94420 + Interview Questions in Chemistry in Class 12

1.	Consider the reaction M_((aq))^(n+)+n e^(-)toM_((s)). The standard reduction potential values of the element M_(1),M_(2) and M_(3) are -0.34V,-3.05 and -1.66V respectively. The order of their reducing power will be:-
Answer» `M_(1) gt M_(2) gt M_(3)` `M_(3) gt M_(2) gt M_(1)` `M_(1) gt M_(3) gt M_(2)` `M_(2) gt M_(3) gt M_(1)` Solution :The reducing power DECREASES as the reduction POTENTIAL increase (BECOMES less negative).

Discussion

2.	Consider the reaction, N_(2) + 3H_(2) rarr 2NH_(3) carried out ata constant temperature and pressure. If DeltaH and Delta U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?
Answer» `DELTA H = 0` `Delta H = Delta U` `Delta H LT Delta U` `Delta H gt Delta U` Solution :For `N_(2) + 3H_(2) rarr 2NH_(3)` `Delta n_(g) = 2 - (1 + 3) = -2` `Delta H = Delta U + Delta n_(g) RT = Delta U - 2RT` :. `Delta H lt Delta U`

Discussion

3.	Consider the reaction HCN_((aq))hArrK_((aq))^(+)+CN_((aq))^(-). At equlibrium, the addition of CH_((aq))^(+) concentration
Answer» Reduce `HNC_((aq))` CONCENTRATION Decrease the `H_((aq))^(+)` ion concentration Increase the equlibrium constant Decrease the equlibrium constant Solution :At equlibrium, the ADDITION of `(CN)^(-)` would decrease the `(H^(+))` ion concentration to PRODUCE more and more HCN to NULLIFY the increase of `CH_(aq)^(-).`

Discussion

4.	Consider the reaction : H_(2)(g),Pt\|H^(+)(aq) , E=0.1" V" The pH of the solution is
Answer» Solution :`H_(2)(g)+2e^(-) to2H^(+)(aq)` `E=E^(@)-(0.0591)/(2)"LOG"[H^(+)]^(2)` `0.1=0-(0.0591)/(2)2LOG[H^(+)]` `0.1=0-0.0591" log " [H^(+)]` `-log[H^(+)]=(0.1)/(0.0591)=1.69` `pH=-log[H^(+)]=1.69=2`

Discussion

5.	Consider the reaction given below : CH_(3)-CHDCONH_(2)+PhCOoverset(15)NH_(2)underset(Delta)overset(Br_(2)+OH^(-))to The products obtained are :
Answer» `CH_(3)CHDNH_(2)` `CH_(3)CHD-overset(15)NH_(2)` `PhNH_(2)` `Ph-overset(15)NH_(2)` Solution :`CH_(3)-CHD-CONH_(2)-Ph-COoverset(15)NH_(2)overset(Br_(2)//OH^(-))tounderset(("Hoffmann bromamide TEST"))"Hoffmann degradation"toCH_(3)-CHD-NH_(2)+Phoverset(15)NH_(2)`

Discussion

6.	Consider the reaction equlibreium, 2SO_(2(g))+O_(2(g))hArr2SO_(3(g)),DeltaH^(@)-=198kJ. On the basis of Le-Chatelier's pronciple, the condition favourable for the forward reaction is
Answer» LOWERING of TEMPERATURE as well as PRESSURE Increasing temperature as well as pressure Lowering the tmpeerature and increasing the pressure Any vaue of temperature and pressure Solution :It is an EXOTHERMIC REACTION hence low temperatue and increasing pressure will favour forward reaction.

Discussion

7.	Consider the reaction equilibrium underset(("Greater volume"))("Ice")hArrunderset(("Lesser volume"))("Water-xkcal")The favourable conditions for forward reaction are
Answer» LOW TEMPERATURE , HIGH PRESSURE and EXCESS of ice low temperature , low pressure and excess of ice high temperature ,low pressure and excess of ice high temperature , high pressure and excess of ice Answer :D

Discussion

8.	Consider the reaction, C(s)+O_(2)(g)toCO_(2)(g)+393.5kJ, the signs of DeltaH,DeltaS and DeltaG respectively are
Answer» `+,-,-` `-,+,+` `-,-,-` `-,+,-` SOLUTION :`DeltaH=-ve"(COMBUSTION REACTION)"` `DeltaS=+ve"(SPONTANEOUS at all TEMPERATURE)"` `DeltaG=-ve""`(`because` reaction is spontaneous)

Discussion

9.	Consider the reaction Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)to2Cr^(3+)+7H_(2)O What is the quantity of electricity in couolombs needed to reduce 1 mole of Cr_(2)O_(7)^(2-)?
Answer» Solution :From the given REACTION, 1 mol FO `Cr_(2)O_(7)^(2-)` ions require `6F=6xx96500C=579000C` of ELECTRICITY for reduction to `Cr^(3+)`.

