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.

State as to why a. A solution of Na_(2)CO_(3) is alkaline? b. Alkali metals are prepared by electrolysis of their fused cholorides? c. Sodium is found to be more useful than potassium?

Answer»

Solution :a. `Na_(2)CO_(3)` is a salt of a weak acid, CARBONIC acid `(H_(2)CO_(3))` and a strong BASE SODIUM hydroxide `(NaOH)`. Therefore, it undergoes hydrolsis to produce strong base `NaOH` and hence its aqueous solution is alkaline in nature.
`Na_(2)CO_(3(s))+H_(2)O_((l))rarrunderset("Strong base")(2NaOH_(aq))+underset("weak acid")(H_(2)CO_(3(aq)))`
b. Since the discharge potential of alkali metals is much higher than that of hydrogen, therefore, when the aqueous solution of any alkali metal chloride is subjected to electrolysis. `H_(2)` instead of the alkali metal is produced at the cathode. Therefore, to prepare alkali metals, electrolysis of their fused chlorides is carried out.
c. Sodium ions are FOUND primarily in the blood plasma and in the interstitial fluid which surrounds the cells while potassium ions are present within the cell fluids. Sodium ions help in the transmission of NEVER signals, in regulating the flow of water across cell membrances and in the transport of sugars and amino acids into the cells. Thus, sodium is found to be more useful than potassium.
Discussion
2.

State any one reason for alkaline earth metals havinga greater tendency to form complexes than alkali metals .

Answer»

Solution :Because of smaller size and HIGHER charge on ALKALINE earth metal cations as compared to the corresponding ALKALI metal cations , alkaline earth metal cations have a GREATER tendency to form complexes .
Discussion
3.

State anti Merkovnikoff's rule and explain with an example.

Answer»

Solution :In the PRESENCE of peroxide, addition of HBr to unsymmetrical ALKENES like PROPENE takes place contrary to the Merkovnikoff.s rule. Thi shappens only with HBr but not with HCl or Hl. This REACTION is know as peroxide or Kharash effect or anti - Merkovnikoff.s rule
`underset("Propene")(CH_(3)-CH=CH_(2))+HBr overset("dibenzyol peroxide")(to)underset("1 Bromopropene")(CH_(3)-CH_(2))-CH_(2)Br`
Discussion
4.

State and give one example for Dobereiner's law of triad and Newland's law of octaves.

Answer»

<P>

Answer :Li,Na, K: Dobereiner's triad
`{:(Li,Be,B ,C,N,O,F),(Na,Mg,Al,Si,P,S,Cl):}}`Newland's OCTAVES
Discussion
5.

State and five ways of enunciating the first law of thermodynamics.

Answer»

Solution :(i) Whenever energy of a particular type disappears equivalent amount of ANOTHER type must be produced.
(ii) TOTAL energy of a system and surroundings remains constant ( or conserved )
(III) " Energy can neither be created nor destroyed, but may be converted from one from to another "
(iv) " The change in the internet energy of a closed system is equal to the energy that passes through its boundary as heat or work ".
(V) Heat and work are TWO ways of changing a system's internal energy " .
Discussion
6.

State and explain Pauli's Exclusion Principle

Answer»

SOLUTION :Pauli.s exclusion principle states that "No TWO electrons in an atom can have the same set of values of all FOUR quantum numbers".
Iilustration : H(Z=1)`1s^1`
One electron is present in hydrogen atom, the four quantum numbers are n = 1, l = 0, m = 0 ands =+ ½. For helium Z = 2. He: `1s^2`. In this one electron has the quantum·number same as that of hydrogen. n = 1, l = 0 m = 0 and `s=+ (1)/(2)`. For other electron, FOURTH quantum number is DIFFERENT. i.e. n = 1, l = 0, m = 0 and `s= -(1)/(2)`
Discussion
7.

State and explain the first law of thermochemistry.

Answer»

Solution :If ENTHALPY of formation of a compound is -x kJ, enthalpy of its decomposition into itselements is +xkJ.
EXAMPLE : `C(S) + O_(2)(g) RARR CO_(2) , Delta H = -xkJ` ACCORDING to the LAW (Lavoisier and Laplace law). `CO_(2(g)) rarr C(S) + O_(2)(g), Delta H = +xkJ`
Discussion
8.

