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 given reaction: N_(2)(g) + 3H_2(g) hArr 2NH_3(g) Equilibrium constant K_c=[NH_3]^2/([N_2][H_2])^3 Some reactions are written below in Column-I and their equilibrium constants in terms of K_c are written in Column-II. Match the following reactions with the corresponding equilibrium constant

Answer»


Solution :For the reaction, `N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))`
Equilibrium `K_c=[NH_3]^2/([N_2][H_2]^3)`
(A) The GIVEN reaction `[2N_(2(g)) + 6H_(2(g)) hArr 4NH_(3(g))]` is twice the above reaction. Hence `K=K_c^2`
(B) The reaction `[2NH_(3(g)) hArr N_(2(g)) + 3H_(2(g))]` is REVERSE of the above reaction.
hence, `K=1/K_c`
(C) The reaction `[1/2N_(2(g)) +3/2 H_(2(g)) hArr NH_(3(g))]` is half of the above reaction
Hence, `K=sqrtK_c=K_c^(1/2)`
Discussion
2.

For the given reaction CH_(4(g))+2O_(2(g)) rarr CO_(2(g)) + 2H_(2)O_((l)) If 64 g of O_(2)is used, then 44 g ofCO_(2) is formed. 8 g of CH_(4)reacts to form 36 g of product. 22 g of CO_(2)is formed from 3.011xx10^(23)molecules of CH_(4). At STP, if 22.4 litres of O_(2(g)) is used, then 11.2 litres ofCO_(2) is formed. Which of the above statements are correct ?

Answer»

`2, 3, 4`
`1, 2, 3`
`1, 3, 4`
1, 3

Solution :(C) 1, 3, 4
(1) Reaction : `CH_(4(g)) + 2O_(2(g)) rarr CO_(2(g)) +2H_(2)O_((l))`
`64 g O_(2)` produces 44 gram `CO_(2)`
`:.` 64 gram produces
`:.` (1) is CORRECT
(2) Reaction : `CH_(4(g))+2O_(2(g)) rarr CO_(2(g))+2H_(2)O_((l))`
`:.` 16 gram `CH_(4)` react to form `(44+36)`
`=80` gram product
`(80xx8)/(16)` gram product `=40` gram product
`:.` (2) is not correct
(3)Reaction : `CH_(4(g))+2O_(2(g)) rarr CO_(2(g)) + 2H_(2)O_((l))`
`:.` 1 mole `CH_(4)` form 1 mole `CO_(2)` gas
`:.6.022 xx 10^(23)` molecules of `CH_(4)` form 44 g `CO_(2)`gas and
`:.3.011xx10^(23)` molecules of `CH_(4)` form
`=(3.011xx10^(23)xx44)/(6.022xx10^(23))=22g` of `CO_(2)`
(3) is correct
(4) Reaction : `CH_(4(g))+2O_(2(g)) rarr CO_(2(g))+2H_(2)O_((l))`
`:.` At STP USED `2 (22.4L) O_(2)` form `22.4 L CO_(2)`
`:.` At STP used `22.4L O_(2)` form
`=(22.4Lxx22.4L)/(2xx22.4 L) CO_(2) = 11.2 ` of `CO_(2)`
(4) is correct
Thus, (1) , (3) and (4) are correct statements
Discussion
3.

For the given reaction how many products are optically active (all isomers) : CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-CH_2-oversetoverset(CH_3)(|)CH-CH_3overset(Br_2//h upsilon)to

Answer»


ANSWER :D
Discussion
4.

For the given reaction 2A(g)+ B(g) hArrC(g) , DeltaH = x kJ which of the following favour the reactants ?

Answer»

LOW pressure
High pressure
Low temperature
Catalyst

Solution :Le-Chatelier.s PRINCIPLE
Discussion
5.

For the given of which molecule resonance is possible ? Why ? CO_(2), NO_(2) , O_(3), CH_(4)

Answer»

SOLUTION :For `CO_(2), NO_(2) , O_(3)` has RESONANCE as they have non BONDING electrons in atoms.
Discussion
6.

