Delta H_(r )^(0)=[Delta H_(r )^(0)(CO)+Delta H_(r )^(0)(H_(2)O)]-[Delta H_(r )^(0)(CO_(2))+Delta H_(r )^(0)(H_(2))]` `= [-111.31+(-242)]-[-393.5+(0)]` `Delta H_(r )^(0)=[-353.31]+393.5` `Delta H_(r )^(0)=40.19` `Delta H_(r )^(0)=+40.19"kJ mol"^(-1)`.

Discussion

24.	Calculate Delta S for 3 moles of a diatomic ideal gas which is heated and compressed from 298K and 1bar to 596K and 4 bar
Answer» `-14.7 "CAL" K^(-1)` `+14.7 "cal" K^(-1)` `-4.9 "cal" K^(-1)` `6.3 "cal" K^(-1)` Solution :`Delta S = nCP "LN" (T_(2))/(T_(1)) + nR "ln" (P_(1))/(P_(2))` `=3 xx (7)/(2) xx2 xx l n (596)/(298) + 3 xx 2 xx l n (1)/(4) = 6.3` cal/K

Discussion

25.	Calculate Delta H_(f) of HCl if bond energy of H-H bond is 437 kJ Cl-CL bond is 244, and H-C is 433 kJ mol^(-1).
Answer» Solution :`(1)/(2)H_(2)(G) + (1)/(2)Cl_(2)(g) rarr HCl(g)` `Delta H = (1)/(2)B_(H-H) + (1)/(2)B_(O-O) = (1)/(2) XX 437 + (1)/(2) xx 244 - 433` `= 218.5 kJ + 122 kJ - 433 kJ = -92.5 kJ mol^(-1)`.

Discussion

26.	Calculate Delta H ^(298) for the reaction :2Li (s) + 2H_(2)O(l) rarr 2Li^(+) ( aq) + 2OH^(-) (aq)+ H_(2)(g) Given that the standard enthalpies of formation of Li^(+)(aq), OH^(-)(aq)and H_(2)O(l) are - 278.5, - 228.9 and - 285.8 kJ mol^(-) respectively( all at 298 K )
Answer» ANSWER :`-443.2kJ`

Discussion

27.	Calculate DeltaH^(@) for the reaction : Na_(2) O + SO_(3) to Na_(2) SO_(4) given the following : A) Na_((s)) + underset((1))(H_(2)O) to NaOH_((s)) + (1)/(2) H_(2 (g)) , Delta H^(@) = -146kJ B) Na_(2) SO_(4 (s)) + underset((1))(H_(2)O) to 2 NaOH_((s)) + SO_(3 (g)) , Delta H^(@) = +418 kJ C) 2 Na_(2) O_((s)) + 2 H_(2 (g)) to 4 Na_((s)) + 2 H_(2) O_((l)) , DeltaH^(@) = +259 kJ
Answer» `+823KJ` `-580.5KJ` `-435KJ` `+531KJ` ANSWER :B

Discussion

28.	Calculate DeltaH in Joules for C_(("graphite")) rarr C_(("Diamond")) by using the following data C_(("graphiter")) + O_(2(g)) rarr CO_(2(g)) , DeltaH^@ =-393.5KJ C_(("Diamond")) + O_(2(g)) rarr CO_(2(g)) , DeltaH^@ =-395.4KJ
Answer» 1900 `-788.9 XX 10 ^(3)` 190000 `788 xx 10 ^(3)` ANSWER :B

Discussion

29.	Calculate Delta H for the reaction 2O_(2)(g) rarr 3O_(2)(g) at 298 K and 1 atmosphere pressure given that Delta U = -287.9 kJ and R = 8.314 JK^(-1) mol^(-1).
Answer» SOLUTION :As per the equation `Delta H = Delta U + Delta n RT`, where `Delta U = -287.9 kJ, Delta n = (3-2) = 1, R = 8.314 JK^(-1)mol^(-1), T = 298K , Delta H = ?` `Delta H = -287.9 kJ + 1 xx 8.314 xx 10^(-3) KJ K^(-1) mol^(-2) xx 298 K` `= -285.4 kJ mol^(-1)` Change in ENTHALPY for the reaction `= -285.4 kJ mol^(-1)`.