State and explain Pauli's Exclusion Principle.

Answer»

Solution :Pauli's exclusion PRINCIPLE states that "No two electrons in an atom can have the same SET of values of all four quantum NUMBER H(Z= 1 )` 1S^(-1)`.
One electron is present in hydrogen atom, the four quantum numbers are : N = 1 , l = 0 , m = 0 and s` = +1//2.`For heliumZ = 2 . He: `1s^(2)`
In this one electron has the quantum number . Same as that of hydrogen n = 1 , l = 0, m = 0 and s ` = - 1//2`. For other quantum number is different ` i.e., n = 1 , l = 0 , m = 0 and ` s` = - 1//2` .
Discussion
9.

State and explain Pauli's exclusion principle .

Answer»

Solution :State and explain Pauli.s EXCLUSION principle states that "No two electrons in an atom can have the same SET of values of all four QUANTUM numbers" lllustration:H(Z=1)`1s^1`
One electron is present in hydrogen Z=2.He: `1s^2`.In this one electron has the quantum NUMBER same as is different .i.e. n=1,l=o,m=o and s =+1/2.For other electron,fourht q uantum number is different i.e.n=1l=0,m=o and s=-1/2.
Discussion
10.

State and explain pauli's exclusion principal.

Answer»

Solution :Pauli.s EXCLUSION principle states that "No two electrons in an atom can have the same set of values of all four quantum NUMBERS". `H(Z = 1) 1s^(-1)`.
One electron is PRESENT in hydrogen atom, the four quantum numbers are : `N =1, l = 0, m = 0 and s = +1//2` . For helium Z = 2. He : `1s^2`
In this one electron has the quantum number. Same as that of hydrogen `n = 1, l = 0 , m = 0 and s = -1//2`. For other quantum number is different i.e., `n = 1, l = 0, m = 0 and s = -1//2`.
Discussion
11.

State and explainHenry's law

Answer»

Solution :Henry.s law: This law states "that the PARTIAL pressure of the gas in vapour phase is DIRECTLY proporational to the mole FRACTION (x) of the gaseous solute in the solution at LOW concentrations.
`P _("solute") prop x_("solute in solution")`
`P _("solute")K _(H)x _("solute in solution")`
`x _("solute")=` mole fraction of solute in the solution
`K _(H)=` empirical constant.
`P _("solute")=` Vapour pressure of the solute (or) the partial pressure of the gas in capour STATE.
The value of `K _(H)` depends on the nature of the gaseous solute and slovent.
Discussion
12.

State and explain Doberiner's Triad (ii) Complete the following equation Na_(2)O_(2) + ul(?) rarr Na_(2)SO_(4)+H_(2)O_(3)

Answer»

Solution :Johann Dobereiner noted that ELEMENTS with similar properties occur in groups of three which he called TRIADS. It was seen that invariably, the atomic weight of the middle number of the TRIAD was nearly equal to the arithmetic mean of the weights of the other two numbers of the triad. For e.g.,

(ii) `Na_(2)O_(2) + underset("Sulphuric acid ")(H_(2)SO_(4)) RARR Na_(2)SO_(4)+ underset("HYDROGEN peroxide")(H_(2)O_(2)`
Discussion
13.

State and explain Dobereiner's "Triad" .

Answer»

Solution :Some elements such as chlorine, bromine and iodine with similar chemical PROPERTIES into the group of three elements called as triads. In triads, the atomic weight of the middle elements nearly equal to the arithmetic MEAN of the atomic weights of the remaining two elements . Any one example
`{:("S.No.","Elements in the TRIAD","Atomic weight of middle element","Average atomic weight of the remaining elements"),(" 1","Li,Na,K","23",""(7+30)/(2)=23),(" 2","Cl,Br,1","80",""(35.5+127)/(2)=81.25),(" 3","Ca,Sr,BA","88",""(40+137)/(2)=88.5):}`
Discussion
14.

State allotropes of SiO_2.

Answer»

QUARTZ
CRISTOBALITE
TRIDYMITE
All

ANSWER :D
Discussion
15.

Stateall fourQuantum number of valenceelectronof CaZ=20

Answer»

Solution :`Ca (Z= 20 ) [Ar}^(18)4s^(2)`
Twovalenceelectronof 4SORBITAL
N= 4,l =0 `m_(1) = 0 ` and`m_(s ) = (1)/(2) ` and `(1)/(2)`
Discussion
16.