For the given equation in closed vessel. NH_4HS_((s)) overset"dissociation"undersetlarrto NH_(3(g)) +H_2S_((g)) What will be the value of K_p ?(P=total pressure )

Answer»

`P^3/27`
`P/2`
`P^2/2`
`P^2/4`

ANSWER :D
Discussion
7.

For the given equation 4NH_(3(g)) + 50_(2(g)) hArr 4NO_((g)) + 6H_(2) O_((l)) Select the correct option for change in Enthalpy.

Answer»

`Delta U + RT`
`Delta U + 5 RT`
`Delta U - RT`
`Delta U - 5 RT`

SOLUTION :`Delta H = Delta U + Deltan_((G)) RT`
`Deltan_((g)) = 4-9 = (-5) therefore Delta H = Delta U - 5 RT`
Discussion
8.

For the give reaction how many products will obtain (all isomers)?

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1
6
4
3

Solution :SIX ISOMERS
Discussion
9.

For the gaseous reaction involving the complete combustion of isobutane

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`DELTAH = DeltaE`
`DeltaH GT DeltaE`
`DeltaH = DeltaE = 0`
`DeltaH LT DeltaE`

ANSWER :D
Discussion
10.

For the gaseous reaction, at 300 K temperature, value of Delta H - Delta U= - 4.98 KJ., Then Delta n_((g))= …..

Answer»

1
2
`-2`
0

Answer :C
Discussion
11.

In a ten litre vessel, the total pressure of a gaseous mixture containing H_(2), N_(2)and CO_(2)is 9.8atm. The partial pressures of H_(2)and N_(2)are 3.7 and 4.2 atm respectively. Then the partial pressure of CO_(2)is

Answer»

50
5
`5.0 xx 10^(2)`
`5.0 xx 10^(-2)`

Solution :`P_(H_(2))+P_(N_(2))+P_(NH_(3))=2.8`
`implies P_(NH_(3))=1.6, K_(P)=(P_(NH_(3))^(2))/(P_(N_(2)).P_(H_(2))^(3))`
Discussion
12.

For the gaseous phase reaction 2A+B harr 2C+D, initially there are 2 mole each of A & B. If 0.4 mol of D is present at equilibrium at a given T & P, in-correct relationship is

Answer»

`P_A LT P_B & P_D lt P_C`
`P_A=P_C & P_B=P_D`
`P_C=2P_D & P_A=3P_B//4`
`P_A gt P_D & P_B gt P_C`

Solution :
but no. of moles of D at equlibrium (x)=0.4
`1.2""1.6""0.8""0.4`
partial pressure of a GAS `ALPHA` its no. of moles
Discussion
13.

For the reaction PCl_(5(g)) harr PCl_(3(g)) + Cl_(2(g)). Which of the graph "is"//"are" correct?

Answer»

`DELTA H lt 0 and Delta S lt 0`
`Delta H GT 0 and Delta S lt 0`
`Delta H = 0 and Delta S lt 0`
`Delta H gt 0 and Delta S gt 0`

ANSWER :D
Discussion
14.

For the gas phase reaction, PCI_(5(g)) hArr PCI_(3(g)) + CI_(2(g)) Which of the following conditions are correct ?

Answer»

`Delta H= 0 and Delta S lt 0`
`Delta H gt 0 and Delta S gt 0`
`Delta h lt 0 and Delta S lt 0`
`Delta H gt0 and Delta S lt`

Solution :For the reaction, `PCI_(5(g)) hArr PCI_(3(g))+ CI_(2(g))`
The reaction given is an example of decomposition reaction and we know that decomposition reaction are endothermic in nature, i.e. `Delta H gt0`.
Further, `Delta N = (1+1)-1=+1`
Hence, more number of molecules are PRESENT in PRODUCTS which shows more randomness i.e. `Delta S gt 0 (Delta S` is positive `)`
Discussion
15.

