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 a good quality cement the ratio of (CaO)L (lime quick) to SiO_(2), Al_(2)O_(3) and Fe_(2)O_(3) is

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SOLUTION :For a GOOD quality cement
`(%CAO)/(%SiO_(2)+%Al_(2)O_(3)%Fe_(2)O_(3))=2`
2.

For a given reaction K_p lt K_c. Increase of pressure favours

Answer»

the backward REACTION
no reaction
the FORWARD reaction
both forward and backward reactions equally

Solution :`K_(P) lt K_(C) implies Delta ng lt 0`
`:.` on increasing pressure, reaction goes in the forward DIRECTION
3.

For a given reaction, DeltaH = 35.5kJ mol^(-1) and DeltaS = 83.6 JK^(-1) mol^(-1) .The reaction s spontaneous at : ( Assume atDeltaHand DeltaS do not vary with temperature )

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`T lt425 K`
`T gt 425 K`
All temperatures
`T gt 298 K`

Solution :`DeltaS= DeltaH - T DeltaS`
For the reaction to be spontaneous , `DeltaG = - ve`
As`DeltaH` and`DeltaS` both are positive,`DeltaG` can be `-ve` only if`T DeltaS gt DeltaH`or`T gt ( DeltaH)/( DeltaS)`
`i.e,T gt ( 35.5xx 1000J)/( 83.6JK^(-1)) ` or `T gt 425 K`
4.

For a given reaction, DeltaH = 35.5 kJ"mol"^(-1) and DeltaS = 83.6 J K^(-1) "mol"^(-1). The reaction is spontaneous at (assume that DeltaH and DeltaS do not vary with temperature.

Answer»

`T gt425K`
all temperatures
`T lt298K`
`T lt425K`

Solution :`DeltaG = DeltaH -TDeltaS`
For a reaction to be at EQUILIBRIUM,
`DeltaG =0`
`DeltaH = TDeltaS`
or `T = (DeltaH)/(DeltaS) = (35.5 xx 10^(3))/(83.6)`
` = 425 K`
Since reaction is endothermic, it will be sponatneous at `T gt 425 K`
5.

For a given reaction at a particular temprature, the equilibrium constant value. Is thevalue of Q also constnat ? Explain.

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Solution :In the chemical REACTION, as the reaction proceeds., there is a containuous CHANGE in the CONCENTRATION of reactants and products and also the Q value until the reaction reaches the EQUILIBRIUM.So even at particular TEMPERATURE, Q is not constant. Even once the equilibrium in achieved then change in concentrationof reactants or products, pressure, volume will change the value of Q.
6.

For a given reaction, at a particular temperature, the equilibrium constant has value. Is the value of Q also constant ? Explain.

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Answer :No. The value of Q is determined by THECONCENTRATION of product and reactants that are not necessarily EQUILIBRIUM concentration. Thus its value is not a constant. Atequilibrium `K_(C) = Q ` .
7.

For a given mass of gas, if pressure is reduced to half and temperature is increased two times, then the volume of gas would become ___ times .

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SOLUTION :`(P_1V_1)/(T_1) = (P_2V_2)/(T_2) , (PV_1)/(T) = ((P//2)V_2)/(2T) IMPLIES V_2 = 4V_1`.
8.

For a given mass of a gas, if pressure is reduced to half and temperature is doubled, then volume V will become

Answer»

`4V`
`2V^2`
`V/4`
`8V`.

SOLUTION :`4V`
9.

For a given gas, T_(C) = 40 K, then T_(i) is

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40 K
270 K
135 K
540 K

Solution :`T_(i) = (2a)/(Rb)`
`T_(C) = (8a)/(27 Rb) = (4)/(27) ((2a)/(Rb))`
`:. T_(C) = (4)/(27) T_(i)`
`T_(i) = (27 xx T_(C))/(4) = (27 xx 40)/(4) = 270 K`
10.

