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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 redox reaction x MnO_(4)^(-) + yH_(2)C_(2)O_(4) + zH^(+)to m Mn^(2+) +n CO_(2)+ p H_(2)O The valeu of x, y, m and n are |
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Answer» 10, 2, 5 ,2 |
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| 2. |
For the redox reaction : MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+)toMn^(2+)+CO_(2)+H_(2)OThe correct coefficients of the reactants in the balanced equation are : |
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Answer» `{:(,MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(," "2," "5,16):}` |
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| 3. |
For the redox reaction MnO_(4)^(-)+C_(2)O_(4)^(-2)+H^(+) rarr Mn^(2+)+CO_(2)+H_(2)O the correct coefficients of the reactants for the balanced reaction are |
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Answer» `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,5,16):}` `(c_(2)O_(4)^(2-) rarr 2 CO_(2) + 2e^(-) xx 5)/(2MnO_(4)^(-) + 5C_(2)O_(4)^(2-) + 16H^(+) rarr 2MN^(2+) + 10CO_(2) + 8H_(2)O)` Thus the coefficient of `MnO_(4)^(-), C_(2)O_(4)^(2-) and H^(+)` in the abovve balanced equation respectively are 2, 5, 16 |
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| 4. |
For the redox reaction: MnO_4 +C_2O_4^(2-) + H^+ rarr Mn^(2+) + CO_2 + H_2O The number of mole of permanganate ion required per mole of oxalate ion for completion of the reaction is |
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Answer» `1/5` |
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| 5. |
For the redox reaction MnO_(4)^(-) + C_(2)O_(4)^(2-) + H^(+) rarr Mn^(2+) + CO_(2)+ H_(2)O The correct coefficients of the reactants for the balanced reaction are |
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Answer» `{:(MnO_4^-,C_2O_4^(2-),H^+),(2,16,5):}` |
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| 6. |
For the redox reaction: Fe^(2+)+ Cr_2O_7^(2-) +H^+ rarrFe^(3+) + Cr^(3+) + H_2O The correct coefficients of the reactants for the balanced reaction are |
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Answer»
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| 7. |
For the red ox reaction : Cr_2O_(7)^(2-) +I^(-) +H^(+) rarr Cr^(3+) + I_2 + H_2O the correct coefficients of the reactants for the balanced equation are |
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Answer» `{:(Cr_(2)O_(7)^(2-),I^(-),H^(+)),(1,3,14):}` `[2I^(-) rarr I_2 +2E^(-)] x 3` `Cr_2O_(7)^(2-) + 6I^(-) +14H^(+) rarr 2Cr^(3+) +3I_2 +7H_2O` |
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| 8. |
For the reation given below, identify X in the reaction CH_3C -= CCH_3 overset(X)rarr CH_3COCOCH_3 |
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Answer» `K_2Cr_2O_7//H_2SO_4` |
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| 9. |
For the reacton C_((s))+CP_(2(g))to2CO_((g)),K_(p)=63 atm at 1000 K. If at equlibrium : Pco=10Pco_(2), then the total pressure of the gases at equlibrium is |
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Answer» `6.3` atm APPLY low of mass ACTION, `K_(p)=((P_(CO))^(2))/(P_(CO_(2)))or 63=((10P_(CO_(2)))^(2))/(P_(CO_(2)))` or `63=(100(P_(CO_(2)))^(2))/(P_(CO_(2)))or 63=100P_(CO_(2))` `P_(CO_(2))36/100=0.63atm` `P_(Total)=P_(CO_(2))+P_(CO)=0.63+6.3=6.93 ` atm. |
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| 10. |
For the reactuon A+Bto Products (-d[A])/(dt)=x.e^(-E_(a)//RT) ,what is ? |
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Answer» COLLISION frequency |
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| 11. |
For the reaction,X_(2)O_(4)(l) rarr2XO_(2)(g) ""DeltaU=2.1 kcal , DeltaS=20 cal K^(-1) at 300 K Hence, DeltaG is |
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Answer» 9.3 KCAL `DeltaG=DeltaH-TDeltaS=-2.700 kcal` |
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| 12. |
For the reactions {:(MnO_(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2+)4H_(2)O","E^(o) = + 1.51 V),(MnO_(2)+4H^(+)+ 2e^(-) rarr Mn^(2+)+2H_(2)O"," E^(o)= + 1.23 V ):} then for the reaction : MnO_(4)^(-) + 4H^(+) + 3e^(-) rarr MnO_(2)+2H_(2)O","E^(o) |
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Answer» `1.70` V |
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| 13. |