Discussion

10.	Consider the reaction :Cr_(2)O_(7)^(2-) + 14H^(+) + 6e^(-) to 2Cr^(3+) + 7H_(2)O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr_(2)O_(7)^(2-) ?
Answer» Solution :One `Cr_(2)O_(7)^(2-)`ion requires 6 electrons for REDUCTION as shown by the above CHEMICAL equation. THUS, 1 mol of `Cr_(2)O_(7)^(2-)` ions require 6 F= 6 x 96500 C = 579000 C of electricity for reduction to `Cr^(3+)`.

Discussion

11.	Consider the reaction: Cr_(2)O_(2)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr_(2)O_(7)^(2-) ?
Answer» Solution :* Reaction: * Therefore, to reduce 1 mole of `Cr_(2)O_(7)^(2-),` 6 mole electrons will be USED. So, the REQUIRED quantity of electricity will be: So, 6 mole electron=6F electricity is used. 6 F electricity=6`xx96500` COULOMB =579000 Coulomb

Discussion

12.	Consider the reaction : Cl_(2)(aq)+H_(2)S(aq)toS(s)+2H^(+)(aq)+2Cl^(-)(aq) The rate equation for this reaction is rate =k[Cl_(2)][H_(2)S] Which of these mechanisms is/are consistent with this rate equation ? A. Cl_(2)+H_(2)StoH^(+)+Cl^(-)+Cl^(+)+HS^(-)("slow"),Cl^(+)+HS^(-)toH^(+)+Cl^(-)+S("fast") B. H_(2)SiffH^(+)+H^(+)HS^(-)("fast equilibrium"),Cl_(2)+HS^(-)to2Cl^(-)+H^(+)+S("slow")
Answer» Neither A nor B A only B only Both A and B Solution :(b) : Rate depends upon the slow step. THUS, slow step should involve 1 molecule of `Cl_(2)` and 1 molecule of `H_(2)S`. Hence, A is the CORRECT mechanism.

Discussion

13.	Consider the reaction : Cl_(2)(aq)+H_(2)S(aq) to S(s)+2H^(+)(aq)+2Cl^(-)(aq) the rate equation for this reaction is rate=k[Cl_(2)][H_(2)S] Which of these mechanism is/ are consistent with this rate equation? A. Cl_(2)+H_(2)S to H^(+)+Cl^(-)+HS^(-) (Slow) Cl^(+)+HS^(-) to H^(+)+Cl^(-)+S (fast) B. H_(2)S harr H^(+)+HS^(-) (fast equilibirium) Cl_(2)+HS^(-) to 2Cl^(-)+H^(+)+S (Slow)
Answer» B only Both A and B neither A nor B A only Solution :Since the slow step is the RATE DETERMINING step hence if we consider option (A) we find Rate`=k[Cl_(2)][H_(2)S]` Now, it we consider option (B) we find Rate`=k[Cl_(2)][HS^(-)]` …(i) For equationn `H_(2)S harrH^(+)+HS^(-)` `K=([H^(+)][HS^(-)])/([H_(2)S]) or, [HS^(-)]=(K[H_(2)S])/([H^(+)])` SUBSTITUTING this value in EQUATION (i) we find Rate`=k[Cl_(2)]k([H_(2)S])/([H^(+)])=(k[Cl_(2)][H_(2)S])/([H^(+)])` THUS slow step should involve 1 molecues of `Cl_(2)` and 1 molecule of `H_(2)S` Hence only, mechanism (A) is consistent with the given rate equation

Discussion

14.	consider the reaction CH_(3)CH_(2)CH_(2)Br+NaCNtoCH_(3)CH_(2)CH_(2)CN+NaBr. the reaction will be fastest in.
Answer» Water Ethanol Methanol N,N'-dimethyl FORMAMIDE (DMF). Solution :N,N'-dimethyl formamide (DMF) is a POLAR aprotic solvent. It favours `S_(N^(2))` REACTION. All other solvents are polar protic in nature. .

