Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

For the reaction X hArr 2Y and Z hArr P+Q occuring at two different pressure P_(1) and P_(2), respectively. The ratio of the two pressure is 1:3. What will be the ratio of equilibrium constant, if degree of dissociation of X and Z are equal.

Answer»

`1:36`
`1:12`
`1:9`
`2:3`

SOLUTION :`{:("for reaction",X,hArr,2Y),("Initial",1,,0),("At equilibrium",1-alpha,,2alpha):}`
Total "moles" `=1-alpha+2alpha=1+alpha`
`P_(X)=((1-alpha)/(1+alpha))P_(1), P_(Y)=((2alpha)/(1+alpha))P_(1)`
`:. K_(p_(1))=((P_(Y))^(2))/(P_(X))=([((2alpha)/(1+alpha))P_(1)]^(2))/(((1-alpha)/(1+alpha))P_(1))=(4alpha^(2)P_(1))/((1-alpha^(2)))` ...(i)
`{:("Similarly for",X,hArr,P,+,Q),("Initial",1,,0,,0),("At equilibrium",1-alpha,,2alpha,,alpha):}`
Total moles `=1-alpha+alpha+alpha=1+alpha`
`P_(Z)=((10alpha)/(1+alpha))P_(2), P_(Q)=(alpha/(1+alpha))P_(2), P_(P)=(alpha/(1+alpha))P_(2)`
`:. K_(p_(2))=(P_(P)xxP_(Q))/P_(Z)=((alpha/(1+alpha))P_(2)xx(alpha/(1+alpha))P_(2))/(((1-alpha)/(1+alpha))P_(2))`
`=(alpha^(2)P_(2))/((1-alpha^(2)))`
DIVIDING equations (i) and (ii), we get,
`K_(p_(1))/K_(p_(2))=(4alpha^(2)P_(1))/((1-a^(2)))xx((1-alpha^(2))/(alpha^(2)P_(2)))=(4P_(1))/P_(2)rArr (4P_(1))/P_(2)=1/3` (given)
`:. P_(1)/P_(2)=1/12`
Discussion
2.

For the reaction X_(2)O_(4(l)) rarr 2XO_(2(g)), Delta U= 2.1 "kcal", Delta S= 20 "cal " K^(-1) at 300K, Hence Delta G is

Answer»

`2.7` KCAL
`-2.7` kcal
`9.3` kcal
`-9.3` kcal

ANSWER :B
Discussion
3.

For the reaction, SrCO_(3(s)) hArr SrO_((s)) + CO_(2(g)) , the value of equilibrium constant K_P = 2.2xx10^(-4)at 1002 K . Calculate K_C for the reaction ?

Answer»

SOLUTION :For the REACTION ,
`SrCO_3(s) hArr SrO(s) + CO_2(g)`
`Deltan_g = 1-0 = 1 `
`thereforeK_p = K_C =(RT)`
`2.2xx10^(-4) = K_C (0.0821) (1002) `
`K_C =(2.2xx10^(-4))/(0.0821xx1002) =2.674xx10^(-6)`
Discussion
4.

For the reaction : SrCO_3 (s) hArr SrO(s)+ CO_2(g), the value of equilibrium constnat K_P = 2.2 xx 10^(-4) " at " 1002 K. Calculate K_C for the reaction.

Answer»

Solution :For the REACTION, `SrCO_3(s) hArr SrO(s) + CO_2(g)`
`Deltan_g = 1- 0 =1`
`:. K_P = K_C (RT)`
`2.2 XX 10^(-4) = K_C (0.082)(1002)`
`K_C = (2.2 xx 10^(-4))/(0.0821 xx 1002) = 2.674 xx 10^(-6)`
Discussion
5.

For the reaction SO_(2(g)) +1/2O_(2(g)) hArr SO_(3(g)), if K_P=K_C(RT)^x, where the symbols have usual meaning then the value of x is : (assuming ideality)

Answer»

`1/2`
1
`-1`
`-1/2`

Solution :`SO_(2(g)) + 1/2O_(2(g)) hArr SO_(3(g))`
`K_P=K_C(RT)^x`
`x=Deltan_g`
=(Number of total moles of gaseous PRODUCT )-(Number of total moles of gaseous REACTANTS)
`=1-(1+1/2)`
`=1-3/2=(-1)/2`
Discussion
6.