For the gas-phase reaction, 2NOhArr N_(2) + O_(2) : Delta H = -43.5keal. Which one of the following is false for the reactionN_(2) (g) + O_(2) (g) hArr 2NO(g)

Answer»

`K_c` is INDEPENDENT of temperature
`K_c` INCREASES as T decreases
`K_c` is decreases as T decreases
`K_c` varies with addition of `NO`

SOLUTION :`K_(C)` depends only on T & `K darr` with decrease in T for ENDOTHERMIC reactions
Discussion
16.

For the gas phase exothermic reaction. A_(2)+B_(2)hArrC_(2), carried out in a closed vessel, the equilibrium moles of a_(2) can be increased by:

Answer»

increasing the temperature
decreasing the pressure
adding INERT GAS at CONSTANT pressure
removing some `C_(2)`

Answer :A::B::C
Discussion
17.

For the four gases A,B,E and D the value of the excluded volume per mole is same. If the order of the critical temperature is T_(B)gtT_(C)gtT_(A)gtT_(E) then the order of their liquefaction pressure at a temperature T(TltT_(E)) will be:

Answer»

<P>`P_(A)ltP_(B)ltP_(E)ltP_(D)`
`P_(B)ltP_(D)ltP_(A)ltP_(E)`
`P_(E)ltP_(A)ltP_(D)lt_(B)`
`P_(D)ltP_(E)ltP_(A)ltP_(B)`

ANSWER :b
Discussion
18.

For the formation of two moles of SO_3(g) from SO_2 and O_2 the equilibrium constant is K_1. The equilibrium constant for the dissociation of one mole of SO_3 " into " SO_2 and O_2is ……………. .

Answer»

`1/K_1`
`K_1^2`
`(1/K_1)^(1/2)`
`K_1/2`

Solution :`2SO_2 + O_2 HARR 2SO_3`
`K_1 = ([SO_3]^2)/([SO_2]^2[O_2])`
DISSOCIATION of 1 mole of `SO_3`
`SO_3 hArr SO_2 + 1/2 O_2`
`K_2 =([SO_2][O_2]^(1/2))/([SO_3])`
`:. K_2 = 1/sqrt(K_1)`
Discussion
19.

For the formation of two moles of SO_(3)(g) from SO_(2) and O_(2), the equalibrium constant is K_(1). The equilibrium constant for the dissociation of one mole of SO_(3) into SO_(2) and O_(2) is

Answer»

`(1)/(K_(1))`
`K_(1)^(2)`
`((1)/(K_(1)))^(1//2)`
`(K_(1))/(2)`

ANSWER :C
Discussion
20.

For the formation of two different organic compounds the only correct combination is

Answer»

<P>(I),(II),(R)
(IV),(i),(P)
(II),(iii),(Q)
(IV),(ii),(R)

Answer :B
Discussion
21.

For the formation of silver mirror the only correct combination is :

Answer»

(IV), (II) ,(R)
(II),(ii),(R)
(III),(i),(R)
(I),(ii),(R)

ANSWER :D
Discussion
22.

The value of Delta_fG^(Theta)for formation of Cr_2 O_3is – 540 kJ "mol"^(-1)and that of Al_2 O_3is – 827 kJ "mol"^(-1) . Is the reduction of Cr_2 O_3possible with Al ?

Answer»

REDUCTION of `Cr_2O_3` by AL will TAKE place 
Oxidation of `Cr_2O_3` by Al will take place 
NEITHER oxidation nor reduction take place 
REACTION is not feasible

Answer :A
Discussion
23.

For the formation of an ionic bond between two atoms, one atom should have ……………..And the other atom should have …………

Answer»

SOLUTION :IONIZATION ENTHALPY , high eletron gain enthalpy
Discussion
24.

For the following three reactions (i), (ii) and (iii) equilibrium constants are given (i)CO_((g)) + H_2O_((g)) hArr CO_(2(g)) + H_(2(g)) , K_1 (ii)CO_(4(g)) + H_2O_((g)) hArr CO_((g)) + 3H_(2(g)) , K_2 (iii)CO_(4(g)) +2H_2O_((g))hArr CO_(2(g)) + 4H_(2(g)) , K_3 Which of the following relation is correct ?