Discussion

15.	Consider the reaction CI_(2) (aq)+H_(2)S(aq) rarr S(s)+2H^(+)(aq) +2CI^(-1)(aq) The rate equation for this reaction is rate = k[CI_(2)] [H_(2)S] Which of these mechanisms is /are consistent with this rate equation ? CI_(2)+H_(2)S rarr H^(+)+CI^(-)+CI^(+)HS^(-) (slow ) CI^(+)+HS^(-)+CI^(-)+S (fas)H_(2)S rarr H^(+)+HS^(-) (fast equilibrium ) CI_(2)+HS^(-) rarr 2CI^(-)+H^(+) +S (slow )
Answer» Neither (i) nor (ii) (i) only (ii) only Both (i) and (ii) Solution :(B) `CI_(2(aq))+H_(2) S_((aq))rarrS_((s))+2H_((aq))^(+)+2CI_((aq))^(-)` `R= K [CI_(2)][H_(2)S]` SINCE RATE of reaction DEPENDS only on the slow step `:.` Mechanism (A) is correct . `CI_(2)+H_(2)S rarr H^(+)+CI^(-)+CI^(+)+HS^(-)` (slow) `CI^(+) +HS^(-)+CI^(-)+S` (fast )

Discussion

16.	Consider the reaction : CH_(3)CH_(2)CH_(2)Br+NaCNrarrCH_(3)CH_(2)CH_(2)CN+NaBr This reaction will be the fastest in :
Answer» WATER ethanol methanol N,N-dimethylformamide (DMF) ANSWER :D

Discussion

17.	Consider the reaction CH_(3)CH_(2)CH_(2)Br+ NaCN to CH_(3)CH_(2)CH_(2)CN +NaBr This reaction will be the fastest in -
Answer» water ethanol methanol N,N'-dimethylformamide (DMF) ANSWER :D

Discussion

18.	Consider the reaction, CH_(3)CH_(2)CH_(2)Br+NaCn to CH_(3)CH_(2)CH_(2)CN+NaBr This reaction will be the faster in
Answer» ethanol methanol N, N'-dimethylformamide (DMF) water Solution :`CH_(3)CH_(2)CH_(2)Br` being a `1^(@)` ALKYL halide reacts with `NaCN` by `S_(N)2` mechahnism which is favoured by polar aprotic SOLVENT, i.e., DMF. Thus, option (c) is correct.

Discussion

19.	consider the reaction: CH_(3)CH_(2)CH_(2)Br + NaCN to CH_(3)CH_(2)CH_(2)CN + NaBr This reaction will be fastest in:
Answer» ethanol methanol N, `N^(')`-dimethyl formamide (DMF) water Solution :c) The reaction PROCEEDS by `S_(N)^(2)` mehanism. It will be favoured by polar aprote SOLVENT i.e., N, `N^(')`-dimethylformamide (DM) which is a polar APROTIC solvent.

Discussion

20.	Consider the reaction: CH_(3)-CH_(2)-CH_(2)-Br+NaCNtoCH_(3)CH_(2)CH_(2)-C-=N+NaBr The correct statement is :-
Answer» The REACTION will be FASTEST in water The reaction will be fastest in N,N-dimethylformamide (DMF) Transition state of `S_(N)2` is tetrahedral and `sp^(3)` hybridized If conc. Of alkyl bromide is tripled and conc. of `OVERSET(THETA)(CN)` is reduced to HALF rate of `S_(N)2` increased by 2 times Answer :B

Discussion

21.	Consider the reaction C_(6)H_(12)O_(6) + 12H_(2)SO_(4) rarr 6CO_(2) + 12SO_(2) + 18H_(2)O How many of the following conclusion are corerct based on this equation? (A) Mass ratio of C_(6)H_(12)O_(6) and H_(2)SO_(4) present is 1 : 12 (B) Mole ratio of C_(6)H_(12)O_(6) and H_(2)SO_(4) present is 1 : 12 (C) Mass ratio of C_(6)H_(12)O_(6) and H_(2)SO_(4) reacted is 1 : 12. (D) Mole ratio ofH_(2)SO_(4) and CO_(2) in reaction mixture is 2 : 1 (E) Mass ratio of H_(2)SO_(4) and CO_(2) in reaction mixrture is 2 : 1 (F) Mole ratio of H_(2)O and C_(6)H_(12)O_(6) present in reaction mixture is 18 : 1 (G) Mole ratio of SO_(2) and CO_(2) formed 2 : 1 (G) Sum of number of moles of H_(2)SO_(4) and SO_(2) in the reaction mixture is always constant. (I) Sum of number of moles of moles of C_(6)H_(12)O_(6) and CO_(2) present in the reaction mixture is always constant.
Answer» SOLUTION :`(G)` & `(H)` are correct STATEMENTS. Rest are WRONG.