For the reaction ,SO_(2) (g) + 1/2 O_(2) (g) hArr SO_(3) (g) , if K_(p) = K_(c) (RT)^(x) where the symbols haveusual meaning, then value of x is ( assuming ideality )

Answer»

1
-1
`-1/2`
`1/2`

Solution :` K_(p) = K_(c) (RT)^(DELTA n_(g)`
` :. X = Delta n_(g) = (n_(p)- n_(r)) _(GASEOUS) = 1 - (1+1/2 ) = -1/2`
Discussion
7.

For the reaction, SO_(2(g))+(1)/(2)O_(2(g)) harr SO_(2(g)). If K_(P)=K_(c)(RT)^(2), when the symbols have usual meaning, the value of x is (assuming ideality)

Answer»

<P>-1
`-(1)/(2)`
`+(1)/(2)`
1

Solution :`K_(P)=K_(C)(RT)^(DELTA n_((g)))`
Discussion
8.

For the reaction SnO_(2(s)) + 2H_(2(g)) hArr 2H_(2) O_((g)) + Sn_((d)) . Calculate K_(p) at 900K, where the equilibrium steam hydrogen mixture was 45% H_(2) by volume.

Answer»

`1.15`
`2.25`
`7.75`
`10`

ANSWER :B
Discussion
9.

For the reaction PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2) (g) the forward reaction at constant temperature is favoured by …………

Answer»

INTRODUCING an INERT gas at CONSTANT VOLUME
introducing ` PCl_3 ` (g) at constant volume
introducing `PCl_5(g)` at constant volume
introducing`Cl_2(g)` at constant volume

SOLUTION :introducing `PCl_5(g)` at constant volume
Discussion
10.

For the reaction PCl_(3)(g) + Cl_(2)(g) rArr PCl_(5)(g) at 250^(@) C, then value of K_(c) is 26 then the value of K_(p)on the same temperature wil be ...........

Answer»


Solution :`Deltan_(G) = 1-2=-1`
`K_(p)-K_(c)(RT)^(Delta)ng= 26(0.0821 xx 523 )^(-1)` = 0.61
Discussion
11.

For the reaction PCl _(5) (g) hArr PCl_(3)(g) + Cl_(2) (g), the forward reaction at constant temperature is favoured by

Answer»

introducing an INERT gas at constant VOLUME
introducing `PCl_(5)`at constantvolume.
introducing an inert gas at constant pressure
increasing the volume of the container

ANSWER :B::C::D
Discussion
12.

For the reaction PCl_(3) (g) + Cl_(2) (g) hArr PCl_(5) (g) the value of K_(c) " at " 250^(@)C " is26. The value of " K_(p)at this temperature will be

Answer»

`0*61`
`0*57`
`0*83`
`0*46`

Answer :A::B::C::D
Discussion
13.

For the reaction of one mole zinc dust with one mole sulphuric acid in a bomb calorimeter, DeltaU and w correspond to

Answer»

`DELTA U gt 0, W=0`
`Delta U LT 0, wlt 0`
`Delta U gt 0,w =0`
`Delta U gt 0, w gt 0`

Answer :A
Discussion
14.

For the reaction of HCl with the following alkenes, predict the correct sequence of reactivity asmeasured by reaction rates : I) ClCH=CH_2 II) (CH_3)_2 C = CH_2 III) OHC CH = CH_2 IV) (NC)_2 C = C(CN)_2

Answer»

`IV GT I gt III gt II`
`I gt IV gt II gt III`
`III gt II gt IV gt I`
`II gt I gt III gt IV`

SOLUTION :`II gt I gt III gt IV `
Discussion
15.

For the reaction NOBr (g)iffNO(g)+(1)/(2)Br_(2)(g) K_(P)=0.15 atm at 90^(@)C. If NOBr, NO and Br_(2) are mixed at this temperature having partial pressures 0.5 atm, and 0.2 atm respectively, will Br_(2) be consumed or formed?

Answer»

<P>

Solution :`Q_(P)=([P_(Br_(2))]^(1//2)[P_(NO)])/([P_(NOBR]])=([0.2]^(1//2)[0.4])/([0.50])=0.36`
`K_(P)=0.15`
`therefore Q_(P)gtK_(P)`
Hence, reaction will shift in BACKWARD direction `therefore Br_(2)` will be consumed
Discussion
16.