For a given exthermic reaction , K_(p) nd K_(p)are the equilibrium constants at temperature T_(1) and T_(2) respectively . Assumbing that heat of reaction is constant in temperature range between T_(1) and T_(2)it is readily observed that

Answer»

<P>` K_(p) gt K_(p)'`
` K_(p) lt K'_(p)`
`K_(p) = K'_(p)`
` K_(p) = 1/(K'_(p)) `

Solution :According to van't Hoff equation ,
` log = (K'_(p))/(K_(p))= -(DeltaH)/(2.303 R) (1/T_(2)-1/T_(1))`
For EXOTHERMIC reaction, `Delta H=-ve`
Also ,as `T_(2)gt T_(1), (1/(T_(2)) - 1/T_(1)) = - ve `
` :. log = (K'_(p))/(K_(p)) = -ve or log K'_(p) - - log K_(p) = -ve`
i.e.,` log .K'_(p) lt log K_(p) or K'_(p)lt K_(p) or K_(p) gt K'_(p)`
11.

For a given exothermic reaction, K_(p) and K_(p) are the equilibirum constants at temperature T_(1) and T_(2) respectively. Assuming that heat of reaction is constant in temperature range between T_(1) and T_(2), it is readily observed that

Answer»

`K_(p) gt K'_(p)`
`K_(p) lt K'_(p)`
`K_(p) = K'_(p)`
`K_(p) = (1)/(K_(p))`

Solution :In EXOTHERMIC reaction, on INCREASING temperature value of `K_(p)` decreases.
`:.``K_(p) gt K'_(p)`(Assumuing `T_(1) lt T_(2)`).
12.

For a general reaction given below, the value of solubility product can be given us {:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):} K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y) Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation [H^(+)] ion, [OH^(-)] ion. It is also useful in qualitative analysis for the idetification and separation of basic radicals Potussium chromate is slowly aded toa solution containing 0.20M Ag NO_(3), and 0.20M Ba(NO_(3))_(2). Describe what happensif the K_(sp) for Ag_(2),CrO_(4), is 1.1 xx 10^(-12) and the K_(sp) of BaCiO_(4), is 1.2 xx 10^(-10),

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The ` Ag_2CrO_4` pecipitates first out of solution and then `BaCrO_4 ` preciptates.
The `BaCrO_4` pecipitates first out of solution and then ` Ag_2CrO_4`preciptates
Both `Ag_2CrO_4 and BaCrO_4`PRECIPITATE simultaneously out of solution
NEITHER `Ag_2CrO_4 " nor " BaCrO_4` precipitates

Solution :` [CrO_4^(2-)]_(Ag^(+))=(Ksp)/( [Ag^(+) ]^(2)) =(1.1xx 10^(-12))/(0.2)=5.5 xx 10 ^(-12) `
` [CrO_4^(2-) ]_(BA^(+2) ) =(Ksp)/([Ba^(+2)])=(1.2xx 10^(-10))/( 0.2 0) = 6 xx 10^(-10) `
` therefore Ag_2 Cr O_4 `ppts first
13.

For a gaseous reaction : A(g)to3B(g)+C(g),DeltaH is positive and the reaction attains equilibrium at 1 bar total pressure and 400 K. Identify the incorrect statement(s) regarding the above reaction:

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On increase of TEMPERATURE,equilibrium will be shifted in FORWARD direction.
When inert gas is introduced into a rigid container containing above equilibria equilibrium shifts towards left.
`Delta_(400)^(circ)=0`,for the above reaction.
If volume of vessel containing the above equilibria is increased without change in temperature then partial PRESSURE kof B decreases as compared to original equilibrium partial pressure of B.

Answer :B::C
14.

For a gaseous phase reaction A_((g)) + 3 B_((g)) hArr 2 C_((g)) , initial pressure is 600 mm at 1 : 3 molar ratio of A and B . If the equilibrium pressure of A is 100 mm , that of B & C respectively are (in mm)

Answer»

300 , 100
150 , 200
150 , 300
300 , 200

Answer :A
15.