Discussion

22.	Consider the reaction C_(3)H_(7)-OH+Et_(3)O^(o+)BF_(4)^(-)rarr C_(3)H_(7)-O-Et+EtOEt+HBF_(4). Which of the following statements si wrong ?
Answer» The NUCLEOPHILE in the reaction is `C_(3)H_(7)OH`. The nucleophile in the reaction is `BF_(4)^(-)`. The leaving GROUP is `Et_(2)O`. `SN^(2)` reaction occurs. Solution :NA

Discussion

23.	Consider the reaction : Briefly explain the choice of major product.
Answer» Solution : Initial reaction of the ALKENE with `H_(3)O^(+)` can form two carbocations. The more stable BENZYLIC tertiary carbocation (shown in the above mechanism) is formed in preference to the less stable PRIMARY carbocation. This is the rate deter MINING step, and thus controls the product DISTRIBUTION. Formation of the more stable carbocation is the mechanistic basis for Markovnikov.s rule.

Discussion

24.	Consider the reaction C_(2)H_(5)Cl+AgCN overset(ErOH)rarr X
Answer» It GIVES propionic acid on hydrolysis It has an ester function It has nitrogen linked to ETHYL carbon It has a cyanide GROUP. ANSWER :D

Discussion

25.	Consider the reaction between chlorine and nitric oxide Cl_(2(g))+2NO_((g))to2NOCl_((g)) n doubling the concentration of both reactants,the rate of the reaction increases by a factor of 8.However,if only the concentration of Cl_(2) is doubled ,the rate increases by a factor of 2.The order of this reaction with respect to NO is....
Answer» 0 1 2 3 Solution :r=K`[Cl_(2)]^(x)[NO]^(y)` …..(i) If we double the concentration of both,rate of reaction will increase by EIGHT time. 8r=K`[2Cl_(2)]^(x)[2NO]^(y)` …….(III)Dividing equation (II) by (iii) `THEREFORE 4=2^(y)therefore 2^(2)=2^(y)therefore y=2`

Discussion

26.	Consider the reaction below at 298K, C("graphite")+2H_(2)(g) to CH_(4) (G) triangle_(f) H^(@) ("Kj/mol") -74.9 S_(m)^(@) ("J/mol-K")+5.6 +130.7 +186.3 Which statement below is correct?
Answer» `triangle_(r)G^(@)` is -50.8 KJ and the REACTION is DRIVEN by ENTHALPY only `triangle_(r)G^(@)` is -50.8 kJ and the reaction is driven by entropy only `triangle_(r)G^(@)` is +50.8 kJ and the reaction is driven by both enthalpy and entropy `triangle_(r)G^(@)`is -50,8 kJ and the reaction is driven by both enthalpy and entropy Answer :A

Discussion

27.	Consider the reaction AtoP. The change in concentration of A with time has been shown graphically. i) Predict the order of the reaction. ii) Derive an expression for the time required for the completion of the reaction.
Answer» Solution :i) The REACTION is of first order. ii) for the DERIVATION of the EXPRESSION

Discussion

28.	Consider the reaction at 300 K H_(2)(g)+Cl_(2)(g) to 2HCl(g)""DeltaH = - 185kJ//molCalculate DeltaU if 3mole of H_(2)completely react with 3 mo,le f of Cl_(2) tofrom HCl.
Answer» 0 `-185 kJ` `555kJ` None of these Solution :`Deltan_(g) = 0rArr DeltaH= DELTAU = - 185 kJ//"MOLE"` For three moles= - `185 XX 3 = -555 `

Discussion

29.	Consider the reaction AtoB .The concentration of both the reactants and the products varies exponentially with time.Which of the following figures correctly describes the changes in concentration of reactant and products with time?
Answer» Solution :CONCENTRATION of reactant.(A) decrease EXPONENTIALLY .Where as that of product (B) increases exponentially with time Hence ,(B)is the correct OPTION.