Answer»

`K_3. K_2^3 = K_1^2`
`K_1sqrtK_2=K_3`
`K_2K_3=K_1`
`K_3=K_1K_2`

Solution :`CO_((G))+H_2O_((g))"" CO_(2(g)) + H_(2(g))`
`K_1=([CO_2][H_2])/([CO][H_2O])`….(1)
`CH_(4(g)) +H2O_((g)) hArr CO_((g)) + 3H_(2(g))`
Discussion
25.

For the following gases equilibrium N_(2) O_(4(g)) hArr 2NO_(2(g)), K_(p)is found to be equal to K_(c)This is attained when temperature is

Answer»

EQUILIBRIUM is possible only in a closed system at a given temperature
Both the opposing PROCESSES occur at the same rate and there is dynamic but stable condition
It CANBE represented as
(X) represents point of dynamic equilibrium

Solution : at x, equilibrium is get to be established
Discussion
26.

For the following three reactions a, b and c equilibrium constant are given a) CO_((g))+H_2O_((g)) harr CO_(2(g))+H_(2(g)) , K_1 b) CH_(4(g))+H_2O_((g)) harr CO_((g))+3H_(2(g)) , K_2 c) CH_(4(g))+2H_2O_((g)) harr CO_(2(g))+4H_(2(g)) , K_3 Which of the following relations is correct?

Answer»

`K_2K_3=K_1`
`K_3=K_1K_2`
`K_3.K_2^3=K_1^2`
`K_1sqrt(K_2)=K_3`

Solution :EQUATION `a+b HARR C, K_(1).K_(2)=K_(3)`
Discussion
27.

For the following reactions the major products are shown: H_(2)C=CH-CH=CH_(2) overset(HBr)underset(0^(@)C) to H_(2)C =CH-underset(Br)underset(|)(CH)-CH_(3)-overset(+25^(@)C) to underset(Br)underset(|)(CH_(2))CH=CHCH_(3) These provide an example of overset(1)(_) control at low temperature and overset(2)(_) control at higher temperature.

Answer»

`{:(1,2),("KINETIC","thermodynamic"):}`
`{:(1,2),("thermodynamic","kinetic"):}`
`{:("kinetic","kinetic"):}`
`{:("thermodynamic","thermodynamic"):}`

SOLUTION :Low TEMPERATURE kinetic control HIGH temperature thermodynamic control.
Discussion
28.

For the following sequential reaction, A overset(K_(1))(rarr)B overset(K_(2))(rarr)C Find out the concentraion of C at tiem t = 1day. Given that K_(1) = 1.8xx10^(-5) s^(-1) and K_(2) = 1.1xx10^(-2) s^(-1)and initial molar concentration of A is 1.8

Answer»


ANSWER :`1.42M`
Discussion
29.

For the following reaction NH_(4)HS_((s)) harr NH_(3(g))+H_(2)S_((g)) the total pressure at equilibrium is 30 atm. The value of K_(P) is

Answer»

15 `ATM^(2)`
225 `atm^(2)`
30 `atm^(2)`
15 atm

Solution :
Discussion
30.

For the following reaction N_(2)+3H_(2)to2NH_(3) equivalent mass of N_(2)=("molar mass of" N_(2))/X What is the value of x.

Answer»


Solution :`OVERSET((0))(N_(2))+3H_(2)rarroverset(-3)(2NH_(3))impliesx=6`
Discussion
31.

For the following reaction N_(2)+3H_(2) rarr 2HN_(3) Identify the compostions which will produce same amount of NH_(3)

Answer»

140 gm `N_(2)` and 35 g `H_(2)`
18 g `H_(2)` and 52 g `N_(2)`
TOTAL 20 moles of mixture having `N_(2)` and `H_(2)` present in stoichiometric ratio (No limiting reagent)
136 gm of mixturehaving mass FRICTION of `H_(2)` =6

Answer :A::C
Discussion
32.