Discussion

30.	Calculate Delta H ^(@)for the reaction CH_(2)= CH_(2) + 3O_(2) rarr 2CO_(2)+ 2H_(2)O Given that the average bond energies of the different bonds are : {:("Bond",C-H,O=O,C=O,O-H,C=C),("Bond energy " ( kJ mol^(-1)),414,499,724,460,619):}
Answer» Solution :`Deta_(r)H [B.E. (C=C) + 4B.E. (C-H) + 3 B.E.(O-O) ] - [2 xx 2 xx B.E. ( C=O) + 2 xx 2 xx B.E.(O-H) ]` ` [619+ 4 ( 414)]+ 3 ( 499) ] - [ 4 ( 724)+ 4 ( 460 ) ] KJ = - 964 kJ`

Discussion

31.	Calculate DetlaG^(@) for the following reaction, Zn_((s)) + Cu_((aq))^(2+) to Zn_((aq))^(2+) + Cu_((s)) . DeltaG_(f)^(@) of Zn_((aq))^(2+) and Cu_((aq))^(+2)is -147.2 kJ mol^(-1)and 65kJmol^(-1) .
Answer» Solution :Standard free energy CHANGE, `DELTA G ^(@) = sum Delta G _(f) ^(@) ` of products `- sum Delta G _(f) ^(@) ` of reactants. `Delta G ^(@) = (- 147 . 2 + 0) - ( 0+ 65)=212.2 KJ mol ^(-1).`

Discussion

32.	Calculate DeltaG^@ for conversion of oxygen to ozone, 3/2O_(2(g)) to O_(3(g)) at 298K, if K_P for this conversion is 2.47 xx 10^(-29)
Answer» <P> Solution :We know `Delta G _(@) = - 2.303 RT LOG K _(P) and R = 8.314 JK ^(-1) MOL ^(-1)` Therefore `Delta G ^(@) =- 2.303 (8.314 JK ^(-1) mol ^(-1))xx (298 K) (log 2.47 xx 10 ^(29))` `= 163000 J mol ^(-1) or 163 kJ mol ^(-1)`

Discussion

33.	Calculate de Broglie wavelength of an electron moving with 1% of the speed of light.
Answer» SOLUTION :`v = 1% " of " 3 xx 10^(8) ms^(-1) = 3xx 10^(6) m s^(-1), m_(e) = 9.11 xx 10^(-31) kg`

Discussion

34.	Calculate consecutive oxidation number of Br in Br_(3)O_(8).
Answer» Solution : Consecutive oxidation NUMBER : +6, +4, +6

Discussion

35.	Calculate average kinetic energy of CO_2 molecules at 27^(@)C.
Answer» SOLUTION :`3741.3 J MOL^(-1)`

Discussion

36.	Calculate (a)wavenumberand (b )frequecyof yellowradiationhavingwavelength5800 A
Answer» SOLUTION :(a) Calculationof wavenumber(v ) , `lambda =500 A` 5800`xx 10^(8) CM` `VEC( v)= (1)/(lambda)= (1)/( 58000xx 10^(10) m)` `=1.7241 xx 10^(4) cm^(-1)` ( B)Calculationof thefrequency(v) : v= `(c )/(lambda)= (3.0 xx 10^(5) MS^(-1))/(5800 xx 10^(10) m)` `= 5 .1724 xx 10^(14)s^(-1)Hz`

Discussion

37.	Calculate (a) wave number and (b) frequency of yellow radiations having wavelength of 5800Å
Answer» SOLUTION :Calculate of wave number: Wave number `bar(v) = (1)/(LAMBDA)` But `LAMDA = 5800Å = 5800 xx 10^(-10)m " (Given)" :. bar(v) = (1)/(5800 xx 10^(-10)m) = 1.72 xx 10^(6) m^(-1)` (b) Calculate of Frequency: Frequency `v = (c)/(lamda)` Substituting `c = 3 xx 10^(8) m//sec and lamda = 5800 xx 10^(-10)m`, we get `v = (3 xx 10^(8) ms^(-1))/(5800 xx 10^(-10)m) = 5.172 xx 10^(14) s^(-1)` or cycles/sec or HZ

