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.

The pressure-volume work for an ideal gas canbe calculated by using the expression W=-int_(V_(i))^(V_(f))P_(ex)dV. The work can also be calculated from the pV - plot by using the area under the curve within the specified limits. When an ideal gas is compressed (i) reversibly or (ii)irreversibly from volume V_(i) to V_(f). Choose the correct option.

Answer»

`W_("reversible")=W_("IRREVERSIBLE")`
`W_("reversible")ltW_("irreversible")`
`W_("reversible") gt W_("irreversible")`
`W_("reversible")=W_("irreversible")+P_(ex)DeltaV`

Solution :
In both the curves, work DONE on the gas is REPRESENTED by the shaded area and the area under the curve is alwaysmore in irreversible compression.
Hence, `W_("irreversible") gt W_("reversible")`
Discussion
2.

The price incurred in producing one kg of magnesium is Rs. 25. What is the price of one mole of metal?

Answer»


ANSWER :0.6 RUPEE
Discussion
3.

The pressure-volume work for an ideal gas can be calculated by using the expression w = -int_(V_(i))^(V_(f)) p_(exdV).The work can be calculated form the pV- plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume V_(i) to V_(f), choose the correct option.

Answer»

<P>w (REVERSIBLE) = w (IRREVERSIBLE)
w (reversible) `lt` w (irreversible)
w (reversible) `GT` w (irresversible)
w(reversible)= w(irreversible) + `p_(EX)DeltaV`

Solution :w (reversible) `lt` w(irreversible) because area under the curve in irreversible compression is more than the area under the curve for reversible compression.
Discussion
4.

The pressure-volume work for an ideal gas can be calculated by using the expression W= - int_(V_i)^( V_f) P_(ex) dV. The work can also be calculated from the pV- plot by using the area under the curve within the specifiedlimits. When an ideal gas is compressed (A) reversibly or (B) irreversibly from volume V_i to V_f. choose the correct option.

Answer»

<P>W (reversible) = W (irreversible)
W (reversible) lt W (irreversible)
W (reversible) gt W (irreversible)
W (reversible) = W (irreversible) `+ P_(ex) . Delta V`

Solution :W (reversible) < W (irreversible) is because area under the CURVE is always more in irreversible compression as can be seen from given figure.

pV-plot when pressure is not constant and CHANGES in finite steps during compression from INITIAL volume, `V_i` to find volume, `V_f`. WORK done on the gas is represented by the shaded area.

pV-plot when pressure is not constant and changes in infinite steps (reversible conditions) during compression from initial volume, `V_i` to final volume, `V_f`. Work done on the gas is represented by the shaded area.
Discussion
5.

The pressure of the atmosphere is 2xx 10^(-6) mm at about 100 mile from the earth and temperature is -180^(@)C. How many moles are three in 1 mL gas at this attitude?

Answer»

SOLUTION :`3.45 XX 10^(-13) MOL`
Discussion
6.

The pressure of an equilibrium mixture of the three gases NO, Cl_(2) and "NOCl" 2NO(g) + Cl_(2) (g ) hArr 2NOCl(g)is suddely decreased by doubling by doubling the volume of the container at constant temperature. When the systemreturns to equilibrium

Answer»

the concentration of `NOCI` will be increased
the VALUE of the EQUILIBRIUM CONSTANT `K_c` will be increased
the NUMBER of moles of `CI_2` will be increased
the number of moles of `NOCI` will be decreased

Solution :Le-chatelier.s principle
Discussion
7.

The pressure of a gas is due to ........... exerted by its molecules per ........... of the walls of the container.

Answer»

SOLUTION :FORCE, UNIT AREA
Discussion
8.

The pressure of a 1 : 4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmoshpere. What would be the partial pressure of dioxygen ?

Answer»

`0.8xx10^(5)` ATM
`0.008 Nm^(-2)`
`8xx10^(4)Nm^(-2)`
0.25 atm

SOLUTION :Moles RATIO of `H_(2)` and `O_(2)` is 1 : 4. Hence partial pressure of dioxygen is
`THEREFORE` Partial pressure of `O_(2)=` Mole fraction of `O_(2)xx` total pressure of mixture
`= (4)/(1+4)xx1=(4)/(5)xx1` atm
`= 0.8 atm = 0.8xx105 Nm^(-2)=8xx10^(4)Nm^(-2)`
Discussion
9.