Discussion

30.	Consider the reaction Altimplies B. The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time?
Answer» ANSWER :B

Discussion

31.	Consider the reaction : AgCl (s) hArr Ag^(+) (aq) + Cl^(-) (aq) At 27^(@)C K_(C) = 1.6 xx 10^(-9) for this reaction. If final solution has a volume of 50 litre then what minimum millimoles of AgCl (s)is needed to achieve equilibrium.
Answer» Solution :`{:(,AgCl (s),hArr,Ag^(+) (aq),+,Cl^(-) (aq),),("mole",-,,x,,x,),("at eqm.",,,,,,):}` `K_(C) = 16 XX 10^(-10) = (x)/(50) xx (x)/(50` `n_(Ag^(+)` at equilibrium `= x = 50 xx 4 xx 10^(-5)` moles `= 2 xx 10^(-3)` moles minimum NUMBER `n_(AgCl)` required to achieve equilibrium `= 2 xx 10^(-3)` moles achieve equilibrum = 2 millimoles

Discussion

32.	Consider the reaction aFeS_(2) + bO_(2) to cFe_(2)O_(3) + dSO_(2)Which is correct for the above reaction
Answer» n-factor for `FeS_(2)` is 11 The RATIO of moles of a:b is 4:11 The ratio of moles of a:b is 11:4 The ratio of moles of C:d is 1:4 Answer :A::B::D

Discussion

33.	Consider the reaction A to B. The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time ?
Answer» SOLUTION :Concentration of reactant, [A] decreases EXPONENTIALLY WHEREAS that of PRODUCT, [B] increases exponentially with time. HENCE, (b) is the correct option.

Discussion

34.	Consider the reaction, @A+B to products. When concentration of B alone was doubled the half life did not change. When the concentraction of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is
Answer» `s^(-1)` `LMOL^(-1)s^(-)` no units `molL^(-)s^(-)` Solution :Since DOUBLING the concentration of B does not change half-life, the REACTION is of ORDER w.r.t B Order of reaction with respect to A=1 because of reaction doubles when concentration of A is doubled keeping concentration of B constant. `therefore` Order of reaction=1+1=2 and units of second order reaction are `Lmol^(-1)sec^(-1)`.

Discussion

35.	Consider the reaction A H_(2)O_(2)+2HIrarrI_(2)+2H_(2)O B HOCI+H_(2)O_(2)rarrH_(3)O^(+)+CI^(-)+O_(2) which of the following statement is correct about H_(2)O_(2) with referenceto thesereactions hydrogen peroxide is ……………………
Answer» 1. an oxidising agent in both a and b 2. an oxidising agent in a nad reducing agentin b 3. a reducing agnet in a and oxidising agentin b 4. a reducing agent in both a and b ANSWER :B

Discussion

36.	Consider the reaction: ""_(92)^(238)U+""_(0)^(1)nto""_(92)^(239)Mto""_(93)^(239)Np+""_(-1)^(0)e Mis :
Answer» U Cf Cm Tc Answer :A

Discussion

37.	Consider the reaction : 4NO_(2)(g) + O_(2)(g) rarr 2N_(2)O_(5)(g), Delta_(f)H = -111 kJ If N_(2)O_(5)(s) is fromed instead of N_(2)O_(5)(g) in the above reaction, the Delta_(f)H is value will be (given, Delta H of sublimation for N_(2)O_(5) is 54 kJ mol^(-1)).
Answer» `+ 54 kJ` `+ 219 kJ` `- 219 kJ` `- 165 kJ` Solution :`4NO_(2) + O_(2) rarr 2N_(2)O_(5) (g) Delta H = -111 kJ` `2N_(2)O_(5) (s) rarr 2N_(2)O_(5) (g) Delta H = 54 xx 2 kJ` SUBTRACTING `4NO_(2) + O_(2) rarr 2N_(2)O_(5) (s) Delta H = - 219 kJ`

Discussion

38.	Consider the reaction : 4 NO_(2 (g)) + O_(2 (g)) to 2 N_(2)O_(5 (g)) , Delta_(r) H = -111 kJ If N_(2) O_(5 (s)) is formed instead of N_(2)O_(5 (g)) in the above reaction , the Delta_(r)H value will be : (given , Delta H of sublimation for N_(2) O_(5) is 54 kJ mol^(-1))
Answer» `+54` KJ `+ 219` kJ `-219` kJ `-165 ` kJ Answer :C

Discussion

39.	Consider the reaction 2NO_((g))+O_(2(g))rarr2NO_(2(g)). If (d[NO_(2)])/(dt)=0.052 M/s then, -(d[O_(2)])/(dt) will be
Answer» 0.052 M/s 0.114 M/s 0.026 M/s `-0.026 M//s` ANSWER :B::C