Starting with sodium chloride how would you proceed to prepare (i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide (iv) sodium carbonate ?

Answer»

Solution :Sodium HYDROXIDE may be OBTAINED by the electrolysis of an aqueous solution of sodium chloride using MERCURY cathode and carbon anode (Castner-Kellner CELL).
Discussion
17.

Starting with sodium chloride how would you proceed to prepare sodium carbonate? Mention the steps only.

Answer»

SOLUTION :Sodium carbonate may be obtained by PASSING `CO_2` through ammoniacal brine (SOLVAY ammonia soda PROCESS) when `NaHCO_3` gets precipitated. This on heating gives `Na_2CO_3`.
`NACL + NH_3 + CO_2 + H_2O to NaHCO_3 darr + NH_4Cl`
`2NaHCO_3 to Na_2CO_3 + CO_2 + H_2O`
Discussion
18.

Starting with sodium chloride how would you proceed to prepare (i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide (iv) sodium carbonate?

Answer»

Solution : (i) Sodium metal : Sodium can be extracted from sodium chloride by Downs process. This process involves the electrolysis of fused NaCl (40%) and CaCl, (60%) at a temperature of 1123K in Downs cell.
STEEL is the cathode and a BLOCK of graphite acts as the anode. Metallic Na and Ca are formed at cathode. Molten sodium is taken out of the cell and collected over KEROSENE.
`underset("(molten)")(NaCl) overset("electrolysis")to Na^(+)+CL^(-)`
At Cathod : `Na^(+)+e^(-) to Na`
At Anode : `Cl^(-) to Cl +e^(-)`
`""Cl+Cl to Cl_(2)`
(ii) Sodium Hydroxide :
(iii) Sodium peroxide : First, NaCl is electrolysed to result in the formation of Na metal (Downs process). This sodium metal is then heated on aluminium trays in air (free of `CO_(2)`)to form its peroxide.
`2Na+O_(2) to O_(2) to Na_(2)O_(2)`
(iv) Sodium Carbonate :
Discussion
19.

Starting with sodium chloride how would proceedto prepare (i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide and (iv) sodium carbonate ?

Answer»

SOLUTION :(i) Sodium metal is manufacturedby electrolysis of a fused mixture of NaCl (40%) and `CaCl_(2)` (60%) in Down's cell at 873 K using iron cathode and graphite anode .Na is liberated at the cathode while `Cl_(2)` is evolved at the anode ,
At cathode : `Na^(+)` (melt) `+ e^(-) to Na (l)`
At anode : `2 Cl^(-)` (melt) `to Cl_(2) (G) + 2 e^(-)`
(ii) Sodium hydroxide is manufactured by electrolysis of an aqueous solution of NaCl (brine ) in CASTNER Kellner cell using mercury cathode and carbon anode . Sodium metal which is discharged at the cathode combines with mercury to form sodium amalgam . `Cl_(2)` gas is evolved at the anode .
`{:("Cathode" : ,, Na^(+) + e^(-) to underset("Sodium")(Na) , ,, 2 Na + x HG to underset("Sodium amalgam")(Na_(2)Hg_(x))), ("Anode" : ,, 2Cl^(-) to Cl_(2) + 2 e^(-) ,,):} `
The sodium amalgam thus obtained is treated with water to form sodium hydroxide and hydrogen gas
`Na_(2)Hg_(x) + 2H_(2)O to 2 NaOH + x Hg+ H_(2)`
(iii) Sodium peroxide is obtained by HEATING sodium in excess of air . The intially formed sodium oxide reacts with more `O_(2)` to form `Na_(2)O_(2)`.
`{:(4Na + O_(2) overset(Delta)(to) 2Na_(2)O), (2 Na_(2)O + O_(2) overset(Delta)(to) 2Na_(2)O_(2)):}`
(iv) Sodium carbonate is obtained by Solvay ammonia process as discussed in Ans
Discussion
20.

Starting with pure 'A' molar mass becomes half of its initial value at equilibrium as shown by a reaction A(g) hArr nB(g).if A is 50% dissociated at equilibrium, find value of n.

Answer»


ANSWER :3
Discussion
21.

Starting with ethyne, how will you prepare pentan-2-one ?