For a gaseous mixtue of 2.41g of helium and 2.79g of neon in an evacuated 1.04 dm^(3) container at 298 K Calculate the partial pressure of each gas and hence find the total pressure of the mixture.

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Answer :Mass of He = 2 . 41 g
No. of moles of He =
`("Mass")/("Molar Mass") = (2.41)/(4) = 0 . 6 0 25 `moles
Mass Of Ne= 2 . 79 g
No. of moles of Ne ` = ("Mass")/("Molar Mass") = (2.79)/(20)`
= 0 . 1395 moles
Volume of the Total no. of moles of the mixtureConysinrt V= 1 . 0 4 `dm^(3)`
TEMPERTURE T= 298 K
Pressure P` = (1)/(V) RT`
ACCORDING to ideal gas equation PV = nRT
` P = ( 0 7 4 2 0 XX 0 . 0 821 xx 298 )/( 1 . 0 4) = 17 . 45` atm
Partial pressure P = molefraction ` xx`Total pressure
` = (nA)/(nA + nB) xx P `
Partial Pressure of Helium ` = P_(He) = (0 . 6025)/(0.7420) xx 17 . 45 `
According to Dalton's law of partial pressure = 3 . 280 atm
`P = P_(1) + P_(2) + P_(3). . . . `
` P_("total") = P_(He) + P_(Ne) = 14. 169 + 3 . 280 = 17 . 449 ` atm
16.

For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is K_C is larger or smaller than K_P ?

Answer»

Solution :For a homogeneous reaction at equilibrium, number of MOLES of products `(n_p)` are GREATER than the number of moles of reactants `(n_R)`, then `Deltan_g = + ve`
`n_P gt n_R`
`:. Deltan_g = +ve `
If `Deltan_g` is + ve, `K_P` VALUE is greater then `K_C`
`K_P = K_C. (RT)^(+ve)`
`:.K_P gt K_C`
Example :`PCl_5 (g) HARR PCl_3(g) + Cl_2(g)`
` 2 - 1 = 1`
`:.K_P = K_C(RT)^1`
`K_P gt K_C`
17.

For a gaseous homogeneous reaction at equilibrium number of moles of products are greater than the number of moles of reactants. Is K_c. is larger or smaller than K_p

Answer»

SOLUTION :For a homogenous reaction at EQUILIBRIUM , NUMBER of MOLES of proudcts `(n_p)` are greater than the number of moles of reaction `(n_g) `, then `Deltan_g = + ve`
18.

For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is K_(C) is larger or smaller than K_(P')

Answer»

Solution :`PCl_(5)(G)hArrPCl_(3)(g)+Cl_(2(g))`
`Deltan_(g)`= No. moles of product - No. of moles of reactnt
`Deltan_(g)=2-1=1`
When `Deltan_(g)=+"ve"`
`K_(P)=K_(C)(RT)^(+"ve")`
`K_(P)gtK_(C')`
19.

For a gas having molar mass M, specific heat at constant pressure can be given as:

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`(GAMMA R)/(M (gamma-1)) `
`(gamma)/(RM)`
`(M)/(R(gamma-1))`
`(gamma RM)/(gamma+1 )`

ANSWER :A
20.

For A _((g)) + B _((s)) to 2 C _((g)) , Delta H and Delta S respectively are 50 KJ and 100 J/K respectively. Then at 228 ^(@)C

Answer»

`DELTA G =02`
`Delta G GT 0`
`Delta G =2`
`Delta G lt 0`

ANSWER :D
21.

For a Friedel - Craft reaction using AlCl_3which compound can be used as solvent, benzene or nitrobenzene ?

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NITROBENZENE but not BENZENE
benzene but not nitrobenzene
both benzene and nitrobenzene
neither benzene nor nitrobenene

Answer :A
22.

For a fixed mass of an ideal gas the correct representation is:

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SOLUTION :Plot of V versus T is a straight line and the slope of the isoba decreases with INCREASE of pressure.
23.