Discussion

40.	Consider the reaction, 2Cl_(2)(g) + 2H_(2)O(g)hArr4HCl(g) + O_(2)(g) DeltaH^(O) = + 113 kJ The four gases, Cl_(2), H_(2)O, HCl and O_(2), are mixed and the reaction is allowed to come to equilibrium. State and explain the effect (increase, decrease, no change) of the operation in the left column (below) on the equilibrium value of the quantity in the right column. Each operation is to be considered separately. Temperature and volume are constant unless stated otherwise. Raising the temperature Concentration of HCl
Answer» ANSWER :INCREASE

Discussion

41.	Consider the reaction, 2Cl_(2)(g) + 2H_(2)O(g)hArr4HCl(g) + O_(2)(g) DeltaH^(O) = + 113 kJ The four gases, Cl_(2), H_(2)O, HCl and O_(2), are mixed and the reaction is allowed to come to equilibrium. State and explain the effect (increase, decrease, no change) of the operation in the left column (below) on the equilibrium value of the quantity in the right column. Each operation is to be considered separately. Temperature and volume are constant unless stated otherwise. Adding catalyst Number of moles of HCl
Answer» ANSWER :no CHANGE

Discussion

42.	Consider the reaction, 2Cl_(2)(g) + 2H_(2)O(g)hArr4HCl(g) + O_(2)(g) DeltaH^(O) = + 113 kJ The four gases, Cl_(2), H_(2)O, HCl and O_(2), are mixed and the reaction is allowed to come to equilibrium. State and explain the effect (increase, decrease, no change) of the operation in the left column (below) on the equilibrium value of the quantity in the right column. Each operation is to be considered separately. Temperature and volume are constant unless stated otherwise. Adding He Number of moles of HCl
Answer» ANSWER :no CHANGE

Discussion

43.	Considerthereaction2NO_((g) ) +O_(2(g) )to2NO_(2(g)).if(d [NO_(2)])/(dt)=0.052 M//sthen-(d[O_(2)])/(dt) will be
Answer» `0.052 M//s` `0.114M//s` `0.026M//s` `-0.026 M//s` SOLUTION :`"RATE"=-(1)/(2)(d[NO])/(DT)=-(d[O_(2)])/(dt)=+(1)/(2) (d[NO_(2)])/(dt)` `-(d[O_(2)])/(dt)=(1)/(2)xx0.052 =0.026 m//s`

Discussion

44.	Consider the reaction, 2Cl_(2)(g) + 2H_(2)O(g)hArr4HCl(g) + O_(2)(g) DeltaH^(O) = + 113 kJ The four gases, Cl_(2), H_(2)O, HCl and O_(2), are mixed and the reaction is allowed to come to equilibrium. State and explain the effect (increase, decrease, no change) of the operation in the left column (below) on the equilibrium value of the quantity in the right column. Each operation is to be considered separately. Temperature and volume are constant unless stated otherwise. Raising the temperature K_(C)
Answer» ANSWER :INCREASE

Discussion

45.	Consider the reaction, 2Cl_(2)(g) + 2H_(2)O(g)hArr4HCl(g) + O_(2)(g) DeltaH^(O) = + 113 kJ The four gases, Cl_(2), H_(2)O, HCl and O_(2), are mixed and the reaction is allowed to come to equilibrium. State and explain the effect (increase, decrease, no change) of the operation in the left column (below) on the equilibrium value of the quantity in the right column. Each operation is to be considered separately. Temperature and volume are constant unless stated otherwise. Decreasing the volume of the container K_(C)
Answer» ANSWER :no CHANGE

Discussion

46.	Consider the reaction, 2Cl_(2)(g) + 2H_(2)O(g)hArr4HCl(g) + O_(2)(g) DeltaH^(O) = + 113 kJ The four gases, Cl_(2), H_(2)O, HCl and O_(2), are mixed and the reaction is allowed to come to equilibrium. State and explain the effect (increase, decrease, no change) of the operation in the left column (below) on the equilibrium value of the quantity in the right column. Each operation is to be considered separately. Temperature and volume are constant unless stated otherwise. Decreasing the volume of the container Partial pressure of Cl_(2)
Answer» ANSWER :INCREASE