Answer»

SOLUTION :`underset"Pent-2-yne"(CH_3-C-=C-CH_2CH_3+H_2O)underset"333 K"overset(60% H_2SO_4, HgSO_4)to [CH_3-oversetoverset"OH"|C=CHCH_2CH_3]overset"Tautomerises"to underset"Pentan-2-one (major ISOMER)"(CH_3-oversetoverset(O)(||)C-CH_2CH_2CH_3)`
Discussion
22.

Starting with NaCl how would you proceed to prepare Na_2O_2

Answer»

Solution :At FIRST Na is prepared form FUSED Nacl as in
Sodium metal is HEATED with EXCESS OXYGEN (air) to form `Na_2O_2` 4Na+O_2 rArr 2Na_2O`2Na_2O+O_2 rArr 2Na_2O_2`
Discussion
23.

Starting with NaCl how would you proceed to prepare Na_2CO_3?

Answer»

SOLUTION :By SOLVAY PROCESS `CO_2` is PASSED through aq.NaCl containing `NH_3` to form `NaHCO_3`. This on heathing yields `Na_2CO_3`
Discussion
24.

Starting with NaCl how would you proceed to prepare Na metal

Answer»

Solution :By the ELECTROLYSIS of molten NaCl (CONTAINING` CaCl_2`)at 873 K using iron CATHODE and graphite ANODE (Down.s method ). Na is deposited at the cathode and `Cl_2` at anond,
Discussion
25.

Startingfrom SiCl_(4), preparethe followingin steps not exceeding the number givein parentheses (giveonly reactions) : (a) Si(I) , (b) Linear siliconcontainingmethyl group(4) (c ) Na_(2)SiO_(2) (3).

Answer»

Solution :(a) `SiCl_(4) + 2Mg overset(DELTA)rarr Si +2MgCl_(2)` or `3SiCl_(4) + 4Al overset(Delta)rarr 4AlCl_(3) + 3Si`
(b) (i) `SiCl_(4) + Mg rarr Si +2MgCl_(2)`, (ii) `Si + 2CH_(3)Cl overset(" Cu POWDER ")underset(570 K)rarr (CH_(3))_(2)SiCl_(2)`
(iii) `(CH_(3))_(2)SiCl_(2) + 2H_(2)O rarr HO-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(Si)-OH`
(IV)
(c ) (i) `SlCl_(4) + 4H_(2)Orarr Si(OH)_(4) + 4HCl`
(ii) `Si(OH)_(4)overset(Delta)rarr SiO_(2) + 2H_(2)O` , (iii) `SiO_(2) + 2 NaOH overset(Delta)rarr Na_(2)SiO_(3) + H_(2)O`
Discussion
26.

Starting from methyl magnesium iodide, how would you prepare (i) Ethyl methyl ether (ii) methyl cyanide (iii) methane.

Answer»

Solution :(i) ETHYL methyl ether: Lower haBogenated ether reacts with grignard reagent to form HIGHER ether.
`underset("Chloro dim ethyl ether")(CH_3 - O - CH_2Cl) + underset("iodide")underset("Methyl magnesium")(CH_3MgI) to underset("ETHYLMETHYL ether")(CH_3 - O - CH_2 - CH_3)`
(II) Methyl cyanide: Grignard reagent reacts with cyanogen chloride to form alkyl cyanide.

(iii) Methane: Grignard reagent reacts with water to give methane as product.
Discussion
27.

Starting from methyl magnesium iodide how would you prepare (i) Ethanol (i) 2-propanol(iii) Tert-butyl alcohol.

Answer»

Solution :(i) Ethanol: Formaldehyde REACTS with `CH_3MgI` to GIVE an ADDITION PRODUCT which on hydrolysis yields ethanol.

(ii) 2-Propanol: ACETALDEHYDE react with `CH_3MgI`to give an addition product which on hydrolysis yields 2-propanol.
Discussion
28.

Starting from CH_(3) MgI , how will you prepare acetaldehyde ?

Answer»

SOLUTION :
Discussion
29.

Starting from CH_(3) Mgl, how will you prepare the following? (a) Acetic acid(B) Acetone

Answer»

SOLUTION :a.Aceticacid: solidcarbondioxidereactswithgrignard reagentto formadditionproduct WHICHON hydrolysisyieldsaceticacid
Discussion
30.