For the following reaction if equal mass of A and B are taken : A+2Bto C Which of the following is/are correct ? (M_A and M_B are molar mass of A and B respectively )

Answer»

If `M_A=2M_B` , then NONE of the REACTANT will be left.
if `M_B LT M_A/2`, then A will be LIMITING reagent
If `M_A=M_B` , then A will be limiting reagent
If `M_B lt M_A/2` , then A will be limiting reagent.

Answer :A::B
Discussion
33.

For the following reaction H_((aq))^(+) + OH_((aq))^(-) rarr H_(2)O_((l)) , Delta H = -Q, where Delta H represents

Answer»

HEAT of formation
Heat of COMBUSTION
Heat of neutralisation
Heat of dilution

ANSWER :C
Discussion
34.

For the following reaction, formation of the prodcuts is favoured by : Ag(g) +4B_(2)(g) hArr2AB_(4)(g)DeltaH lt 0

Answer»

low TEMPERATURE and HIGH PRESSURE
high temperature and low pressre
low temperature and low pressure
high temperature and high pressure

Answer :A
Discussion
35.

For the following question , enter the correct numerical value, ( in decimal - notation , truncated / rounded - off to the second decimal place , e.g., 6*25, 7*00 -0*33, 30*27, -127*30) using the mouse and theonscreenvirtual nemeric keypad in the placedesigned to enter the answer. The approach of the following equilibrium was observed kinetically from both direction : PtCl_(4)^(2-) + H_(2)O hArr Pt (H_(2)O) hArr Pt (H_(2)O) Cl_(3)^(-) + Cl^(-) At25^(@)C , it was found that(d[PtCl_(4)^(-)])/dt=(3*90 xx10^(-5)s^(-1) ) [PtCl_(4)^(2-)] - (2*1 xx 10^(-3) mol^(-1)s^(-1)) [Pt(H_(2)O ) Cl_(3)^(-)] The value of the equilibrium constant when fourth Cl^(-) ion is complexed , i.e., for the step involving backward reaction will be ..........

Answer»


Solution :Forforward reaction , ` ((dx)/(dt))_(f) = k_(f) [PtCl_(4)^(2-)]`
For backward reaction,
`((dx)/(dt))_(B) = k_(b) [ Pt (H_(2)O)Cl_(3)^(-)][Cl^(-)]`
NET ` (dx)/(dt) = k_(f) [ Pt Cl_(4)^(2-)] -k_(b) [Pt (H_(2)O)Cl_(3)^(-)][Cl^(-)]`
Comparingwith the given expression
` k_(f) = 3*90 xx 10^(-5) s^(-1)`
and `k_(b) = 2*1 xx 10^(-3) L " mol "^(-1)s^(-1)`
` :. K_("forward ") =1/K_("forward") = 1/(1*86 xx 10^(-2))=53*85`
Discussion
36.

For thefollowingquestionsenterthe correctnumericalvalue(in decimal -notation , truncated // rounded - offto the seconddecimalplace e.g.,6.25 , 7.00 -0.33,- 30,30.27 , - 127 .30 ) usingthe mouseand theon- screen virtualnumerickeypadin theplacedesignatedto enterthe answer. The amountof energyreleasedwhen 1 xx 10^(10) atoms ofchlorinein vapour stateare convertedto C1^(-) ionsaccordingto theequation, C1 (g) + e^(-)to C1^(-)( g) " is" 57.86 xx10^(-1) J Calculate theelectron gainenthalpyof chlorineatom intermsof eV peratom.

Answer»


Solution :The AMOUNTOF energyreleasedwhen `1xx 10^(10)` atoms of chlorinein vapourstateareconvertedinto `C1^(-)`ionsaccordingto theequation.
`C1 (g) E^(-) toC1^(-)( g)" is"57. 86 xx10^(-1) J`
`:. ` The electrongain enthalpyof chlorinei.e.,the amountof energyreleasedwhen 1mole`(6.023 XX 10^(23))` atomsof chlorineare convertedinto `C1^(-)` ionsaccordingto theaboveequation willbe
`= (57. 86 xx 10^(-10))/(1 xx 10^(10))xx 6. 023xx 10^(23)`
`=- 348 .49 xx 10^(3)j// mol = - 348 .49 kJ// mol`
Now1 EV `//` atom = 96 .49 kJ `mol^(-1)`
`:. ` Electron gainenthalpyof chlorine`= - (348 .49)/( 96. 49) `
`=- 3.61eV // atom`
Discussion
37.