The pressure of a 1:4 mixtureof dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen ?

Answer»

`0.8xx10^(5)` atm
`0.008 N m^(-2)`
`8xx10^(4) N m^(-2)`
0.25 atm

Solution :PARTIAL pressure of `O_(2)`=Mole fraction of `O_(2)`xxTotal pressure of MIXTURE
`=(4)/(5)xx1 "atm"=0.8xx10^(5)" N "m^(-2)=8xx10^(4)" N "m^(-2)`.
Discussion
10.

The pressure of 1 atmosphere is equal to

Answer»

760 cm
`10^(5) Nm^(-2)`
`10^(4) "dyne" cm^(-2)`
1 bar

Solution :(d) 1 ATMOSPHERE=`1.01xx10^(5) Nm^(2)`=1 bar
Discussion
11.

The pressure exerted by the gas is directly proportional to

Answer»

ROOT mean square SPEED 
AVERAGE speed 
Mean square speed 
All of these 

ANSWER :C
Discussion
12.

The pressure exerted by a mass of 'x' mg resting on the area of 1.00 cm^2 is 1.00 Pa, then 'x' is

Answer»

10.300 
10.308 
1030 
10.2 

ANSWER :D
Discussion
13.

The pressure exerted by a mass of 'x' mg resting on the area of 1.00 cm^2is 1.00 Pa, then 'x' is

Answer»

10.3
`10.308`
`1030`
`10.2`

ANSWER :D
Discussion
14.

The pressure exerted by 6.0 g of methane gas in a 0.03 m^(3) vessel at 129^(@)C is (Atomic masses : C=12.01, H=1.01 and R=8.314 JK^(-1)mol^(-1))

Answer»

215216 Pa
13409 Pa
41648 Pa
31684 Pa

Solution :MOLAR MASS of `CH_(4)=12.01+4xx1.01`
`=16.05" G "mol^(-1)`
`:. 6" g of "CH_(4)=(6)/(16.05)"mole"`,
`T= 129^(@)C=129+273" K"=402" K"`
PV=nRT
or `P=(nRT)/(V)=(6)/(16.05)xx(8.314xx402)/(0.03)=41648" Pa"`
Discussion
15.

The pressure exerted by 6.0 g of methane gas in a 0.03 m^(3) vessel at 129^(@)C is (Atomic masses : C = 12.01, H = 1.01 and R = 8.314 JK^(-1)mol^(-1))

Answer»

31684 PA
215216 Pa
13409 Pa
41648 Pa

Solution :`P=(NRT)/(V)`
`= (6)/(16.02)=(8.314xx402)/(0.03)=41648` Pa
Discussion
16.

The pressure exerted by 6.0 g fo methane gas in 0.03m^3 vessel at 129^@ C is (Atomic masses : C = 12.01, H = 1.01 and R = 8.314JK^(-1) mol^(-1))

Answer»

215216 Pa
13409Pa
41648 Pa
31684 Pa

Answer :C
Discussion
17.

The pressure exerted by 12 g of an ideal gas at temperature t^(@)C in a vessel of volume V litre is one atm. When the temperature is increased by 10 degree at the same voume, the pressure increases by 10%. Calculate the temperature t and volume V (Molecular weight of the gas=-120).

Answer»

Solution :Applying PV=nRT
In the 1 st CASE,`1xxV=(12)/(120)xxRxx(273+t)""`……….(i)
In the 2nd case,`1.1xxV=(12)/(120)xxRxx(283+t)""`………..(ii)
DIVIDING (i) by (ii), `(1)/(1.1)=(273+t)/(283+t)`
On SOLVING it gives `t=-173^(@)C`.
Putting in (i), `"" V=0.1xx0.082xx100=0.82" litre"`
Discussion
18.

The presence of which one of the following groups on benzene nucleus activates it towardselectrophilic substitution?

Answer»

`-CN`
`-CHO`
`-COOR`
`-O-COR`

SOLUTION :`-O-COR` group on benzene NUCLEUS activates it towards electrophilic substitution
Discussion
19.

The presence of -OH groups on adjacent carbon atoms can be detected by the reaction of the compound.