Starting from CH_(3)MgI, how will you prepare the following? (i) Acetic acid (ii) Acetone (iii) Ethyl acetate (iv) Iso propyl alcohol (v) Methyl cyanide.

Answer»

Solution :(i) Acetic acid: Solid carbon dioxide reactws with GRIGNARD reagent to FORM addition product which on hydrolysis yields carboxylic acids.

(ii) Acetone:

(iii) Ethyl acetate: Ethylchloroformate reacts with Grignard reagent to form esters.

(iv)Iso propyl alcohol: Aldehydes other than FORMALDEHYDE, react with Grignard reagent to GIVE addition product which on hydrolysis yields secondary alcohol.

(v) Methyl cyanide: Grignard reagent reacts with cyanogen chloride to from alkyl cyanide
.
Discussion
31.

Starting from CH_3MgI. How will you prepare the following ? (i)Acetic acid (ii) Acetone (iii) Ethyl acetate (iv) Isopropyl alcohol (v) Methyl cyanide

Answer»

Solution :(i) ACETIC ACID from grignard REAGENT:

(II) ACETONE îrom Grignard reagent
Discussion
32.

Startin with the correctly balanced half rection write the overall ionic reaction in the following changes (i) chloride ion is oxidised to CI_(2) by MnO_(4)^(-) (in acid solution) (ii) Nitrous acid (HNO_(2)) reduces MnO_(4)^(-) (in acid solution ) (iii) Nitrous acid (HNO_(2)) oxidises I^(-) to I_(2) (in acid solutoin ) (iv) chlorate ion (CIO_(3)^(-)) oxidises Mn^(2+) to MnO_(2) (s) (in acid solution) (v) chromite ion (CrO_(3)^(-)) is oxidised by H_(2)O_(2) (in strongly basic medium ) also find out the change in the oxidatoin number of the underline atoms

Answer»

Solution :(i) `2MnO_(4)^(-)+16H^(+) + 10 CI^(-) rarr5CI_(2)+2 Mn^(2+) to +2 in Mn^(2+)`
oxidation NUMBER of Mn changes form +7 in `MnO_(4)^(-)to + 2Mn^(2+)`
(ii)`2MnO_(4)^(-)+ 6H ^(+) + 5NO_(2)^(-) rarr 5NO_(3)^(-) +3H_(2)O + 2 Nn^(+)`
Oxidatoin number of N changes from +3 in `NO_(3)^(-) "ion to" +5 in NO_(3)^(-)` ion
(iii) `2I^(-) + 4H + 2 NO_(2)^(-) rarr I_(2) + 2NO+2H_(2)O`
oxidation number of N changes form +3 is `NO_(2)^(-)` to +2 in NO
(iv) `3MN^(2+) +CIO_(3)^(-) +6H^(+)rarr3Mn^(4+)+CI^(-)+3H_(2)O`
oxidation numbr of CIchanges from + 5 in `CIO_(3)^(-) "to -1 in" CI^(-)`
(v) `2CrO_(3)^(-)+ h_(2)O_(2)+2OH^(-)rarr2CrO_(4)^(-)+2H_(2)O`
oxidation number of Cr changes from +5 in `CrO_(3)^(-)` to +6 in `CrO_(4)^(2-)]`
Discussion
33.

Startingfrom‘a’molesof H_2and‘b’molesofI_2anequilibriumH_2+I_2

Answer»

`(4x^2)/(AB)`
`(4x^2)/((a-x)(b-x))`
`(2x)/((a-x)(b-x))`
`(4x^2)/((a-2x)(b-2x))`

SOLUTION :
`K_(C)=([HI]^(2))/([H_(2)][I_(2)])=((2x)^(2))/((a-x)(b-x))`
Discussion
34.

Stannous chloride has high melting point (is a solid) whilestannic chloride has low melting point (is a liquid). Why?

Answer»

Solution :Stannous chloride has ionic nature.
Hence, `SnCl_(2)` is HIGH MELTING solid.
Due to higher OXIDATION state of Sn in STANNIC chloride, it is covalent. Hence, `SnCl_(4)`. is low melting substance.
Discussion
35.