For the following question, enter the correct numerical value, ( in decimal - notation , truncated //rounded - offto the second decimal place, e.g., 6.50, 7.00, - 0.33, 30.27, - 127.30 )using the mouse and the onscreen virtual numeric keypadin the place designated to enter the answer.

Answer»


Solution :The surface of copper gets TARNISHED by the formation of copper oxide. `N_(2)` gas was passed to prevent the oxide formation during heating of copper at 1250 K. However ,the `N_(2)`gas contains 1 mole `%` of water vapour as impurity . The water vapour oxidize copper as per the reaction given below `:`
`2Cu(s) + H_(2)O(g) rarrCu_(2)O(s) + H_(2)(g)`
`p_(H_(2))` is the minimumpartal pressure of `H_(2)` ( in bar) NEEDED to prevent the oxidation at 1250 K. The value of In `(p_(H_(2))` is `"................"`
( Given `:` TOTAL pressure `= 1 ` bar, R ( universal gas constant `) = 8 JK^(-1) , mol^(-1)`
In `( 100 = 2.3, Cu(s)` and `Cu_(2)O(s)` are MUTUALLY immiscible
At `1250 K: 2 Cu(s) + (1)/(2) O_(2)(g) rarr Cu_(2)O(s), DeltaG^(@) = - 78000 J mol^(-1)`
`H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)O(g) , DeltaG^(@) = - 178000 J mol^(-1)`
G is the Gibb's energy )
Discussion
38.

For the following isomerisation reaction cis-butene-2hArrtrans-butene-2 K_(p)=732 Which of the following statement is true at point A?

Answer»

<P>`QGT(p)`
`QltK_(p)`
`Q=K=1`
`Q=K=1.732`

Solution :AT point `A, Q=tan60^(@)C`
`Q=1.732`
`THEREFORE Q=K=1.732`
Discussion
39.

For the following isomerisation reaction : "Cyclohexane" harr "hexene-1", K = 1.732 Which of the following statements holds good at point "A"?

Answer»

`Q GT K`
slope = `tan 60^(@) = K_(P)` c) `P_("(hexene-1)") = 1.732 xx P_("(Cyclohexane)")`
Q = K = 1.732
Q=K=1.732

Solution :at A, slope = `tan 60^(@)=1.732=K_(P)`
i.e., `(P_("hexene")-1)/(P_("cyclohexane"))=1.732`
Discussion
40.

For the following gaseous phase eqilibrium PCl_(5(s)) hArrPCg_(3(g))+Cl_(2(g))Kp is found to be equal to K_(x) K_(x) is equilibrium constant when concentratiion are taken in interms of mole fraction). This attained when

Answer»

1 ATM
0.5 atm
2 atm
4 atm

Solution :`K_(P)=K_(X)(P)^(Delta n_((G)))`
Discussion
41.

For the following equilibrium,K_c=6.3xx10^14 at 1000 K NO(g)+O_3(g)iffNO_2(g)+O_2(g) What is K_c for the reverse reactions?

Answer»

SOLUTION :`K_c` for the REVERSE REACTION =`1/K_c=1/(6.3xx10^4)=0.159xx10^-14=1.59xx10^-15`
Discussion
42.