Answer»

CONC. `H_(2)SO_(4)`
Conc. `HNO_(3)`
`HIO_(4)`
None of these

Solution :The presence of `-OH` groups on adjacent carbon atoms can be detected by `HIO_(4)` or LEAD tetra-acetate.
Discussion
20.

The presence ofNH_(4)Cl " in " NH_(4)OH solution will ............. the degree of dissociation of NH_(4)OH.

Answer»


ANSWER :DECREASE
Discussion
21.

The presence of ionic salts in a liquid

Answer»

Decreases the viscosity of the liquid 
Increases the viscosity of the liquid 
Does not EFFECT the viscosity of the liquid 
NONE of the above is correct Intermolecular FORCES 

Answer :B
Discussion
22.

The presence of Hg^(2+) ions is detected as its sulphide in group IIA of qualitative analysis. The black HgS is one of the least soluble precipitate known (K_(sp) = 4 xx 10^(-54)). Aqua regia and sodium sulphide (2M) dissolve the black precipitate. potassium iodide when added slowly to the solution contaning Hg^(2+) ions, a red precipitate is formed which dissolves in excess KI to form a colourless soluble complex. when NH_(4)CI solution is added to the above colourless soluble complex containing some NaOH, a brown precipitate is formed. when to a warm alkaline aqueous solution of Hg(NO_(3))_(2) a solution a ammonium chloride is added, a white precipitate is obtained. Identify the incorrect statement with respect to Hg(II) salts.

Answer»

`Hg^(2+)` ions in the presence of DILUTE (M) hydrochloric acid gives initially a white PRECIPITATE of `Hg(II)` chloro sulphide with `H_(2)S` which then reacts with further quantity of `H_(2)S` to form black precipitate of `HgS`.
The black precipitate of `HgS` is insoluble in ammonium sulphide
`HgO.Hg(NH_(2))NO_(3)` precipitate sublimes at atmospheric PRESSURE on heating
`Hg^(2+)` ions react with dilute solution of `KCN` and form red precipitate

Solution :`HgSdarr + Na_(2)S rarr Na_(2)[HgS_(2)]` (soluble)
`K_(2)HgI_(4)+NH_(4)CI +KOH rarr HgO.Hg(NH_(2))I`
`Hg^(2+) +CO^(2+)+4SCN^(-) rarr underset(("Deep blue crystalline ppt"))(Co[Hg(SCN)_(4)]darr)`
`2HG^(2+) +NO_(3)^(-) +4NH_(3)+H_(2)O rarr HgO.underset(("White ppt"))(Hg(NH_(2))NO_(3)darr +3NH_(4)^(+)`
Discussion
23.

The presence of Hg^(2+) ions is detected as its sulphide in group IIA of qualitative analysis. The black HgS is one of the least soluble precipitate known (K_(sp) = 4 xx 10^(-54)). Aqua regia and sodium sulphide (2M) dissolve the black precipitate. potassium iodide when added slowly to the solution contaning Hg^(2+) ions, a red precipitate is formed which dissolves in excess KI to form a colourless soluble complex. when NH_(4)CI solution is added to the above colourless soluble complex containing some NaOH, a brown precipitate is formed. when to a warm alkaline aqueous solution of Hg(NO_(3))_(2) a solution a ammonium chloride is added, a white precipitate is obtained. The white precipitate formed is due to the formation of:

Answer»

mercury(II) oxide
mercury (I) amido nitrate
mercury (II) oxide and mercury (II) amido nitrate
none of the above

Solution :`HgSdarr + Na_(2)S rarr Na_(2)[HgS_(2)]` (SOLUBLE)
`K_(2)HgI_(4)+NH_(4)CI +KOH rarr HgO.Hg(NH_(2))I`
`Hg^(2+) +Co^(2+)+4SCN^(-) rarr underset(("Deep blue crystalline ppt"))(Co[Hg(SCN)_(4)]DARR)`
`2HG^(2+) +NO_(3)^(-) +4NH_(3)+H_(2)O rarr HgO.underset(("White ppt"))(Hg(NH_(2))NO_(3)darr +3NH_(4)^(+)`
Discussion
24.