For the reactions, (i) H_(2)(g)+Cl_(2)(g)rarr 2HCl(g)+ xKJ (ii) H_(2)(g)+Cl_(2)(g)rarr 2HCl(l)+ yKJ Which one of the following statement is correct : |
| Answer» Answer :B | |
| 14. |
For the reactions: (i) H_(2)(g) + Cl_(2)(g) to 2HCl (g) + xkJ (ii) H_(2)(g) + Cl_(2)(g) to 2HCl(g) + ykJ Which one of the following statements is correct? |
| Answer» ANSWER :B | |
| 15. |
For the reactions A rarr B, DeltaH = + 24 kJ//mol and B rarr C, DeltaH = - 18kJ//mol, the decreasing order of enthalpy of A, B, C follows the order |
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Answer» A, B, C `implies H_(B)-H_(A)=+24 "...(i)"` `BrarrC, DeltaH=-18 kJ//mol` `implies H_(C)-H_(B)=-18` `implies H_(B)-H_(C)=+18"...(II)"` From EQS.(i) and (ii), we have `H_(C)-H_(A)=6` `thereforeH_(B)gtH_(C)gtH_(A)`. |
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| 16. |
For the reactionN_2 (g) + 3H_2 (g) to 2NH_3 (g) How is the rate of formation of ammonia related to the rate of disappearance of H_2 ? |
| Answer» SOLUTION :RATE of REACTION =`-1/3 (DELTA[H_2])/(DELTAT)=1/2(Delta[NH_3])/(Deltat)` Or `(Delta[NH_3])/(Deltat)=-2/3(Delta[H_2])/(Deltat)` | |
| 17. |
For the reaction:[Cu(NH_(3))_(4)]^(2+)+H_(2)OhArr[Cu(NH_(3))_(3)H_(2)O]^(2+)+NH_(3) The net rate of reaction at any time is given by: rate =2.0xx10^(-4)[[Cu(NH_(3))_(4)]^(2+)][H_(2)O]-3.0xx10^(5)[[Cu(NH_(3))_(3)H_(2)O]^(2+)][NH_(3)] The correct statement is (are) |
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Answer» Rate constannt for forward reaction `=2XX10^(-4)` `K_(c)=(K_(f))/(K_(b))=(2xx10^(-4))/(3xx10^(-5))=0.666xx10^(-9)=6.66xx10^(-10)` |
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| 18. |
For the reaction:[Cu(NH_3)_4]^(2+)+H_2Orarr[Cu(NH_3)_3H_2O^(2++)NH_3the net rate of reaction at any time is given by :net rate =2.0xx10^(-4)[[Cu(NH_3)_4]^(2+)-3.0xx10^5[[Cu(nh_3)_3H_2O]^(2+)].[NH_3]Then correct statement is (are): |
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Answer» RATE CONSTANT for forward reaction`=2xx10^4` |
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| 19. |
For the reactionA + B ⇌ C + D the initial cocentration pf A and B are equal but the equiliberium concentration of C is twice that of equiliberium concentration of A. Find the value of the equiliberium constant . |
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Answer» 4 |
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| 20. |
For the reaction,C_2H_5OH+HX overset(ZnX_2)rarr C_2H_5X +H_2O the reactivity order for halogen, acid is : |
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Answer» HBR `GT` HI `gt`HCI |
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| 21. |
For the reaction2NO_2+F_2rarr2NO_2F , following mechanism has been provided: NO_2+F_2overset(slow)rarrNO_2F+F NO_2+Foverset(FAST)rarrNO_2F Thus rate expression of the above reaction can be written as: |
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Answer» `r=k[NO_2]^2[F_2]` |
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| 22. |
For the reaction Zn(s)+Cu^(2+) to Zn^(2+)(aq)+Cu(s), Nernest eqaution at 25^(@)C is |
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Answer» `E=E^(@)+(0.059)/(2)log""([Zn_((aq))^(2+)])/([Cu_((aq))^(2+)])` |
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| 23. |
For the reaction ,XA+YBtoZC, if (-d[A])/(dt)=(-d[B])/(dt)=(1.5d[C])/(dt), then the correct statement among the following is…….. |
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Answer» the VALUE of X=Y=Z=3 |
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| 24. |
For the reaction X_2 Y_4 (l) to 2XY_2(g)at 300 K the values of Delta U and Delta Sare 2 kcal and 20 cal K^(-1)respectively. The value of Delta G for the reaction is |
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Answer» -3400 CAL ` Delta H = 2000 + 2 XX 2 xx 300` `= 3200 cal` `Delta G = Delta H - T Delta S = 3200 - (300 xx 20) = - 2800 cal` |
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| 25. |
For the reaction X2 + 2Y– → Y2 + 2X–, X and Y may not be |
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Answer» If X = F than Y = CL, BR or I ( |
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| 26. |
For the reaction X+Y to Z, it is found that doubling the concentration of X doubles the rate and doubling the concentration of Y again doubles the reaction rate. What is the overall order of the reaction? |
| Answer» Answer :B | |
| 27. |