For a equilibrium mixture in a closed vessel. N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)) the value of equilibrium on constant depends.

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Temperature
Total PRESSURE of the SYSTEM
INITIAL CONCENTRATION of `N_2` and `H_2`
Volume of REACTION vessel

Answer :A
24.

For a cyclic process, the change in internal energy of the system is ____

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SOLUTION :EQUAL to ZERO
25.

For a container contining A(g),B(g),C(g) "&" D(g) with rigid walls, an experiment is carried upon. This experiment involves increase in temperature of container in stepsof 1^(@)C and system is allowed to attain equilibrium, followed by calculation of K_(1) "&" K_(2) at each step, where K_(1) "&" K_(2) are equilibrium constants for reaction (1) & (2) respectively. A(g)+2B(g)hArrC(g)+D(g) ....(1) C(g)+D(g)hArrA(g)+2B(g) .....(2) Select the graph showing correct relationship-

Answer»




SOLUTION :N//A
26.

For a conjugate acid -base pair , K_a and K_b are related as

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`K_a.K_b =1`
` K_a.K_b= K_W`
` K_a.K_b=14`
` K_a. K_b= 7`

SOLUTION :` K_b " of "A^(-)=K_h " of "A ^(-) `
`K_h =(K_W)/( K_a)=K_b RARR K_a.K_b =KW `
27.

For a concentrated solution ofa weak electrolyte, A_(x) B_(y) of concentration 'C', the degree of dissociation alpha is given by

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`alpha=sqrt(K_(eq)//C (x+y))`
`alpha = sqrt(K_(eq)C//(XY))`
`alpha = (K_(eq)//C^(x+y-1)x^(x)y^(y))^(1//(x+y))`

Solution :`{:(,A_(x)B_(y),hArrxA^(y+),+YB^(x-)),("INITIAL",c,""0," "0),("At eqm.",c"("1-alpha")",xcalpha,ycalpha):}`
`K_(eq)=((xcalpha)^(x)(ycalpha)^(y))/(c(1-alpha))~~((xcalpha)^(x)(ycalpha)^(y))/(c)`
`""("Taking"1-alpha~~1)`
`or" "K_(eq)=(x^(x)y^(y)c^(x+y)alpha^(x+y))/(c)=x^(x)y^(y)c^(x+y-1)alpha^(x+y)`
`:.alpha=(K_(eq)//c^(x+y-1)x^(x)y^(y))^(1//(x+y))`
28.

For a compound LiAl(SiO_(3))_(2),What is thecharge on the SiO_(3) unit. What is the arrangement of oxygenatoms aroundthe siliconatom ?

Answer»

Solution :Since lithiumhas one unit `+ve`chargeand aluminium has THREE units+ve charge, therefore, TOTAL+ve chargeon cations is 4. Sincethe mineral`LiAl(SiO_(3))_(2)` is electricity neutral, therefore, total -ve chargeon TWO `SiO_(3)` unitsis `-4`or each `SiO_(3)` unit mustcarry two units-ve chargeand must berepresentedas `SiO_(3)^(2-)`.
We know that all silicates consist of `SiO_(4)^(4-)`tetrahedra.
But from theformula of thegiven silicate, it appears that there areonly 3oxygenper SILICON atom. This canbe explainedif two of theoxygenatoms ofeach `SiO_(4)^(4-)`unitshare with each othergiving a simple chainstructure for the silicate, `LiAl(SiO_(3))_(2)` as shown below:
29.

For a compound to be aromatic it must have (4n + 2) pi electrons.

Answer»


ANSWER :T
30.

For a closed system consisting of a reaction N_(2) O_(4(g)) rarr 2NO_(2(g)), the pressure "_____________".

Answer»

REMAINS constant
decreases
increases
becomes zero.

Answer :C
31.