Discussion

38.	Calculate a value of effective magnetic moment mu_(eff) of Ce^(3+) ion.
Answer» Solution :`mu_(EFF) = g sqrt(J(J+1))` ....(i) where `g = 1 + [(S(S+1)-L(L+1)+J(J+1))/(2J(J+1))]` ...(ii) `Ce^(3+)` as `f^(1)` CONFIGURATION `:. S = 1 xx (1)/(2) = (1)/(2)` `L = 3` `J=L - S` (Subshell is less than half-filled) `= 3-(1)/(2) = (5)/(2)` from (ii) `g= 1 + ([((1)/(2) xx (3)/(2))-(3 xx 4)+((5)/(2) xx (7)/(2))])/(2 xx (5)/(2) xx (7)/(2))= (6)/(7)` from (i) `mu_(eff) = gsqrt(J(J+1))BM` `= (6)/(7) sqrt((5)/(2) xx (7)/(2)) = 2.54 BM`

Discussion

39.	Calculate (a) hydrolysis constant, (b) degree of hydrolysis and (c) pH of 0.01 M sodium acetate, if K_a of CH_3 COOH is 1.9 xx10^(-5)
Answer» SOLUTION :`5.26 XX 10^(-10), 2.29xx 10^(-4) and8.36 `

Discussion

40.	Calculate (a) DeltaG^0and (b) the equilibrium constant for the formation of NO_2 from NO and O_2 at 298 K, NO_((g)) + 1/2O_(2(g)) hArr NO_(2(g)) where , DeltaG_f^ө(NO_2)= 52.0 kJ "mol"^(-1) DeltaG_f^ө(NO)=87.0 kJ mol^(-1) DeltaG_f^ө (O_2) = 0.0 kJ "mol"^(-1)
Answer» Solution :Calculation of `DeltaG^0` free energy change of reaction : `DeltaG_r^0 = sumDelta_f G_"(PRODUCTS)"^ө -sumDelta_f G_"(reactant)"^ө` `=Delta_f G^ө (NO_2)-(Delta_f G^ө, (NO)+1/2 Delta_f G^ө(O_2))` `=52.0-[87.0+1/2(0)]` `=-35 "kJ mol"^(-1)` `=-35xx10^3 "J mol"^(-1)` Calculate of equilibrium CONSTANT `K_c` : `DeltaG_r^0=-2.303 RT LOG K_c` `therefore log_10 K_c=-(Delta_r G^0)/(2.303 RT)` `therefore log_10 K_c=-((-35xx10^3 "J mol"^(-1)))/((2.303)(8.314 "J mol"^(-1) K^(-1))(298 K))` =6.1341 `therefore K_c`= Antilog 6.1341 `=1.3618xx10^6 approx 1.36xx10^6`

Discussion

41.	Calculate (a)Delta G^(@) and (b) the equilibriumconstantfor theformation of NO_(2) " from"NO and NO_(2)and O_(2) at 298 K NO (g) +1/2 O_(2) (g) hArr NO_(2) (g)whereDelta _(f) G^(@) (NO_(2)) = 520 "kJ//mol ", Delta_(f) G^(@) (NO) = 870"kJ//mol," Delta _(f) G^(@) (O_(2)) = 0 "kJ//mol.
Answer» SOLUTION :` Delta _(r) G^(@) = sum Delta _(f)G^(@) ("Products") - sumDelta_(f) G^(@)("REACTANTS") ` ` = Delta _(f) G^(@) (NO_(2)) - [Delta_(f) G^(@)(NO) +1/2 Delta_(f) G^(@)(O_(2)) ] = 520 - ( 870+1/2 xx 0) = -350 "kJ "MOL"^(-1)` (b) ` Delta G^(@)= 2303" RT log K Hence," -(-35000) = 2303 xx 8 314 xx 298 xx "log K"` or ` " log K"= 6* 1341 or K = 1* 361 xx 10^(6) `