Standard vaporisation enthalpy of benzene at its boiling point is 230.8 kJ mol^(-1) . For how long would a 100 W electricheater have to operate to vaporize a 100 g sample of benezene at its boiling temperature? ( power = energy // time , 1 W = 1 J s^(-1))

Answer»

Solution :1 moleof benezene , `C_(6) H_(6) (78 G)` requires energy for VAPORIZATION `= 30.8kJ`
`:. 100g` benezenewill require energy `= (30.8) /( 78) xx 100 J = 39.5 kJ`
100 Wheater gives energy of 100 J per second
`:. ` Time required for GETTING 39.5 kJ of energy`= (39500 J )/(100J ) = 395 sec = 6` min 35 sec.
Discussion
36.

Standard reduction p[otentials of the half rections are given bvelow: F_(2)(g)+2e^(-)rarr2F_(aq)^(-):"" E^(@)=+2.85V CI_(2)(g)+2e^(-)rarr2CI(aq)^(-):"" E^(@)=+1.36V Br(2)(g)+2e^(-)rarr2Br(aq)^(-):"" E^(@)=+1.06V I_(2)(s)+2e^(-)rarr2I_(aq)^(-):"" E^(@)=+0.53V The stronges oxidising and reducing agents respectively are

Answer»

`F_(2) and I^(-)`<BR>`Br_(2) and CI^(-)`
`CI_(2) and Br^(-)`
`CI_(2) and I_(2)`

Solution :More the VALUE of NEGATIVE reduction potential stronger in the reducing AGENT. Therefore, `1^(-)` ion is the stongest reducing agent .More the value of p[ositive reduction potential stronger is the oxidising agent .Therefore,`F_(2)` is the strongest oxidising agent.
Discussion
37.

Standard reduction potential of the half reaction are give below F_(2)(g)+2e^(-)rarr2F^(-)(aq),e^()=2.85 V CI_(2)(g)+2e^(-)rarr2CI^(-)(aq),E^(@)=+1.36 V Br_(2)(g)+2e^(-)rarr2Br^(-)(aq),E^(@)=+1.06 v I_(2) (g) +2e^(-)rarr2I^(-)(aq),e^(@)=+0.53 v The strongestoxidising andreducing agent respectively are

Answer»

`F_(2)` and `I^(-)`
`Br_(2)` and `CI^(-)`
`CI_(2)` and `Br^(-)`
`CI_(2)` and `I_(2)`

Solution :`F_(2)` with the highest REDUCING potential (+2.85 V) is the stongestoxidising AGENT and `T^(-)` with the highest oxidation potential (-0.53 V) is the strongest reducing agent
Discussion
38.

Standard reduction potential is least for Li. Why?

Answer»

SOLUTION :LITHIUM cation is small in size with high hydration ability. `Li(s) + aq RARR Lit(aq) +e^(-)`.
The reaction PROCEEDS LEADING to the formation of stable hydrated cation. Hence reduction potential is least.
Discussion
39.

Standard potential (E^(@)) for some half reactions are given below (i) Sn^(4+) + 2e^(-) to Sn^(2+) , E^(@) = 0.15 V (ii) 2Hg^(2+) + 2e^(-) to Hg_(2)^(2+), E^(@) = 0.92 V (iii) PbO_(2) + 4H^(+) +2e^(-) to Pb^(2+) + 2H_(2)O, E^(@) = +1.45 V based on the above, which one of the following statements is correct ?

Answer»

`Sn^(4+)` is a stronger oxidising agent than `Pb^(4+)`
`Sn^(2+)` is a stronger reducing agent than `Hg_(2)^(2+)`
`Pb^(2+)` is a stronger oxidizingagent than `Pb^(4+)`
`Pb^(2+)` is a stronger reducing agent than `Sn^(2+)`

Solution :Electrochemical SERIES, more the reduction potential, stronger the oxidizing POWER and VICE VERSA
Discussion
40.

Standard molar entropies of Mg_((s)) , C_((s)), MgO_((s)) & CO_((g)) respectively are S_1,S_2,S_3, & S_4 J//K//mol. Then DeltaS_("sys") for MgO_((s)) + C_((s)) rarr Mg_((s)) + CO_((g))

Answer»

`S_1 + S_2 + S_3 + S_4`
`(S_1 + S_4) - (S_2 + S_3)`
`(S_2 + S_3) - (S_1 + S_4)`
`(S_4 - S_3)`

ANSWER :B
Discussion
41.