The presence of Hg^(2+) ions is detected as its sulphide in group IIA of qualitative analysis. The black HgS is one of the least soluble precipitate known (K_(sp) = 4 xx 10^(-54)). Aqua regia and sodium sulphide (2M) dissolve the black precipitate. potassium iodide when added slowly to the solution contaning Hg^(2+) ions, a red precipitate is formed which dissolves in excess KI to form a colourless soluble complex. when NH_(4)CI solution is added to the above colourless soluble complex containing some NaOH, a brown precipitate is formed. when to a warm alkaline aqueous solution of Hg(NO_(3))_(2) a solution a ammonium chloride is added, a white precipitate is obtained. What product is formed when black precipitate dissolves in sodium sulphide (2M) solution?

Answer»

`NA[HgS_(2)]`
`Na_(2)[HgS_(2)]`
`Na_(3)[HgS_(2)]`
None

Solution :`HgSdarr + Na_(2)S RARR Na_(2)[HgS_(2)]` (soluble)
`K_(2)HgI_(4)+NH_(4)CI +KOH rarr HgO.Hg(NH_(2))I`
`Hg^(2+) +Co^(2+)+4SCN^(-) rarr underset(("Deep blue crystalline ppt"))(Co[Hg(SCN)_(4)]darr)`
`2HG^(2+) +NO_(3)^(-) +4NH_(3)+H_(2)O rarr HgO.underset(("White ppt"))(Hg(NH_(2))NO_(3)darr +3NH_(4)^(+)`
Discussion
25.

The presence of Hg^(2+) ions is detected as its sulphide in group IIA of qualitative analysis. The black HgS is one of the least soluble precipitate known (K_(sp) = 4 xx 10^(-54)). Aqua regia and sodium sulphide (2M) dissolve the black precipitate. potassium iodide when added slowly to the solution contaning Hg^(2+) ions, a red precipitate is formed which dissolves in excess KI to form a colourless soluble complex. when NH_(4)CI solution is added to the above colourless soluble complex containing some NaOH, a brown precipitate is formed. when to a warm alkaline aqueous solution of Hg(NO_(3))_(2) a solution a ammonium chloride is added, a white precipitate is obtained. The brown precipitate formed is due to the formation of:

Answer»

oxy-dimercuric ammonium iodide
potassium tetraiodomercurate (II)
ammonium tetraiodomercurate (II)
(a) and (b) both

Solution :`HgSdarr + Na_(2)S rarr Na_(2)[HgS_(2)]` (soluble)
`K_(2)HgI_(4)+NH_(4)CI +KOH rarr HgO.HG(NH_(2))I`
`Hg^(2+) +Co^(2+)+4SCN^(-) rarr UNDERSET(("Deep BLUE crystalline ppt"))(Co[Hg(SCN)_(4)]darr)`
`2HG^(2+) +NO_(3)^(-) +4NH_(3)+H_(2)O rarr HgO.underset(("White ppt"))(Hg(NH_(2))NO_(3)darr +3NH_(4)^(+)`
Discussion
26.

The presence of Hg^(2+) ions is detected as its sulphide in group IIA of qualitative analysis. The black HgS is one of the least soluble precipitate known (K_(sp) = 4 xx 10^(-54)). Aqua regia and sodium sulphide (2M) dissolve the black precipitate. potassium iodide when added slowly to the solution contaning Hg^(2+) ions, a red precipitate is formed which dissolves in excess KI to form a colourless soluble complex. when NH_(4)CI solution is added to the above colourless soluble complex containing some NaOH, a brown precipitate is formed. when to a warm alkaline aqueous solution of Hg(NO_(3))_(2) a solution a ammonium chloride is added, a white precipitate is obtained. Hg^(2+) ions give deep blue crystalline precipitate with:

Answer»

`KI`
`Co(SCN)_(2)`
`SnCI_(2)`
`NH_(3)`

Solution :`HgSdarr + Na_(2)S rarr Na_(2)[HgS_(2)]` (soluble)
`K_(2)HgI_(4)+NH_(4)CI +KOH rarr HgO.Hg(NH_(2))I`
`Hg^(2+) +Co^(2+)+4SCN^(-) rarr underset(("Deep blue crystalline ppt"))(Co[Hg(SCN)_(4)]darr)`
`2Hg^(2+) +NO_(3)^(-) +4NH_(3)+H_(2)O rarr HgO.underset(("White ppt"))(Hg(NH_(2))NO_(3)darr +3NH_(4)^(+)`
Discussion
27.