Discussion

42.	Calculale the d_((N-C)) in (CH_(3))_(3)N molecule by using Eqs. (1.18) and (1.19) above and show from which equation d_((N-C)) closely resembles with the experimental value of d_((M\|N-C))=1.47 Å. (Given: r_(N) = 0.75 Å, r_(C ) = 0.77Å, chi_(A)=3.0, chi_(C )=2.5
Answer» Solution : We get `d_(N - C) = r_(N) + r_(C) = (0.75 + 0.77) Å = 1.52 Å` we get `d_(N - C) = r_(N) + r_(C) - 0.09 (chi_(N) - chi_(C))` `= 0.75 + 0.77 - 0.09 (3.0 - 2.5)` `= 1.475 Å` The VALUE of `d_(N - C)` CLOSELY resembles the EXPERIMENTAL value.

Discussion

43.	Calcualte the work down when 11.2 g of iron dissolves in hydrochloric acid in (i) a closed vessel (ii) an open beaker at 25^(@)C ( Atomic meass ofFe = 56u)
Answer» Solution :Iron reacts with HCL acid to produce `H_(2)` gas as `Fe(s) + 2HCL(aq) rarr FeCl_(2) (aq)+ H_(2)(g)` Thus, 1 mole of Fe, i.e., 56 g Fe produce `H_(2)` gas `= 1 `mole `:. 11.2 g` Fe will produce `H_(2)` ga `= (1)/( 56) xx 11.2 = 0.2 `mole. (i) If the reaction is carried out in a closed vessel , `Delta V = 0` `:.w = - P_(ext)DELTAV = 0` (II) If the reaction is carried out in open beaker ( external pressure being1 atm) Final volume occupied by 0.2 moles of `H_(2) ` at `25^(@)C` and 1 atm pressure can be calculated as follows `:` `PV = nRT` `:. V = ( nRT)/( P) = ( 0.2 mol xx 0.0 821L atm K^(-1) xx 298 K)/( 1 atm) = 4.89 L` `:. DeltaV = V _("final") - V_("initial") = 4.89 L` `w= - P_(ext)DeltaV = -1 atm xx 4.89 L atm` `= - 4.89 L atm` `= - 4.89 xx 101.3 J = - 495.4 J`

Discussion

44.	Calculae the entropy change fo n-hexane when one mole of it evaporates at 341.7 K (Delta_(vap) H = 29.0 kJ mol^(-1))
Answer» ANSWER :`84.86 JK^(-1) MOL^(-1)`

Discussion

45.	Calculate the volume occupied by 2 moles of NO_(2) at STP.
Answer» ANSWER :`1.75 XX 10^(22)` MOLECULES

Discussion

46.	Calcualte the number of chiral center in the molecule Ethyl 2,2-dibromo-4-ethyl-6-methoxy cyclohexane carboxylate.
Answer» ANSWER :[3]

Discussion

47.	Calcualte the mass percentage of nitrogen in hydrazinium Sulphate (N_(2)H_(5))_(2)SO_(4). {:("Molar mass", (g mol^(-1))), ((N_(2)H_(5))_(2)SO_(4), 162.2):}
Answer» 10.8 17.3 34.5 51.2 Answer :C

Discussion

48.	Calcualte the enthalpy change when infinitely dilute solution of CaCI_(2) and Na_(2)CO_(3) are mixed. Delta_(f)H^(Theta) for Ca^(2+)(aq), CO_(3)^(2-)(aq), and CaCO_(3)(s) are -129.80, -161.65, and -288.50 kcal mol^(-1) respectively.
Answer» Solution :Infinitely DILUTE SOLUTIONS are completely ionised. `{:(Caunderset(darr).^(2+)CI_(2)^(2-) +Na_(2)^(2+)CO_(3)^(2-)),(CaCO_(3)darr+2NaCI):}` `{:(Caunderset(darr).^(2+)+2CI^(Theta)+2Na^(darr)+CO_(3)^(2-)),(Ca^(2+)+CO_(3)^(2-)+2Na^(o+)+CO_(3)^(2-)):}` `Ca^(2+) +CO_(3)^(2-) rarr CaCO_(3)(s)` `-DELTAH^(Theta) = Delta_(f)H^(Theta)CaCO_(3) -[Delta_(f)H^(Theta)ca^(2+)+Delta_(f)H^(Theta)CO_(3)^(2-)]` `= - 288.5 -(-129.80-161.65)` `DeltaH^(Theta) = 2.95 kcal`