For the reaction NH_(4)COO NH_(2)(g) hArr 2NH_(3)(g) + CO_(2)(g) If equilibrium pressure is 3 atm. Find the value of Kp

Answer»


ANSWER :4
Discussion
17.

ForthereactionN_(2)O_(4)(g) hArr 2 NO_(2) (g), the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct ?

Answer»

<P>The REACTION is endothermic
The reaction is exothermic
If `NO_(2) (g) and N_(2) O_(4)(g)` are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more `N_(2)O_(4) (g)` will be formed,
The entropy of the system increases.

Solution :(a) As the value of K increases with increaseof temperature and `K=(k_(f))/(k_(b))`, this means that `k_(f)` increases, i.e., formard reaction is favoured. Hence, reaction is endothermic.
(C) At 400 K, `Q=(p_(NO_(2))^(2))/(p_(N_(2)O_(4)))=((20)^(2))/(2) = 200 `. Thus `Q gt K`. Equilibrium will shift backward to form more `N_(2)O_(4)`.
(d) As reaction is accompanied by INCREASE in the number of moles , entropy increases.
Discussion
18.

For the reaction N_(2)O_(4(g))hArr 2NO_(2(g)) .which of the following factors will have no cffect on the value of equilibrium constant ?

Answer»

Temperature
Intial CONCENTRATION of `N_2O_4`
presence of catalyst
pressure

Solution :K is not AFFECTED by ANYTHING EXCEPT T
Discussion
19.

For the reaction N_2O_(4(g)) hArr 2NO_(2(g)) ,the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is/are correct ?

Answer»

The reaction is endothermic
The reaction is exothermic
If `NO_(2(g))` and `N_2O_(4(g))` are mixed at 400 K at PARTIAL pressures 20 bar and 2 bar respectively, more `N_2O_(4(g))` will be formed.
The entropy of the system increases.

Solution :For the reaction, `N_2O_(4(g)) hArr 2NO_(2(g))`
At 400 K, K=50
At 500 K, K=1700
(a) If the value of K increase with increase of temperature and `K_f/K_b`, Because of that `K_f`increases, i.e., forward reaction is TAKE PLACE. So, reaction is endothermic.
(c) As, number of moles of gaseous products are greater than the number of moles of gaseous reactants. So, higher pressure favours the backward reaction. More `N_2O_(4(g))` will be obtained, if `P_"product" GT P_"reactant"`
(d) As reaction is accompanied by increase in the number of moles, entropy increases.
Discussion
20.

For the reaction N_2O_4 harr 2NO_(2(g)), the degree of dissociation at equilibrium is 0.2 at 1 atm. Then K_p will be

Answer»

<P>`1//2`
`1//4`
`1//6`
`1//8`

Solution :
`K_(P)=([P((2alpha)/(1+alpha))]^(2))/(P((1-alpha)/(1+alpha)))=P xx (4alpha^(2))/(1-alpha^(2))`
`K_(P)=1 xx (4 xx (0.2)^(2))/(1-(0.)^(2))=(1)/(6)`
Discussion
21.

For the reaction N_(2)H_(4)(g)toN_(2)H_(2)(g)+H_(2)(g), DeltaH_("reaction")^(@)=109KJ//mol in_(N-N)=163KJ//mol in N-H=391KJ//mol in_(H-H)=KJ//mol then ,then bond dissocition energy of N=N is:

Answer»

500KJ/mol
436KJ/mol
600KJ/mol
400Kj/mol

Answer :d
Discussion
22.

For the reaction : N_(2)(g)+3H_(2)(g)=2NH_(3)(g) Equilibrium constant K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))

Answer»


SOLUTION :NA
Discussion
23.

For the reaction N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g), predict whether the work is done on the system or by the system.

Answer»

Solution :Volume is DECREASING therefore, work is DONE on the SYSTEM.
Discussion
24.

For the reaction, N_(2(g)) + 3H_(2(g)) rarr 2NH_(3(g)). Heat of reaction at constant volume exceeds the heat of reaction at constant pressure by the value of xRT. The value of x is______

Answer»


SOLUTION :`DELTA E - Delta H = - Delta ng RT = 2RT`
Discussion
25.