The presence of colloidal particles of dust in air imparts blue colour to the sky. This is due to

Answer»

ABSORPTION of light
reflection of light
refraction of light
scattering of light.

Answer :D
Discussion
28.

The presence of delocalized pi - electrons in benzene indicates that it is

Answer»

less stable than cyclohexatriene
more stable than cyclohexatriene
more basic than cyclohexatriene
Both (b) and (c)

Solution :Delocalization of p elecctrons in BENZENEIS responsiblefor stabilit . So , alternate single and double bonds containing compounds show RESONANCE effect , which GETS more stability
Discussion
29.

The presence of carbon in an organic compound can be shown by

Answer»

heating the compound wih sodium
heating the compound with CUPRIC oxide
heating the compound on bunsen flame
heating the compound with magnesium

Solution :Compound when heated with Cuo reduces CuO to Cu and oxidises C to `CO_(2)` which turns LIME water milky.
`C+2CuO to 2Cu+CO_(2)`
`CO_(2)+CA(OH)_(2)to underset(("Milky"))(CaCO_(3))+H_(2)O`
Discussion
30.

The prepare a pure sample of normal hexane using sodium metal as one reactant, the other reactant or reactants will be

Answer»

Isobutyl CHLORIDE and ethyl chloride
METHYL BROMIDE and n-pentyl bromide
n-Propyl bromide
Ethyl bromide and n-Butyl bromide

Answer :C
Discussion
31.

The prefix femto stands for

Answer»

`10^(9)`
`10^(-12)`
`10^(-15)`
`10^(5)`

ANSWER :C
Discussion
32.

The prefix and the suffix for CHO group are..............and.............respectively when the carbon atom of the CHO group is not included in the parent chain.

Answer»


ANSWER :FORMYL, carbaldehyde
Discussion
33.

The preferred sites of protonation in the following compounds are (I) Ph-CH_(2)-overset(1)overset(NH)overset(||)C-overset(2)overset(NH_(2))""(II)Ph-CH_(2)-overset(3)overset(O)overset(||)C-overset(4)(N)H_(2)

Answer»

1 and 3
2 and 4
1 and 4
2 and 3

Solution :PROTONATION at site 1 and 3 is SUPPORTED by RESONANCE stabilization.
Discussion
34.

The preferred sites of protonation in the following compounds are

Answer»

1 and 3
2 and 4
1 and 4
2 and 3

Answer :A
Discussion
35.

The predominant yellow line in the spectrum of a sodium vapour lamp has a wavelength of 590 nm. What minimum accelerating potential is needed to excite this line in an electron tube having sodium vapours?

Answer»


ANSWER :2.11 VOLT
Discussion
36.

The predominant form of histamine present in human blood is (pK_(a) , Histidine=6.0)

Answer»




ANSWER :B
Discussion
37.

The precipitate of CaF_(2) (K_(sp) = 1.7 xx 10^(-10)) is obtained when equal volumes of the following are mixed ?

Answer»

`10^(-4) M Ca^(2+) + 10^(-4) M F^(-)`
`10 ^(-2) M Ca^(2+) + 10 ^(-3) MF^(-)`
`10^(-5) M Ca^(2+) + 10^(-3) M F^(-)`
`10^(-3) M Ca^(2+) + 10^(-5) M F^(-)`

ANSWER :B
Discussion
38.

The precent composition of an organic compound A is carbon 85.71% and hydrogen 14.29. its vapour density is 14. Consider the following reactions sequence A overset(Cl_(2) //H_(2)O)rarr B underset((2) H_(3)O^(+))overset((1) KCN, EtOH)rarr C Identify C

Answer»

`CH_(3)- underset(OH)underset(|)(CH) - CO_(2)H`
`HO - CH_(2) - CH_(2) - CO_(2)H`
`HO - CH_(2)CO_(2)H`
`CH_(3)CH_(2)CO_(2)H`

Answer :C
Discussion
39.

The precentage of empty space in a body centred cubic arragngement is …….

Answer»

74
68
32
26

Answer :D
Discussion
40.