Discussion

49.	Calcualate the resonance energy of N_(2)O fromthe following data : Delta_(f)H^(@) of N_(2)O= 82kJ mol^(-1) , Bond energies of N -= N , N = N , O=O and N=O bonds are 946, 418, 498and 607 kJ mol^(-1) respectively.
Answer» SOLUTION :The equation for the formation of one MOLE of `N_(2)O` will be `N -= N(g) + (1)/(2)O= O(g) rarr N = N = O` Calculate value of `Delta_(F)H^(@)` for this reaction will be `Delta_(f)H^(@) = `[ B.E.`(N-=N) + (1)/(2) `B.E. ( `O=O )] - [B.E(N=N) + B.E. ( N=O)]` `= [ 946 +(1)/(2) ( 498 ) ] - [418 + 607] kJ mol^(-1) = 170 kJ mol^(-1)` Resonance energy`=` Observed `Delta_(f)H^(@)` - Calculated`Delta_(f)H^(@) = 82- 170 = -88kJ mol^(-1)`

Discussion

50.	Calcualate the value of Z_("eff") on 3delectronsof Sc .
Answer» SOLUTION :N/a

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Calculate E^@ For the cell : Al|Al^(3+)(1M)||Cu^(2+)(1M)Cu Given E^@(Al^(3+)|Al)=-1.66V and E^@(Cu^(2+) |Cu)=0.34V

Calculate distance of closet approach by an alpha -particle of KE=2.5 MeV being scattered by gold nucleus (Z=79).

Calculate E^(@)for the cell : AI|AI^(3+)(1 M)||Cu^(23+)(1M)|Cu Given : E_(AI^(3+)//AI)^(@) and E_(Cu^(2+)//Cu)^(@) as -1.66 V and + 0.34 V respectively

Calculate density of HCl gas having density 8kg//m^(3) at -40^(@)C temperature.

Calculate Delta_(r)H^(@) for the reaction, H_(2)(g) + 1/2O_(2)(g) to H_(2)O(g) Given that bond enthalpies of H-H bond , O = O bond and O - H bond are 433 kJ mol^(-1), 492 kJ"mol"^(-1) and 464 kJ"mol"^(-1)

Calculate DeltaS for ice at 275 K temperature.

Calculate Delta_(r)G for thereaction at 27^(@)C H_(2)(g)+2Ag^(+)(aq)hArr2Ag(s)+2H^(+)(aq) Given : P_(H2)=0.5 bar,[Ag^(+)]=10^(-5)M, [H^(+)]=10^(-3)M,Delta_(r)G^(@)[Ag^(+)(aq)]=77.1kJ//mol

Calculate Delta_(r) G^(@) for conversion of oxygen to ozone. 3 //2 O_(2)(g) rarrO_(3)(g) at298 K , if K_(p) for this conversion is 2. 47 xx 10^(-29).

Calculate Delta_(r )G^(@) for the reaction for which the value of K_(p) is 1.5 xx 10^(10) at 300 K.

Calculate Delta_(r ) G^(@) for the conversion of oxygen to ozone, (3)/(2)O_(2) (g) rarr O_(3) at 298 K, given partial pressure equilibrium constant is 2.47 xx 10^(-29).

Calculate Delta_r G^Ө for conversion of oxygen to ozone , 3//2 O_(2(g)) to O_(3(g)) at 298 K . If K_P for this conversion is 2.47xx10^(-29) .