For a chemical reaction the values of DeltaH and DeltaS at 300 K are - 10 kJ mol^(-1)and-203 K^(-1)mol^(-1) respectively. What is the value of DeltaG of the reaction? Calculate the DeltaG of a ran. at 600K assuming DeltaH and DeltaS values are constantDeltaHand DeltaS values are constant. Predict the nature of the reaction.

Answer»

Solution :Given:
`DeltaH = -10 kJ mol^(-1) = -10000 J mol^(-1)`
`DeltaS = - 20 JK^(-1)mol^(-1)`
`T = 300 K`
`DeltaG = ?`
`DeltaG = DeltaH-TDeltaS`
`DeltaG = - 10 kJ mol^(-1)- 300 K xx (-20xx10^(-3) ) kJ K^(-1)mol^(-1)`
`DeltaG = (-10+6) kJ mol^(-1)`
`DeltaG = (- 4 kJ mol^(-1)`
At 600 K
`DeltaG = -10 kJ mol^(-1)- 600 K xx (-20 xx10^(-3) ) KJ K^(-1)mol^(-1)`
`DeltaG = (-10 + 12) kJ mol^(-1)`
`DeltaG = + 2 kJ mol^(-1)`
The value of `DeltaG` is NEGATIVE at 300K and the reaction is spontaneous, but at 600K the value `DeltaG` becomes positive and the reaction is non-spontaneous.
32.

For a chemical reaction the values of DeltaH and DeltaS at 300 K are - 10 kJ "mol"^(-1) and -20 J "deg"^(-1) "mol"^(-1) respectively. What is the value of DeltaG of the reaction? Calculate the DeltaG of a reaction at 600K assuming DeltaH and DeltaS values are constant. Predict the nature of the reaction.

Answer»

Solution :`DeltaH=-10 "kJ mol"^(-1) =-10000 "J mol"^(-1)`
`DeltaS=-20 J K^(-1) "mol"^(-1)`
T=300 K
`DELTAG`=?
`DeltaG=DeltaH-TDeltaS`
`DeltaG=-10 "kJ mol"^(-1) -300 K xx (-20 xx10^(-3))kJ K^(-1) "mol"^(-1)`
`DeltaG=(-10+6) "kJ mol"^(-1)`
`DeltaG=-4 "kJ mol"^(-1)`
At 600 K
`DeltaG=-10 "kJ mol"^(-1) -600 K xx(-20 xx10^(-3))kJ K^(-1) "mol"^(-1)`
`DeltaG=(-10+12) "kJ mol"^(-1)`
`DeltaG=+2 kJ mol^(-1)`
The VALUE of `DeltaG` is NEGATIVE at 300K and the reaction is spontaneous, but at 600K the value `DeltaG` BECOMES positive and the reaction is non-spontaneous.
33.

For a certain reaction, Delta H^(0) " & " Delta S^(0) respectively are 400kJ & 200 J/mol/K. The process is non-spontaneous at

Answer»

2100 K
2010 K
1990 K
2020 K

Answer :C
34.

For a balanced reversible gaseous reaction : A(g)+ B(g) hArr 3C(g)+2D(g), which is non-spontaneous at low temperature? Identify the corrrect option(s). [Assume DeltaH^(circ)and DeltaS^(circ) of reaction to be independent of temperature.]

Answer»

It will be non-spontaneous even it higher temperatures.
If at equilibrium temperature INCREASES,concentration of 'C' and 'D' will ALSO INCREASE at new equilibrium.
If volume of the container is SUDDENLY increased at equilibrium WITHOUT changing temperature,then concentration of 'C' and 'D' also increases at new equilibrium.
If inert gas is added at constant volume at same temperature, the total pressure will increase.

Answer :B::D
35.

Show that A ∪B =A ∩ B impliesA = B

Answer»

`[-2.303 xx 2 xx 300]KJ `
`[2.303 xx 2 xx 10^(-3) xx 300]KJ`
`[-2.303 xx 2 xx 300] CAL`
`[2.303 xx 2 xx 10^(-3) xx 300] K.Cal`

ANSWER :C
36.