For the following equilibrium N_(2)O_(4) hArr 2NO_(2) K_(c)=0.67. If we start with 3 moles of NO_(2) and 1 mole of N_(2)O in1L flask, then NO_(2)present at equilibrium is :

Answer»

1.5 mol
2.0 mol
0.5 mol
1.0 mol

Solution :`{:(,N_(2)O_(4),hArr,2NO_(2),),("Initial",1" mol",,2" mol",),("At eqm.",m(1-x)"mol",,(3+2x)"mol",):}`
`K_(c)=((3+2x)^(2))/(1-x)=0.67`
`(3+2x)^(2)=0.67(1-x)`
`4x^(2)+9+12x=0.67-0.67x`
`4x^(2)+12.67x+8.33=0`
`x=(-12.67+-sqrt((12.67)^(2)-16xx8.33))/(8)`
`=(-12.67+-sqrt(160.53-133.28))/(8)`
`=(-12.67+-sqrt(27.25))/(8)`
`=(-12.67+-5.22)/(8)`
Taking `+ve` SIGN
`x=(-17.89)/(8)=-2.24`
Taking `+ve` sign
`x=-0.93`
As `x CANCEL(=)-2.24"":.x=-0.93`
`:. `Moles of `N_(2)O_(4)` at eqm `=1-x`
`=1+0.93=2("Approx.")`
Discussion
43.

For the following equilibrium , K_(c)= 6.3 xx 10^(14) at 1000 K NO(g)+O_(3)(g) hArr NO_(2)(g) + O_(2)(g) Both the forward and reverse reactions in the equilbrium are elementary bimolecular reactions what is K_C for the reverse reaction?

Answer»

Solution :For the REVERSE reaction `K_(C)= (1)/(K_C)= (1)/(6.3xx 10^(14)) = 1.59 xx 10^(-15)`
Discussion
44.

For the following equilibrium , K_(c) = 6*3 xx 10^(14)" at "1000 K NO(g) +O_(3) (g) hArr NO_(2) (g) + O_(2) (g) Both the forward and reverse reactions in the equilibrium in the equilibrium are elementry bimolecular reactions. What is K_(c) for the reverse reaction ?

Answer»

SOLUTION :For the REVERSE REACTION , ` K' _(c) = 1/K_(c) = 1/(6*3 XX 10^(14) )= 1*59 xx 10^(-15) `
Discussion
45.

For the following equilibrium, K_c = 6.3 xx 10^14 at1000 K. NO_((g)) + O_(3(g)) = NO_(2(g)) + O_(2(g)) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c for the reverse reaction ?

Answer»

SOLUTION :(`K_c` of reverse REACTION of any reaction )= `1/(K_c "of forward reaction")`
`THEREFORE K._c=1/(6.3xx10^14)=1.587xx10^(-15)`
Discussion
46.

For the following equilibrium, H_(2) O(l) hArr H_(2) O(g)the increase in the pressure causes

Answer»

FORMATION of more `H_(2)O_((l))`
formation of more `H_(2)O_((G))`
increase in the BOILING point of `H_(2)O_((l))`
DECREASE in boiling point of `H_(2)O_((l))`

Solution :Le-Chatelier.s principle
Discussion
47.

for the following conversionthe product formed is

Answer»




SOLUTION :
Discussion
48.

For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond is indicated : (a)C_4H_8 (one double bond ) (b) C_5H_8 (one triple bond )

Answer»

SOLUTION :(a)Isomers of `C_4H_8` having ONE double bond are :
(i)`underset"But-1-ene"(overset4CH_3overset3CH_2-overset2CH=overset1CH_2)`

(iv)`underset"2-Methylprop-1-ene"(CH_3-oversetoverset(CH_3)|C=CH_2)`
(B)(i)`underset"Pent-1-yne"(CH_3CH_2CH_2overset2C-=overset1CH)` , (ii)`underset"Pent-2-yne"(CH_3CH_2-C-=overset2C-overset1CH_3)` , (iii)`underset"3-Methylbut-1-yne"(CH_3overset3- oversetoverset(CH_3)|(CH)-overset2C-=overset1CH)`
Discussion
49.

For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated: (a)C_(4)H_(8)(one double bond) (b) C_(5)H_(8) (one triple bond)

Answer»

SOLUTION :
Discussion
50.

For the following bond cleavages, use curved arrows to show the electron flow and classify each as homolysis or heterolysis. Identify the reactive intermediates produced in gtC=O+hatOH rarr C=O+H_2O

Answer»


ANSWER :`(##ANE_PKE_CHE_0XI_C12_E02_032_S01##)`
Discussion