Discussion

47.	Consider the reaction, 2Cl_(2)(g) + 2H_(2)O(g)hArr4HCl(g) + O_(2)(g) DeltaH^(O) = + 113 kJ The four gases, Cl_(2), H_(2)O, HCl and O_(2), are mixed and the reaction is allowed to come to equilibrium. State and explain the effect (increase, decrease, no change) of the operation in the left column (below) on the equilibrium value of the quantity in the right column. Each operation is to be considered separately. Temperature and volume are constant unless stated otherwise. Adding O_(2) Number of moles of H_(2)O
Answer» ANSWER :INCREASE

Discussion

48.	Consider the reaction, 2Cl_(2)(g) + 2H_(2)O(g)hArr4HCl(g) + O_(2)(g) DeltaH^(O) = + 113 kJ The four gases, Cl_(2), H_(2)O, HCl and O_(2), are mixed and the reaction is allowed to come to equilibrium. State and explain the effect (increase, decrease, no change) of the operation in the left column (below) on the equilibrium value of the quantity in the right column. Each operation is to be considered separately. Temperature and volume are constant unless stated otherwise. Adding O_(2) Number of moles of HCl
Answer» ANSWER :DECREASE

Discussion

49.	Consider the reaction, 2Cl_(2)(g) + 2H_(2)O(g)hArr4HCl(g) + O_(2)(g) DeltaH^(O) = + 113 kJ The four gases, Cl_(2), H_(2)O, HCl and O_(2), are mixed and the reaction is allowed to come to equilibrium. State and explain the effect (increase, decrease, no change) of the operation in the left column (below) on the equilibrium value of the quantity in the right column. Each operation is to be considered separately. Temperature and volume are constant unless stated otherwise. Decreasing the volume of the container Number of moles of Cl_(2)
Answer» ANSWER :INCREASE

Discussion

50.	Consider the reaction, 2Cl_(2)(g) + 2H_(2)O(g)hArr4HCl(g) + O_(2)(g) DeltaH^(O) = + 113 kJ The four gases, Cl_(2), H_(2)O, HCl and O_(2), are mixed and the reaction is allowed to come to equilibrium. State and explain the effect (increase, decrease, no change) of the operation in the left column (below) on the equilibrium value of the quantity in the right column. Each operation is to be considered separately. Temperature and volume are constant unless stated otherwise. Increasing the volume of the container Number of moles of H_(2)O
Answer» ANSWER :DECREASE

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Consider the reaction M_((aq))^(n+)+n e^(-)toM_((s)). The standard reduction potential values of the element M_(1),M_(2) and M_(3) are -0.34V,-3.05 and -1.66V respectively. The order of their reducing power will be:-

Consider the reaction, N_(2) + 3H_(2) rarr 2NH_(3) carried out ata constant temperature and pressure. If DeltaH and Delta U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

Consider the reaction HCN_((aq))hArrK_((aq))^(+)+CN_((aq))^(-). At equlibrium, the addition of CH_((aq))^(+) concentration

Consider the reaction : H_(2)(g),Pt|H^(+)(aq) , E=0.1" V" The pH of the solution is

Consider the reaction given below : CH_(3)-CHDCONH_(2)+PhCOoverset(15)NH_(2)underset(Delta)overset(Br_(2)+OH^(-))to The products obtained are :

Consider the reaction equlibreium, 2SO_(2(g))+O_(2(g))hArr2SO_(3(g)),DeltaH^(@)-=198kJ. On the basis of Le-Chatelier's pronciple, the condition favourable for the forward reaction is

Consider the reaction equilibrium underset(("Greater volume"))("Ice")hArrunderset(("Lesser volume"))("Water-xkcal")The favourable conditions for forward reaction are

Consider the reaction, C(s)+O_(2)(g)toCO_(2)(g)+393.5kJ, the signs of DeltaH,DeltaS and DeltaG respectively are

Consider the reaction Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)to2Cr^(3+)+7H_(2)O What is the quantity of electricity in couolombs needed to reduce 1 mole of Cr_(2)O_(7)^(2-)?

Consider the reaction :Cr_(2)O_(7)^(2-) + 14H^(+) + 6e^(-) to 2Cr^(3+) + 7H_(2)O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr_(2)O_(7)^(2-) ?

Consider the reaction: Cr_(2)O_(2)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr_(2)O_(7)^(2-) ?

consider the reaction CH_(3)CH_(2)CH_(2)Br+NaCNtoCH_(3)CH_(2)CH_(2)CN+NaBr. the reaction will be fastest in.