For , A + B hArr C. the equilibrium concentration of A and B at a temperäture are 15 mol lit^(-1) When volume is doubled the reaction has equilibrium concentration of A as 10 mol lit^(-1) then

Answer»

`K_(c) = 2` mole `lit^(-)`
concentration of C in original equilibrium is 45 M
`K_(c) = 0.2 mol^(-1)` litre
as increasing volume the reaction PROCEEDS in the backward direction

Solution :
Since, on increasing volume, pressure decreases and the reaction proceeds in the direction where it shows an increase in mole i.e backward reaction.
con. II EQULIBRIUM `[(15)/(2)+x][(15)/(2)+x][(a)/(2)-x]`
`:. (15)/(2) +x=10 :. x=(5)/(2)`
Now `Kc=([C])/([A][B])=([(a)/(2)-(5)/(2)])/([(15)/(2)+(15)/(2)][(15)/(2)+(15)/(2)])` ...(1)
`Kc=([C])/([A][B])=(a)/(15 xx 15)` ...(2)
Kc are same
`:. (((a-5))/(2))/(10 xx 10)=(a)/(15 xx 15) implies a=45M`
Now, `Kc=(45)/(15 xx 15)=0.2 "mole"^(-1)` lit
37.

For A + B hArr C + D,Delta H = -Q KJ , K_(f) & K_(b) respectively are 0.25 & 5000 at 400 K . Now ,K_(c) for same process at 500 K may be

Answer»

`4 XX 10^(-5)`
`4.5 xx 10^(-4)`
`5 xx 10^(-5)`
`5.2 xx 10^(-5)`

ANSWER :A
38.

For a 3s- orbital Psi (3s) = 1/(9sqrt(3)) ((1)/(a_0))^(3//2) (6 - 6sigma + sigma^2)e^(-sigma//2), where sigma = (2rZ)/(3a_0) . What is the maximum radial distance of node from nucleus?

Answer»

`((3+ SQRT(3))a_0)/(z)`
`(a_0)/(z)`
`3/2 ((3+SQRT3)a_0)/(z)`
`(2a_0)/(z)`

Solution :At radial NODE `varphi^2 =0 , sigma^2 - 6 sigma + 6 =0`
`sigma = 3 + sqrt3 , r =3/2 ((3+sqrt3)/(2))a_0`
39.

For 8g of helium gas (assume ideal behaviour), what is the slope of the straight line in PV VS T graph

Answer»

1 lt. ATM. `K^(-1)`
0.0821 lt. atm. `K^(-1)`
0.1642 lt. atm .`K^(-1)`
2 lt. atm . `K^(-1)`

Solution :`PV = nRT`.
PV = VS T slop = NR = `8/4 xx 0.0821`
`= 2 xx 0.0821 = 0.1642`.
40.

From quantization of angular momentum, one gets for hydrogen atom. the radius of the n^(th) orbit as r_(n) = (n^(2))/(m_(e)) ((h)/(2 pi))^(2) ((4 pi^(2) epsilon_(0))/(e^(2)) )for a hydrogen like atom of atomic number 'Z'.

Answer»


SOLUTION :`Psi^2 = 0 IMPLIES27 - (18 r)/(a_0) - (18)/(a_0) + (2r^2)/(a_0^2) = 0 , (d)/(1.73 a_0) = 3`
41.

For 2H_(2)O_(2)to2H_(2)O+O_(2) What is the oxidation number of Oxygen in (2)?

Answer»


ANSWER :-2
42.

For 2HIhArrH_(2)+I_(2), the equilibrium constant is K. What is the equilibrium constant for HIhArr1/2H_(2)+1/2I_(2) at the same temperature?

Answer»

SOLUTION :For `2HIhArrH_(2)+I_(2),""K=([H_(2)][I_(2)])/([HI]^(2))`
For `HIhArr1/2H_(2)+1/2I_(2)""K^(1)=([H_(2)]^(1//2)[I_(2)]^(1//2))/([HI])`
`thereforeK^(1)=sqrtK`
43.