For the reaction, N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g), Delta H = -95.2 kJ and Delta S = -198.1 JK^(-1). Calculate the temperature at which Gibb's energy change of the reaction (Delta G) becomes equal to zero.

Answer»

Solution :`DELTA G = Delta H - T Delta S`. When `Delta G = 0, Delta H = T Delta S`
or `T = (Delta H)/(Delta S) ""T = (-95.2 xx 10^(3)J)/(-198.1 JK^(-1)) = 480.6 K`.
Discussion
26.

For the reaction N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)), the standard equilibrium constant K_P is 5.8xx10^5 at 298 K temperature . If the concentration of gases indicate by mol L^(-1) then find the value of standard equilibrium constant from the following. (R=0.08314 L bar K^(-1) "mol"^(-1) )

Answer»

`3.5xx10^6`
`3.8xx10^7`
`3.56xx10^(+8)`
`3.99xx10^9`

Solution :`{:(,N_2+,3H_2hArr, 2NH_3),("Mole",1,3,2),("Total of REACTANTS" ,,"4 mol", "2 mol products"):}`
`Deltan_((G))` = mole of products 0 mole of reactants
=(2-4)=-2
`K_p=K_c(RT)^(Deltan)`
`therefore K_c=K_p/(RT)^(Deltan)`
`=(5.8xx10^5)/(0.08314xx298)^(-2)`
`=(5.8xx10^5)(0.08314xx298)^2`
`=3.561xx10^8`
Discussion
27.

For the reaction N_(2)(g) + 3H_(2)(g) rarr 2NH_(2)(g), DeltaH =- 95.4 Kjand Delta S=- 198 .300 JK^(-1) . Calculate the temperature in centrigrade at which it attains equilibrium.

Answer»


ANSWER :`208.1^(@)C`
Discussion
28.

For the reaction N_(2) O_4 (g) hArr 2NO_(2) (g) K_(C)= 0.21 at 373 K . The concentration of N_(2) O_(4) and NO_(2) are found to be 0.125 mol dm^(-3) and 0.5 mol dm^(-3) respectively at a given temperature . Predict the direction of the reaction.

Answer»

At equilibrium
REVERSE DIRECTION
forward direction
Both reverse and forward direction

Solution :`Q = ([NO_(2)]^(2))/([N_(2) O_(4)]) = (0.5 xx 0.5)/(0.215) = 2`
`K_(C) = 0.21 , Q =2 `
`Q gt K_(C)`Hence the reaction will proceed in the reverse direction
Discussion
29.

For the reaction N_(2) (g) + O_(2) (g) hArr 2 NO (g), " the value of " K_(c)" at "800^(@)C " is " 0*1. When the equilibrium concentration of both the reaction is 0*5 mol, what is the value of K_(p) at this temperature ?

Answer»

`0*5`
`0*1`
`0*01`
`0*025`

ANSWER :A::B::C::D
Discussion
30.

For the reaction ,N_(2) (g) + 3 H _(2) (g) hArr 2 NH_(3) (g),the partial pressures ofN_(2)andH_(2) are 0.80 and 0.40 atmosphere respectively at equilibrium . The total pressure of the system is 2.80 atmospheres. What is K_(p) for the above reaction ?

Answer»

Solution :The reaction is : `N_(2) (G) + 3H_(2) (g) hArr 2 NH_(3) (g)` ,
We are given that at equilibrium , `p_(N_2) = 0.80 " atmosphere ", p_(H_2) = 0.40 " atmosphere "`
` p_(N_2) + p_(H_2) + p _(HN _(3)) = 2. 80 " atmosphere " :. p_(NH_(3)) = 2.80- (0.80 + 0.40) = 1.60 " atmosphere ".`
APPLYING the law of chemical equilibrium , we get (taking pressures with respect with respect to standard STATE pressure of 1 atm)
`K_(p) =(P_(NH_(3))^(2))/(P_(N_2) xx P_(H_(2))^(3)) = (1.60)^(2)/(0.80 xx(0.40)^(3))=50.0`
Discussion
31.

For the reaction N_(2)+3H_(2)hArr2NH_(3) at 773K, the value of K_(p)=1.4xx10^(-15). Calculate K_(c) (Given R = 8.314 JK^(-1)mol^(-1)).