Calculate DeltaH_(f)^(0) for the reactions. CO_(2_((g)))+H_(2_((g)))rarrCO_(2_((g)))+H_(2)O_((g)) " given that "DeltaH_(f)^(0) " for " CO_(2_((g))),CO_((g))and H_(2)O_((g)) " are " -393.5,-111.31 and -242 " KJ.mol"^(-1)respectively .

Calculate DeltaH of the reaction H - underset(Cl) underset(|) overset(H) overset(|) (C ) - Cl(g)rarr C(g) +2H(g) + 2Cl(g) Theaverage bond energies of C-H bond and C-Cl bond are 416kJ and 325 kJ mol^(-1) respectively.

Calculate DeltaG^( Theta ) for conversion of oxygen to ozone,(3)/(2) O_(2(g)) to O_(3(g)) at 298 K. If K_(p) for this conversion is 2.47 xx 10^(-29).

Calculate DeltaG^0 for conversion of oxygen to ozone 3//2O_2 hArr O_(3(g)) at 298 K, if K_p for the conversion is 2.47 xx 10^(-29)in standard pressure units.

Calculate DeltaG^@ for the following cell reaction

Calculate DeltaG^(@) for conversion of oxygen to ozone 3//20_(2)(g)toO_(3)(g) at 298K if Kp=2.47xx10^(-29)

Calculate DeltaG and DeltaG^(@) for the reaction :A + B hArr C + Dat 27^(@)C .Equilibrium constnat(K) for this reaction = 10^(2)

Calculate delta H_(r)^(0) for the reaction CO_(2)(g) + H_(2)(g) rarr CO(g) + H_(2)O(g) given that Delta H_(r)^(0) for CO_(2) (g), CO (g) and H_(2)O (g) are - 393.5, -111.31 and - 242 KJ mol^(-1) respectively.

Calculate DeltaH_f^@ for the reaction CO_2(g) +H_2(g) to CO(g) + H_2O(g) given that DeltaH_f^0 for CO_2(g),CO(g) and H_2O(g) are -393.5 , -111.31 and -242 "kJ mol"^(-1) respectively.

Calculate Delta S for 3 moles of a diatomic ideal gas which is heated and compressed from 298K and 1bar to 596K and 4 bar

Calculate Delta H_(f) of HCl if bond energy of H-H bond is 437 kJ Cl-CL bond is 244, and H-C is 433 kJ mol^(-1).

Calculate Delta H ^(298) for the reaction :2Li (s) + 2H_(2)O(l) rarr 2Li^(+) ( aq) + 2OH^(-) (aq)+ H_(2)(g) Given that the standard enthalpies of formation of Li^(+)(aq), OH^(-)(aq)and H_(2)O(l) are - 278.5, - 228.9 and - 285.8 kJ mol^(-) respectively( all at 298 K )

Calculate DeltaH in Joules for C_(("graphite")) rarr C_(("Diamond")) by using the following data C_(("graphiter")) + O_(2(g)) rarr CO_(2(g)) , DeltaH^@ =-393.5KJ C_(("Diamond")) + O_(2(g)) rarr CO_(2(g)) , DeltaH^@ =-395.4KJ

Calculate Delta H for the reaction 2O_(2)(g) rarr 3O_(2)(g) at 298 K and 1 atmosphere pressure given that Delta U = -287.9 kJ and R = 8.314 JK^(-1) mol^(-1).

Calculate Delta H ^(@)for the reaction CH_(2)= CH_(2) + 3O_(2) rarr 2CO_(2)+ 2H_(2)O Given that the average bond energies of the different bonds are : {:("Bond",C-H,O=O,C=O,O-H,C=C),("Bond energy " ( kJ mol^(-1)),414,499,724,460,619):}

Calculate DetlaG^(@) for the following reaction, Zn_((s)) + Cu_((aq))^(2+) to Zn_((aq))^(2+) + Cu_((s)) . DeltaG_(f)^(@) of Zn_((aq))^(2+) and Cu_((aq))^(+2)is -147.2 kJ mol^(-1)and 65kJmol^(-1) .