Consider the reaction : CH_(3)CH_(2)CH_(2)Br+NaCNrarrCH_(3)CH_(2)CH_(2)CN+NaBr This reaction will be the fastest in :

Consider the reaction CH_(3)CH_(2)CH_(2)Br+ NaCN to CH_(3)CH_(2)CH_(2)CN +NaBr This reaction will be the fastest in -

Consider the reaction, CH_(3)CH_(2)CH_(2)Br+NaCn to CH_(3)CH_(2)CH_(2)CN+NaBr This reaction will be the faster in

consider the reaction: CH_(3)CH_(2)CH_(2)Br + NaCN to CH_(3)CH_(2)CH_(2)CN + NaBr This reaction will be fastest in:

Consider the reaction: CH_(3)-CH_(2)-CH_(2)-Br+NaCNtoCH_(3)CH_(2)CH_(2)-C-=N+NaBr The correct statement is :-

Consider the reaction C_(3)H_(7)-OH+Et_(3)O^(o+)BF_(4)^(-)rarr C_(3)H_(7)-O-Et+EtOEt+HBF_(4). Which of the following statements si wrong ?

Consider the reaction : Briefly explain the choice of major product.

Consider the reaction C_(2)H_(5)Cl+AgCN overset(ErOH)rarr X

Consider the reaction below at 298K, C("graphite")+2H_(2)(g) to CH_(4) (G) triangle_(f) H^(@) ("Kj/mol") -74.9 S_(m)^(@) ("J/mol-K")+5.6 +130.7 +186.3 Which statement below is correct?

Consider the reaction AtoP. The change in concentration of A with time has been shown graphically. i) Predict the order of the reaction. ii) Derive an expression for the time required for the completion of the reaction.

Consider the reaction at 300 K H_(2)(g)+Cl_(2)(g) to 2HCl(g)""DeltaH = - 185kJ//molCalculate DeltaU if 3mole of H_(2)completely react with 3 mo,le f of Cl_(2) tofrom HCl.

Consider the reaction AtoB .The concentration of both the reactants and the products varies exponentially with time.Which of the following figures correctly describes the changes in concentration of reactant and products with time?

Consider the reaction Altimplies B. The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time?

Consider the reaction : AgCl (s) hArr Ag^(+) (aq) + Cl^(-) (aq) At 27^(@)C K_(C) = 1.6 xx 10^(-9) for this reaction. If final solution has a volume of 50 litre then what minimum millimoles of AgCl (s)is needed to achieve equilibrium.

Consider the reaction aFeS_(2) + bO_(2) to cFe_(2)O_(3) + dSO_(2)Which is correct for the above reaction

Consider the reaction A to B. The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time ?

Consider the reaction, @A+B to products. When concentration of B alone was doubled the half life did not change. When the concentraction of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

Consider the reaction A H_(2)O_(2)+2HIrarrI_(2)+2H_(2)O B HOCI+H_(2)O_(2)rarrH_(3)O^(+)+CI^(-)+O_(2) which of the following statement is correct about H_(2)O_(2) with referenceto thesereactions hydrogen peroxide is ……………………

Consider the reaction: ""_(92)^(238)U+""_(0)^(1)nto""_(92)^(239)Mto""_(93)^(239)Np+""_(-1)^(0)e Mis :

Consider the reaction : 4NO_(2)(g) + O_(2)(g) rarr 2N_(2)O_(5)(g), Delta_(f)H = -111 kJ If N_(2)O_(5)(s) is fromed instead of N_(2)O_(5)(g) in the above reaction, the Delta_(f)H is value will be (given, Delta H of sublimation for N_(2)O_(5) is 54 kJ mol^(-1)).

Consider the reaction : 4 NO_(2 (g)) + O_(2 (g)) to 2 N_(2)O_(5 (g)) , Delta_(r) H = -111 kJ If N_(2) O_(5 (s)) is formed instead of N_(2)O_(5 (g)) in the above reaction , the Delta_(r)H value will be : (given , Delta H of sublimation for N_(2) O_(5) is 54 kJ mol^(-1))

Consider the reaction 2NO_((g))+O_(2(g))rarr2NO_(2(g)). If (d[NO_(2)])/(dt)=0.052 M/s then, -(d[O_(2)])/(dt) will be

Considerthereaction2NO_((g) ) +O_(2(g) )to2NO_(2(g)).if(d [NO_(2)])/(dt)=0.052 M//sthen-(d[O_(2)])/(dt) will be