Answer»

Solution :`K_(p)=K_(c)(RT)^(DELTAN),Deltan=2-4=-2`
`1.4xx10^(-15)=K_(c)(8.314xx773)^(-2)`
`K_(c)=1.4xx10^(-15)xx(8.314xx773)^(2)=5.782xx10^(-8)`
Discussion
32.

For the reaction, N_(2) + 3H_2 rarr2NH_(3)if E_1and E_2equivalent masses of NH_3 and N_2respectively then E_1 -E_2 is

Answer»

1
2
3
4

Solution :`OVERSET(0)(N_2)RARROVERSET(-3)(2NH_3),overset(-3)(2NH_3)rarroverset(O)(N_2)`
`17/6-(28)/3,(34-28)/6=6/6 =1`
Discussion
33.

For the reaction, N_(2) + 3H_(2) hArr2NH_(3) in a vessel, equal moles of N_(2) and H_(2) are mixed to attain equilibrium.

Answer»

`[N_(2)]=[H_(2)]`
`[N_(2)] gt [H_(2)]`
`[N_(2)] lt [H_(2)]`
`[H_(2)] gt [NH_(3)]`

Solution :
Discussion
34.

For the reaction, M^(x+) + MnO_(4)^(-)rarrMO_(3)^(-)+ Mn^(2+) + 1//2 O_2if one mole of MnO_4^(-)oxidises 1.67 moles of M^(x+) " to " MO_(3)^(-)then the value of x in the reaction is?

Answer»


Solution :`overset(+7) (MnO_4^(-))+5e^(-) rarrMn^(2+)`
Since 1 mole of `MnO_(4)^(-)`ACCEPTS 5 MOLES of electrons , therefore , 5 moles electrons are lost by 1.67moles of `M^(x+)`
`:. 1`Mole of `M^(x+)` will lose electrons = 5/1.67 = 3 moles ( approx.)
Since `M^(x+)`changes to `MO_(3)^(-)`( where O.N.of M = +5) by accepting 3 electrons
`:. x = +5-3 =+2`
`:. Mn^(7+)+5e rarr Mn^(2+)`
Discussion
35.

For the reaction , MnO_(2) + 4HCl overset(100%)rarr MnCl_(2)+ Cl_(2)+2H_(2)O, 8.7 gm MnO_(2) is dissolved in 500 ml of HCl solutioncontaining7.3 gm HCl per litre (Mn=55)

Answer»

HCl is the LIMITING reagent.
`MnO_(2)` is the limiting reagent
0.025 moles of `MnCl_(2)` will form
560 ml `Cl_(2)` GAS will liberate at `0^(@)` C and 1 atm.

Answer :A::C::D
Discussion
36.

For the reaction, KO_(2) + H_(2)O+CO_(2) rarrKHCO_(3) +O_2the mechanism of reaction suggests that:

Answer»

Acid-base reaction
Disproportionation reaction
HYDROLYSIS
REDOX change

Solution :`4KO_2+2H_(2)O RARR 4KOH+3O_2`
(Hydrolysis and disproportionation)
`4KOH +4CO_2 rarr 4KCO_3` (Acid - base reaction )
Discussion
37.

For the reaction in equilibrium: 2NOBr_((g)) hArr 2NO_((g)) + Br_(2(g)) If P_(Br_(2)) is (P)/(9)t equilibrium and P is totalpressure, If (P)/(Kp) = x^(2) then what is x value.

Answer»

<BR>

Solution :`2NOBr_((g)) harr 2NO_((g))+Br_(2)`
pressure at equilibrium `P-(3P)/(9), (2P)/(9), (P)/(9)`
`P_(T)=(6P)/(9)+(2P)/(9)+(P)/(9)=P`
`K_(P)=(P^(2)NO.PBr_(2))/(P^(2)NOBR)=(((2p)/(9))^(2) xx P)/(((6P)/(9))^(2)) IMPLIES K_(P)=(P)/(81)`
Given `(P)/(K_(P))=x^(2), K_(P)=(P)/(x^(2))=(P)/(81) implies x=9`
Discussion
38.