For thereaction X to Y +Zif theinitial concentration of Xwas reduced form2M to 1M in 20 min and from 1M to 0.25M in 40min , find the order . |
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Answer» Solution :Half-life, `(t_(1//2))alphaa^(1-n)` Here, half-life is INDEPENDENT on the INITIAL concentration. The order of the REACTION is `1`. |
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| 28. |
For the reaction X-Y, the concentration of X are 1.2 M, 0.6M, order of reaction is: |
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Answer» Zero `t_(1//2) = 1HR`. For the nth order REACTION. `(t_(1//2))_(1)/(t_(1//2))_(2) = ([A_(0)]_(2)^(n-1))/([A_(0)]_(1)^(n-1))` `1=(6/3)^(n-1)` or `(2)^(0) = (2)^(n-1)` n-1=0 or n=1 Order of reaction is one. |
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| 29. |
For thereaction, theconcentrationof thereactantwasreducedfrom0.1 Mto 0.05 M in6 hrs . Andfrom0.05 Mto 0.025in 12 hrs. the orderof reaction is |
| Answer» Answer :B | |
| 30. |
For the reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by |
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Answer» With INCREASE in TEMPERATURE, the VALUE of K for endothermic reaction increases because unfavourable CHANGE in entropy of the surroundings decreases. |
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| 31. |
For the reaction 2NO(g)+O_(2)(g)to2NO_(2(g)) volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to O_(2) and second order with respect to NO, the rate of reaction will |
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Answer» DIMINISH to one - EIGHTH of its INITIAL value |
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| 32. |
For the reaction system, 2NO(g)+O_(2)(g) to 2NO_(2)(g) volume issuddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to O_(2) and second order with respect to No, the rate of reaction will |
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Answer» diminish to one-EIGHTH of its initial value `thefore` Newrate`=k[2O_(2)][2NO]^(2)=8k[O^(2)][NO]^(2)` The new RATE increase to eight times of its initial. |
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| 33. |
For the reaction, SO_(2(g))+(1)/(2)O_(2(g))iffSO_(3(g)) if we write K_(p)=K_(c)(RT)^(x), then x becomes |
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Answer» <P>`-1` |
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| 34. |
For the reaction : SO_(2)(g)+(1)/(2)O_(2)(g)hArr SO_(3)(g)If K_(p)=K_(c ) (RT)^(x) where the symbols have usual meaning, then the value of x is : (assuming ideality) |
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Answer» <P>`-1` `Delta n_(g)=1-(1+(1)/(2))=-(1)/(2)` `therefore K_(p)=K_(C )(RT)^(-1//2)` `therefore x = -1//2` |
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| 35. |
For the reaction SO_2(g)+1/2O_2(g) |
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Answer» 1 For the GIVEN reaction, `/_\n=x=1-(1+1/2)=-1/2` |
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| 36. |
For the reaction SO_(2)+1/2O_(2)hArrSO_(3), if we write K_(p)=K_(c)(RT)^(x), then x becomes |
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Answer» <P>`-1` |
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| 37. |
For the reaction RtoP ,the concentration of a reactant changes from 003 M to 0.02 M in 25 minutes.Calculate the average rate of reaction using units of time both in minutes and seconds. |
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Answer» Solution :Given Reaction R`to`P for this reaction AVERAGE rate `r_(AV)=-(Delta[R])/(Deltat)` `Delta[R]=[R_(2)]-[R_(1)]` +`4.0xx10^(-4) mol L^(-1) m^(-1)` The CALCULATION of rate in 1 second `Deltat=(t_(2)-t_(1))` `=-0.01 mol L^(-1)` So, average rate `r_(av)=-(Delta[R])/(Delta)` `=-(-0.01 mol L^(-1))/(150s)` `+6.667xx10^(-5) mol L^(-1) S^(-1)` OR `r_(av)=(r_(av)mol L^(-1) m^(-1)xx1M)/(60 s^(-1))` `=(4.0xx10^(-4))/(60) mol L^(-1)s^(-1)` `=6.667xx10^(-5) mol L^(-1) s^(-1)` |
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| 38. |
For the reaction RtoP, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. |
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Answer» Solution :Average rate `=-(Delta[R])/(DELTAT)=-([R]_(2)-[R_(1)])/(t_(2)-t_(1))=-(0.02" M"-0.030" M")/(25" MIN")=-(-0.01" M")/(25" min")=4xx10^(-4)" M min"^(-1)` or `=-(-0.01" M")/(25xx60s)=6.66xx10^(-6)" Ms"^(-1)` |
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| 39. |