The potential of an electrode is represented as Pt. (0.25 atm)//H^(+) [0.5]

Answer»

0
0.018 V
0.059 V
0.118 V

Solution :`E_(H2)=E_(H2)^(@)+(0.059)/(2)"LOG"([H^(+)]^(2))/([H_(2)])`
`=0+(0.059)/(2)"log"((0.5)^(2))/((0.25)^(2))=`Zero
Discussion
41.

The potential energy of an electron in the hydrogen atom is -6.8 eV. Indicate in which excited state, the electron is present ?

Answer»

first
second
third
fourth

Answer :A
Discussion
42.

The potentialenergy difference between the staggered and eclipsed conformation of ethaneis ………………

Answer»

SOLUTION :12.5 KJ`//`MOL
Discussion
43.

The postulate of Bohr theory that electrons jump from one orbit to the other, rather than flow is according to

Answer»

The QUANTISATION CONCEPT
The wave nature of electron
The probability expression for electron
HEISENBERG uncertainty principle

ANSWER :A
Discussion
44.

The possition number of 'Cl' in the major product of the reactio (locant)

Answer»


SOLUTION :`-NHCOCH_(3)` is BETTER ACTIVATING GROUP than `CH_(3)` group.
Discussion
45.

The positive value of DeltaS indicates that

Answer»

the SYSTEM becomes LESS disordered
the sytem becomes more disordered
the system is in EQUILIBRIUM position
the system TENDS to reach at at equilibrium position.

Answer :B
Discussion
46.

The position of the element with electronicconfiguration1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3) 4s^(2) is placed in the group ….. .

Answer»

3
5
2
11

Answer :B
Discussion
47.

The position of both, an electron and a helium atom, is known within 1.0 nm. Further, the momentum of the electron is known within 5.0 xx 10^(-26) kg ms^(-1). The minimum uncertainty in the measurement of momentum of helium atom is

Answer»

`50 KG ms^(-1)`
`5.0 xx 10^(-26) kg ms^(-1)`
`80 kg ms^(-1)`
`80 xx 10^(-26) k ms^(-1)`

SOLUTION :`Deltax Delta p =` const. As `Delta X` is same, `Delta p` will be same
Discussion
48.

The polymerisation of ethylene to linear polyethylene is represented by the reaction, nCH_(2)= CH_(2) rarr (- CH_(2) - CH_(2) -) where n has a largeintegral value. Given that the average enthalpies of bond dissociation forC=C andC-C at298 K are+590 and+331kJ mol^(-1) respectively, calculate the enthalpy of polymerisation per mole of ethyleneat298 K.

Answer»

Solution :In the given polymerisation reaction,ONE molecule of ETHYLENE involves breaking of C`=` C double bond and formation of threeC-C single bonds.However,in the complete polymer chain,the NUMBER of single C-C bondsformedin two per C=C double bond broken
Energy required in breaking of one moleofC=C doublebonds `= 590 kJ`
Energy released in the formation of two moles of`C-C` single bonds `= 2 xx 331 = 662kJ`
`:.`Net energy released per moleof ethylene`= 662 - 590 = 72 kJ , i.e.,DELTAH =- 72kJmol^(-10`
Discussion
49.

The pollutants which come directly in the air from sources are called primary pollutants Primary pollutants are sometimes converted into secondary pollutants. Which of the following belongs to secondary air pollutants?

Answer»

CO
Hydrocarbon
Peroxyacetyl NITRATE
NO

Solution :Peroxyacetyl Nitrate (PAN) formed from the PRIMARY pollutants `NO_2, O_3` and hydrocarbons is the secondary pollutant
Discussion
50.

The pollutants which come directly in the air from sources are called primary pollutants. Primary pollutants are sometimes converted into secondary pollutants. Which of the following belongs to secondary air pollutants ?

Answer»

CO
Hydrocarbon
Peroxyacetyl nitrate
NO

Solution :Hydrocarbons present in atmosphere combine with oxygen atom which produced by the photolysis of NO, to FORM highly reactive INTERMEDIATE which is known as free radical. Free radical INITIATES a SERIES of reactions. Peroxyacetyl nitrate is formed, which can be known as secondary pollutant.
Hydrocarbon + O `to RCO^@` (Free radicals)
`RCO^* + O_2 to RCO_3^*`
`RCO_3^* + NO_2 to RCO_3 NO_2`
Discussion