For the reaction in equilibrium A harr B ([B])/([A])=4.0 xx 10^(8)""(-d[A])/(dt)=2.3 xx 10^(6)S^(-1)[A]""(-d)/(dt)[B]=K[B] Thus, K is

Answer»

`1.1 xx 10^(-15) s^(-1)`
`5.8 xx 10^(-3)s^(-1)`
`1.7 xx 10^(2)s^(-1)`
`9.2 xx 10^(14)s^(-1)`

Solution :`A HARR B, H_(c)=([B])/([A])=4.0 xx 10^(8)`
`(-d)/(dt)(A)` = Rate of forward
`= 2.3 xx 10^(6)S^(-1)[A]`
= rate constant of forward
Also `K_(c)=(K_(f))/(K_(b)) implies K_(b)=(K_(f))/(K_(c))=(2.3 xx 10^(6))/(4.0 xx 10^(8))`
`= 5.75 xx 10^(-2)s^(-1)`
Discussion
39.

For the reaction : I^(-) +ClO_3^(-)+H_2SO_4rarr Cl^(-)+HSO_4^(-)+I_2 The correct statement (s) in the balanced equation is/are

Answer»

stoichiometic coefficient of `HSO_4^(-)` is 6
iodide is oxidised
SULPHUR is reduced
`H_2O` is one of the products.

Solution :The BALANCED chemical equation is :

(a) Stoichiometric coefficient of `HSO_4^(-) ` is 6.
(B) O.N.of `I^(-)` INCREASES from -1 to 0 and hence it is oxidised .
`H_2O` is one the products.
Discussion
40.

For the reaction, I^(-)+CIO_(3)^(-)+H_(2)SO_(4)rarrCI^(-)+HSO_(4)^(-)+I_(2) the correct statement (s) in the balanced equation is//are

Answer»

stoichiometric coeffiecient of `HSO_(4)^(-)` is 6
iodide is oxidised
suphur is reduced
`H_(2)O` is one of the products

Solution :The oxidation half reaction is:
`2I^(-)rarrI_(2)+2E^(-)`
The REDUCTION half rection is :
`6H^(+)+CIO_(3)^(-)+6E^(-)rarrCI^(-)+3H_(2)O`
Multiply eqn.(i) by 3 and add to eqn.(i)
`6H^(+)+CIO_(3)^(-)+6e^(-)rarrCI^(-)+3H_(2)O`
`6I^(-)rarr3I_(2)+6e^(-)`
`(6I^(-)+6H^(+)+CIO_(3)^(-)rarr3I_(2)+3H_(2)O+CI^(-))/((a) "The stoichiometry of" HSO_(4)^(-) is 6`
The `H^(+) "IONS are provided by"H_(2)SO_(4)"or"HsO_(4)^(-)` ion
(b) `I^(-)` ions are oxidisded to `I_(2)`
(d) `H_(2)O` is one of the products formed in the REDOX rection.
Discussion
41.

For the reaction I^(-)+CIO_(3)^(-)+H_(2)SO_(4)rarr CI^(-)+HSO_(4)^(-)+I_(2) The correct statement (s) in the balanced equatoin is / ar

Answer»

stoichiometriccoefficient of `HSO_(4)^(-)` is 6
IODIDE is oxidized
suphur is REDUCED
`H_(2)O` is one of the products

Solution :Balanced equation of the reaction is

SINCE O.N of S remains the same i.e +6 in `H_(2)SO_(4)` and `HSO_(4)^(-)` therefore S is not reducedthe remaining three statement (a,b,d) are CORRECT
Discussion
42.

For the reaction I_(2(g)) hArr I_(2(s)), Delta H= -ve. Then choose the correct statement from the following (A) The process is spontaneous at all temperature (B) The process is accompained by an increase in entropy (C ) The process is accompained by a decrease in entropy (D) The process is accompained by a decrease in enthalpy

Answer»

Only a, B and C
Only b and d
Only c and d
Only a,c and d

ANSWER :C
Discussion
43.

For the reaction, H_(2)(g)to 2HI(g) K_(2)=50.0 at 721 K. What is the value of Delta^(circ) for this reaction (per mole ofH_(2)) at 721K?

Answer»

`-32.3kJ`
`-23.5kJ`
`-10.2kJ`
`-0.231kJ`

ANSWER :B
Discussion
44.

For the reaction H_(2)(g)+I_(2)(g)hArr 2HI(g), the standard free energy is Delta G^(Θ) gt 0. The equilibrium constant (K) would be.........