For the reaction R-X+OH^(-)toR-OH+X^(-) the rate reaction is given as, rate=4.7xx10^(-5)[R-x][OH^(-)]+.024xx10^(-5)[RX] what percentage of R-X react by S_(N^(2)) mechanism when [OH^(-)]=0.001 molar. |
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Answer» |
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| 40. |
For the reaction R-X+OH^(-)toR-OH+X^(-1), the rate of reaction is given as, rate =4.7xx10^(-5)[R-X][OH^(-)]+0.24xx10^(-5)[RX]. What percentage of R-X react by S_(N)2 mechanism when [OH^(-)]=0.001 molar? |
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Answer» 1.9 |
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| 41. |
For the reaction R to P,the concentration of a reactant change from 0.03 M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time and seconds. |
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Answer» SOLUTION :Average rate =`-(Delta[R])/(DELTAT)=-([R_2]-[R_1])/(Deltat)=-(0.02-0.03)/25` `=0.01/(25 MIN)=4xx10^(-4)moL^(-1)min^(-1)` Average rate`=0.01/(25xx60s)=6.66xx10^(-6)molL^(-1)s^(-1)` |
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| 42. |
For the reaction R to P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate average rate of reaction. |
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Answer» SOLUTION :Formula : AVERAGE rate` = - (Delta|R|)/(DELTAT) = (-[-0.02- 0.03])/(25) = 4 xx 10^(-4) "MOL "L^(-1) min^(-1)` |
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| 43. |
For the reaction R to P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. |
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Answer» Solution :Average rate `=-(DELTA[R])/(Deltat)=-([R]_(2)-[R_(1)])/(t_(2)-t_(1))` `= -(0.02M-0.030M)/(25"minutes")= -(0.01M)/("25 minutes")=4xx10^(-4)" MOL litre"^(-1)" minutes"^(-1)` `=(-0.1M)/(25xx60s)=6.66xx10^(-6)" mol litre"^(-1)" SECOND"^(-1)`. |
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| 44. |
For the reaction R to P, a graph of [R] against time is found to be a straight line with negative slope. What is the order of reaction ? |
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Answer» SECOND ORDER Thus, plot of [A] vs t is straight line with NEGATIVE slope. |
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| 45. |
For the reaction R to P, half-life (t_(1//2)) is observed to be independent of initial concentrtion of the reactants. What is the order of the reaction? |
| Answer» SOLUTION :FIRST ORDER REACTION. | |
| 46. |
For the reaction R rarr P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 min. Calculate the average rate of reaction using units of time in seconds. |
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Answer» `6.66 xx 10^(-5) Ms^(-1)` AVERAGE rate of reaction `("Change in concentration of reactant or PRODUCT")/("Time taken")` `=(-DELTA[R])/(t)=-([R_(2)]-[R_(1)])/(t)=-((0.02-0.03)M)/((25xx60)s)` `=6.6 xx 10^(-6)Ms^(-1)` |
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| 47. |
For the reaction R rarrP,the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes . Calculate te average rate of reaction using units of time both in minutes and seconds . |
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Answer» Solution :Average rate `=(DELTA[R])/(Deltat)=([R]_2-[R]_1)/(t_2-t_1)` `=(0.02M-0.03M)/(25min)=(-0.01M)/(25min)=4XX10^(-4)"MIN"^(-1)` `=(-0.01M)/(25xx60)=(overset(..)uoverset(..)uoverset(..)u)xx-6-1` |
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| 48. |
For the reaction R - P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using of time both in minutes and second. |
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Answer» Solution :Average rate `= -(DELTA(R ))/(Delta t)=-([R]_(2)-[R]_(1))/(t_(2)-t_(1))` `=-(0.02 M - 0.03 M)/("25 min")=(-0.01 M)/("25 min")` `= 4xx10^(-4)"M min"^(-1)` and `=-(-0.01 m)/(25xx60)=6.66xx10^(-6)MS^(-1)` |
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| 49. |
For the reaction Pt// H_2 (1 atm) // H^(+) (aq) // // Cl^(-) (aq) // AgCl // Ag , K_(c) (equilibrium constant ) is represented as |
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Answer» `K_(c) = ([CL^(-)][AGCL])/([H^(+)][H_2])` |
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| 50. |
For the reaction, PCl_(5)(g)to PCl_(3)(g)+Cl_(2)(g), The forward reaction at constant temperature is favoured by |
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Answer» Introducing an inert GAS at constant volume |
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