Standard molar enthalpy of formation, Delta_(f)H^(c-) is just a special case of enthalpy of reaction, Delta_(r)H^(c-) . Is the Delta_(r)H^(c-) for the following reaction same as Delta_(f)H^(c-) ? Give reaction for your answer. CaO(s) +CO_(2)(g) rarr CaCO_(3)(s), Delta_(f)H^(c-)= - 178. 3 kJ mol^(-1)

Answer»

Solution :No, because enthalpy of FORMATION is the enthalpy changewhen 1 MOLE of the compound is formed from the elements , i.e., for the reaction `Ca(s) + C(s) + (3)/(2) O_(2)(g) rarr CaCO_(3)(s)`. Thus, this reaction is DIFFERENT from the givien reaction. Hence, `Delta_(r) H^(@)cancel(=) Delta_(f)H^(@)`.
Discussion
42.

Standard molar enthalpy of formation, Delta_(f)H^(@) is just a special case of enthalpy of reaction, Delta_(r)H^(@). Is the Delta_(r)H^(@) for the following reaction same as Delta_(f)H^(@)? Give reason for your answer. CaO(s) + CO_(2)(g) to CaCO_(3)(s), Delta_(f)H^(@) = -178.3 kJ"mol"^(-1).

Answer»

SOLUTION :No, because `CaCO_(3)` is FORMED from other compounds and not from its constituent ELEMENTS.
Discussion
43.

Standard molar enthalpy of formation, Delta_(f) H^( Theta ) is just a special case of enthalpy of reaction, Delta_(r) H^( Theta ) . Is the Delta_(r) H^( Theta ) for the following reaction same as Delta_(f) H^( Theta ) ?Give reason for your answer. CaO_((s)) + CO_(2(g)) to CaCO_(3(s)) , Delta_(f) H^( Theta ) = - 178 "kJ mol"^(-1)

Answer»

Solution :No, the `Delta_(r) H^( Theta )` for the given reaction is not same as `Delta_(r) H^( Theta )`.
"The standard enthalpy change for the formation of one MOLE of a compound from its ELEMENTS in their most stable states (reference states) is called standard molar enthalpy of formation," `Delta_(r) H^( Theta )`.
`Ca_((s)) + C_((s)) + (1)/(2) O_(2(g)) to CaCO_(3(s)) , Delta_(F) H^( Theta )`
This reaction is different from the given reaction.
Hence, `Delta_(f) H^(0) ne Delta_(f) H^( 0) `
Discussion
44.

Standard molar enthalpy is lowest for

Answer»

Diamond CARBON 
GRAPHITE carbon
Water vapour 
Liquid water 

ANSWER :D
Discussion
45.

Standard Gibbs energy of reaction (Delta_(r)G^(@)) at a certain temperature can be completed as Delta_(r)G^(@)=Delta_(r)H^(@)-Tdelta_(r)S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows: Delta _(r) H_(T_(2))^(@)-Delta_(r)C_(p)^(@)(T_(2)-T_(1)) Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)In ((T_(2))/(T_(2))) Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) Delta_(r)^(@)G^(@)=-RTInK_(eq) Consider the following reaction : CO(g)+2H_(2)(g)toCH_(3)OH(g) Given Delta_(r)H^(@)(CH_(3)Oh,g]=-201KJ//mol Delta_(r)H^(@)(CO,g)=-114KJ//mol s^(@)(CH_(3)OH,g)=240J//mol-k, S^(@)(H_(2)g)=198J//mol-KC_(p.m)^(@)(h_(2))=28.8JK^(-1)mol^(-1) C_(p.m(CO)=29.4J//mol-K C_(p.m)^(@)(CH_(3_)OH)=44J//mol-K and In ((320)/(300))=0.06,"all data at"300K. Delta_(r)H^(@)at 320K is :

Answer»

`-288.86KJ//mol`
`-289.1KJ//mol`
`-87.86KJ/mol`
NONE of these

Answer :C
Discussion
46.