Answer»

K= 0
`K gt 1`
`K = 1 `
`K LT 1`

Solution :`DeltaG^(@)=-RT " ln" K. DeltaG^(@_ gt 0` MEANS `Delta g^(@) ` is + ve. This can be soonly if ln K is - ve , i.e., `K lt 1`.
Discussion
45.

For the reaction H_(2(g)) + I_(2(g)) hArr 2HI_((g)) ,the standard free energy is DeltaG^ө gt 0. The equilibrium constant (K) would be

Answer»

K=0
K `GT` 1
K=1
`K LT 1`

Solution :`DeltaG^ө` and K are RELATED as `DeltaG^ө` =-RT ln `K_c`. When one `DeltaG^ө gt 0`
`THEREFORE DeltaG^ө` is positive . Thus in `K_c` is negative i.e., `K_c lt1`
Discussion
46.

For the reaction H_2(g)+Br_2(g) leftrightarrow 2HBr(g) at 1024K, K_pis 1.6 times 10^5. IF 10 bar of hydrogen bromide is introduced into the vessel at 1024K to establish equilibrium, what is the equilibrium pressure of hydrogen bromide?

Answer»

SOLUTION :9.98 BAR
Discussion
47.

For the reaction , H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)(l),DeltaC_(p)=32 JK^(-1), DeltaHat 27^(@)C = - 285.8 kJ mol^(-1). What will be the value of DeltaHat 127^(@)C ?

Answer»

`- 289 . 0 kJ mol^(-1)`
` - 282 . 6 kJ mol^(-1)`
` -317 .0 kJ mol^(-1)`
` -253.8 kJ mol^(-1)`

Solution :According to Kirchhoff's equation,
`( DeltaH_(2) - DeltaH _(1))/(T_(2) - T_(1))= DeltaC_(p)`
or `DeltaH_(2) - DeltaH _(1)= DeltaC_(p) ( T_(2) - T_(1))`
or `DeltaH =DeltaH_(1)+ DeltaC_(p) ( T_(2)-T_(1))`
`=- 285.8 + ( 32 XX 10^(-3)) ( 400- 300)`
`=-282.6kJ mol^(-1)`
Discussion
48.

For the reaction H_2+I_2 hArr 2HI, DeltaH=12.40 Kcal the heat of formation of HI is _____

Answer»

Solution :`H_2+ I_2 to 2HI "" DeltaH`=12.40 K.cal
`1//2 H_2 +1//2 I_2 to HI =DeltaH//2`
`THEREFORE 12.40/2=6.20 "K.cal mol"^(-1)`
Discussion
49.

For the reaction ,H_(2) + I_(2) hArr 2 HI , K= 47*6. If the intial number of moles of each reactant and product is 1 mole , then at equilibrium

Answer»

`[I_(2)] = [H_(2)],[I_(2)] GT [HI]`
`[I_(2)]lt [H_(2)], [I_(2)] = [HI]`
`[I_(2)] = [H_(2)] , [I_(2)] lt [HI]`
` [I_(2)] gt [H_(2)] , [I_(2)] = [HI]`

Solution :` K= ([HI]^(2))/([H_(2)][I_(2)])`
As 1 mole of`H_(2) " reactswith1 moleofI_(2)`,EVEN after equilibrium`[H_(2)] = [I_(2)]`.
HENCE, `K= ([HI]^(2))/([I_(2)]^(2)) `
or ` ([HI])/([I_(2)])= sqrt(K) = sqrt(47.6)`
i.e., `[HI] gt [I_(2)] or [I_(2)]lt [Hi] `
Discussion
50.

For the reaction H_(2) + I_(2) hArr 2 HI, " if intially 25 mL of "H_(2) and 20 " mL of "I_(2) are present ia a container in a container and at equilibrium, 30 mL of HI is foprmed , then calculate equilibrium constant.

Answer»

SOLUTION :`{:(,H_(2),+,I_(2),hArr,2 HI),(" Initial:",25 mL,,29 mL,,0),(" At EQM. :",25-x,,20-x,,2 x):}`
` :. " At equilibrium ",H_(2) = 25 - 15 10" mL," I_(2) = 20-15 " mL" and HI = 30 " mL".`
As equal volumes contain equal number of mole , THEREFORE, volumes can be used in place of moles. HENCE,
`K= [HI]^(2)/([H_(2)][I_(2)])=(30)^(2)/((10)(5))=900/50= 18`
Discussion