Calculate DeltaG^@ for conversion of oxygen to ozone, 3/2O_(2(g)) to O_(3(g)) at 298K, if K_P for this conversion is 2.47 xx 10^(-29)

Calculate de Broglie wavelength of an electron moving with 1% of the speed of light.

Calculate consecutive oxidation number of Br in Br_(3)O_(8).

Calculate average kinetic energy of CO_2 molecules at 27^(@)C.

Calculate (a)wavenumberand (b )frequecyof yellowradiationhavingwavelength5800 A

Calculate (a) wave number and (b) frequency of yellow radiations having wavelength of 5800Å

Calculate a value of effective magnetic moment mu_(eff) of Ce^(3+) ion.

Calculate (a) hydrolysis constant, (b) degree of hydrolysis and (c) pH of 0.01 M sodium acetate, if K_a of CH_3 COOH is 1.9 xx10^(-5)

Calculate (a) DeltaG^0and (b) the equilibrium constant for the formation of NO_2 from NO and O_2 at 298 K, NO_((g)) + 1/2O_(2(g)) hArr NO_(2(g)) where , DeltaG_f^ө(NO_2)= 52.0 kJ "mol"^(-1) DeltaG_f^ө(NO)=87.0 kJ mol^(-1) DeltaG_f^ө (O_2) = 0.0 kJ "mol"^(-1)

Calculate (a)Delta G^(@) and (b) the equilibriumconstantfor theformation of NO_(2) " from"NO and NO_(2)and O_(2) at 298 K NO (g) +1/2 O_(2) (g) hArr NO_(2) (g)whereDelta _(f) G^(@) (NO_(2)) = 52*0 "kJ//mol ", Delta_(f) G^(@) (NO) = 87*0"kJ//mol," Delta _(f) G^(@) (O_(2)) = 0 "kJ//mol.

Calculale the d_((N-C)) in (CH_(3))_(3)N molecule by using Eqs. (1.18) and (1.19) above and show from which equation d_((N-C)) closely resembles with the experimental value of d_((M|N-C))=1.47 Å. (Given: r_(N) = 0.75 Å, r_(C ) = 0.77Å, chi_(A)=3.0, chi_(C )=2.5

Calcualte the work down when 11.2 g of iron dissolves in hydrochloric acid in (i) a closed vessel (ii) an open beaker at 25^(@)C ( Atomic meass ofFe = 56u)

Calculae the entropy change fo n-hexane when one mole of it evaporates at 341.7 K (Delta_(vap) H = 29.0 kJ mol^(-1))

Calculate the volume occupied by 2 moles of NO_(2) at STP.

Calcualte the number of chiral center in the molecule Ethyl 2,2-dibromo-4-ethyl-6-methoxy cyclohexane carboxylate.

Calcualte the mass percentage of nitrogen in hydrazinium Sulphate (N_(2)H_(5))_(2)SO_(4). {:("Molar mass", (g mol^(-1))), ((N_(2)H_(5))_(2)SO_(4), 162.2):}

Calcualte the enthalpy change when infinitely dilute solution of CaCI_(2) and Na_(2)CO_(3) are mixed. Delta_(f)H^(Theta) for Ca^(2+)(aq), CO_(3)^(2-)(aq), and CaCO_(3)(s) are -129.80, -161.65, and -288.50 kcal mol^(-1) respectively.

Calcualate the resonance energy of N_(2)O fromthe following data : Delta_(f)H^(@) of N_(2)O= 82kJ mol^(-1) , Bond energies of N -= N , N = N , O=O and N=O bonds are 946, 418, 498and 607 kJ mol^(-1) respectively.

Calcualate the value of Z_("eff") on 3delectronsof Sc .

Calculate (a)Delta G^(@) and (b) the equilibriumconstantfor theformation of NO_(2) " from"NO and NO_(2)and O_(2) at 298 K NO (g) +1/2 O_(2) (g) hArr NO_(2) (g)whereDelta _(f) G^(@) (NO_(2)) = 520 "kJ//mol ", Delta_(f) G^(@) (NO) = 870"kJ//mol," Delta _(f) G^(@) (O_(2)) = 0 "kJ//mol.