For 2H_(2)O_(2)to2H_(2)O+O_(2) What type of Redox reaction is it?

Answer»

SOLUTION :DISPROPORTIONATION or DISPROPORTIONATE REACTION.
44.

For 1s orbital Hydrogen atom radial wave function is given as: R(r) =1/(sqrtpi) ((1)/(a_0))^(3//2) e^(-r//a_0) (where a_0 = 0.529 A^@) .The ratio of radial probability density of finding electron at r = a_0 to the radial probability density of finding electron at the nucleus is given as (x.e^(-y)). Calculate the value of (x + y).

Answer»


Solution :`[(R(r)]^2 r=0)/([R(r)]^2 "atnucles")= ([1/(sqrt(PI))(1/(a_0))^(3/2) ][e^((r)/(a_0))]^2)/([1/(sqrtpi) (1/(a_0))^(3/2)]^2 [e^((r)/(a_0))]^2 )= e^(-2)`
r=0`therefore x + y =3`
45.

For 2A + 2B hArr2C + 2D K_(c) = (1)/(16) then K_(c)for C + DhArr A + B is _______

Answer»


SOLUTION :`K_(c)=(1)/(sqrt((1)/(16)))=4`
46.

For 11.2L of an ideal gas at STP the value of PV is

Answer»

0.25 RT 
RT 
0.5 RT 
0.75 RT 

ANSWER :C
47.

For 1-methoxy-1,3-butadiene, which of the following resonating structure is the least stable?

Answer»

`H_(2)OVERSET(Ө)(C)-overset(OPLUS)(C)H-CH=CH-O-CH_(3)`
`H_(2)overset(oplus)(C)-CH=CH-CH=overset(Ө)(O)-CH_(3)`
`H_(2)C=CH-overset(oplus)(C)H-overset(Ө)(C)H-O-CH_(3)`
`H_(2)C=CH-overset(Ө)(C)H-CH=overset(oplus)(O)-CH_(3)`

Answer :C
48.

For 1 - methoxy- 1,3- butadience ,which of the following resonating structure is the least stable ?

Answer»

`H_(2)overset(+) (C )-overset(+)CH-Ch=CH-O-CH_(3)`
`H_(2)overset(+)(C) -CH =CH-CH=overset(+)O -CH_(3)`
`H_(2)overset(+)C-CH-overset(+)CH-overset(+)CH-O-CH_(3)`
`H_(2)C=CH-overset(+)CH-CH=overset(+)(O)-CH_(3)`

Answer :C
49.

Following two equilibria are established seperately in 2 different containers of unequal volume. PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) COCl_(2)(g)hArrCO(g)+Cl_(2)(g) Now the containers are connected together by a thin tube of negligible volume. Select incorrect statements. (Assume T constant)

Answer»

Degree of dissociation of both `PCl_(5)(g)"&"COCl_(2)(g)` will decrease
Degree of dissociation of both `PCl_(5)(g)"&"COCl_(2)(g)` will increase
Degree of dissociation of `PCl_(5)(g)` may increase, decrease or REMAIN the same IRRESPECTIVE of effect on degree of dissociation of `COCl_(2)`.
Degree of dissociation of `PCl_(5)(g)` may increase, decrease or remain the same, but the effect WOULD be same as that on degree of dissociation of `COCl_(2)`.

Solution :N//A
50.

Following two equilibrium is simultaneously established in a container. PCI_5(g)hArrPCI_3(g)+CI_2(g) CO(g)+CI_2(g)hArrCOCI_2(g) If some Ni(s) is introduced in the container forming Ni(CO)_4(g) then at new equilibrium.

Answer»

`PCI_3` concentration will INCREASE.
`PCI_3` concentration will decrease.
`CI_2 concentration will REMAIN same.
`PCI_3` concentration will increase.

Answer :B