Standard Gibbs energy of reaction (Delta_(r)G^(@)) at a certain temperature can be completed as Delta_(r)G^(@)=Delta_(r)H^(@)-Tdelta_(r)S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows: Delta _(r) H_(T_(2))^(@)-Delta_(r)C_(p)^(@)(T_(2)-T_(1)) Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)In ((T_(2))/(T_(2))) Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) Delta_(r)^(@)G^(@)=-RTInK_(eq) Consider the following reaction : CO(g)+2H_(2)(g)toCH_(3)OH(g) Given Delta_(r)H^(@)(CH_(3)Oh,g]=-201KJ//mol Delta_(r)H^(@)(CO,g)=-114KJ//mol s^(@)(CH_(3)OH,g)=240J//mol-k, S^(@)(H_(2)g)=198J//mol-KC_(p.m)^(@)(h_(2))=28.8JK^(-1)mol^(-1) C_(p.m(CO)=29.4J//mol-K C_(p.m)^(@)(CH_(3_)OH)=44J//mol-K and In ((320)/(300))=0.06,"all data at"300K. Delta_(r)s^(@)at 320K is:

Answer»

155.18 J/mol-K
150.02J/mol-K
172J/mol-K
none of these

Answer :d
Discussion
47.

Standard Gibbs energy of reaction (Delta_(r)G^(@)) at a certain temperature can be completed as Delta_(r)G^(@)=Delta_(r)H^(@)-Tdelta_(r)S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows: Delta _(r) H_(T_(2))^(@)-Delta_(r)C_(p)^(@)(T_(2)-T_(1)) Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)In ((T_(2))/(T_(2))) Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) Delta_(r)^(@)G^(@)=-RTInK_(eq) Consider the following reaction : CO(g)+2H_(2)(g)toCH_(3)OH(g) Given Delta_(r)H^(@)(CH_(3)Oh,g]=-201KJ//mol Delta_(r)H^(@)(CO,g)=-114KJ//mol s^(@)(CH_(3)OH,g)=240J//mol-k, S^(@)(H_(2)g)=198J//mol-KC_(p.m)^(@)(h_(2))=28.8JK^(-1)mol^(-1) C_(p.m(CO)=29.4J//mol-K C_(p.m)^(@)(CH_(3_)OH)=44J//mol-K and In ((320)/(300))=0.06,"all data at"300K. Delta_(r)H^(@) at 300K for the reaction is :

Answer»

`-87KJ/mol`
`87KJ//mol`
`172J//mol-K`
NONE of these

Answer :a
Discussion
48.

Standard Gibbs energy of reaction (Delta_(r)G^(@)) at a certain temperature can be completed as Delta_(r)G^(@)=Delta_(r)H^(@)-Tdelta_(r)S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows: Delta _(r) H_(T_(2))^(@)-Delta_(r)C_(p)^(@)(T_(2)-T_(1)) Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)In ((T_(2))/(T_(2))) Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) Delta_(r)^(@)G^(@)=-RTInK_(eq) Consider the following reaction : CO(g)+2H_(2)(g)toCH_(3)OH(g) Given Delta_(r)H^(@)(CH_(3)Oh,g]=-201KJ//mol Delta_(r)H^(@)(CO,g)=-114KJ//mol s^(@)(CH_(3)OH,g)=240J//mol-k, S^(@)(H_(2)g)=198J//mol-KC_(p.m)^(@)(h_(2))=28.8JK^(-1)mol^(-1) C_(p.m(CO)=29.4J//mol-K C_(p.m)^(@)(CH_(3_)OH)=44J//mol-K and In ((320)/(300))=0.06,"all data at"300K. Delta_(r)S^(@)at 300 K for the reaction is :

Answer»

152.6J/K -MOL
181.6J?K-mol
`-16J/K--mol`
NONE of these

Answer :C
Discussion
49.

Standard free energy change of a reaction is +150 kJ mol^(-1). Calculate K_(p) at 30^(@)C.

Answer»

Solution :`DELTA G^(@) = -2.303 RT log_(10) K_(p)`
`log_(10) K_(p) = (Delta G^(@))/(-2.303 xx R xx T)`
`log_(K_(p)) = (150000)/(-2.303 xx 8.314 xx 300)`
`log K_(p) = -26.1136`
`K_(p) = "ANTILOG"- 26.1136`
`K_(p) = "antilog" BAR(27).8864`
`= 7.698 xx 10^(-27)`
Discussion
50.

Standard free energy change(Delta G^(@)) of a reaction is related to its equilibrium constant as: K= 10^(x) wherex = "…................."

Answer»

SOLUTION :`- (DELTAG^(@))/( 2.303RT) `
Discussion