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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 a reaction Ato Product, half life is 50 min, when the initial concentration of A is 4 molar Half life of same reaction is 80 min, when the initial concentration of A is 2 molar. Then |
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Answer» Order is 1 `C_(0)to4""2` `((t_(1/2))_(1))/((t_(1/2))_(2))=((C_(2))/(C_(1)))^((n-1)),(40/80)=(2/4)^((n-1)),(1/2)^(2)=(1/2)^(n-1)` `1=n-1,n=1+1,n=2` `r=K.C^(2),M/t=K.M^(2)impliesK=M^(-1)"time"^(-1)="mole"^(-1),"lit"-"time"^(-1)` |
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| 2. |
For a reaction at equilibrium, the partial pressure of B is found to be one fourth of the partial pressure of A. The value of Delta G^(theta) of the reaction A rarr B is |
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Answer» `-RT log 4` :. `Delta G = - 2.303 RT log. (1)/(4)` `= 2.303 RT log 4` |
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| 3. |
For a reaction at TKDeltaH gt0 and DeltaS gt0. If the reaction attains equilibrium at a temperature of T_(1)K, (assume DeltaH and DeltaS are independent of temperature) then the reaction is spontaneous |
| Answer» Answer :B | |
| 4. |
For a reaction at 25^(@)C enthalpy change (DeltaH) and entropy change (DeltaS) are -11.7 xx 10^(3) "J mol"^(-1) and -105 "J mol"^(-1) K^(-1) respectively. Find out whether this reaction is spontaneous or not ? |
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Answer» |
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| 5. |
For a reaction at 25^@C enthalpy change (triangleH) and entropy change (triangleS) are-11.7xx10^3J mol^-1 and -105 J mol^-1 K^-1 respectively. The reaction is: |
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Answer» Spontaneous |
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| 6. |
For a reaction at 25^(@)C enthalpy change and entropy changes are -11.7 xx 10^(3) J mol^(-1)and -105 J mol^(-1)K^(-1) respectively. What is the Gibbs free energy |
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Answer» 15.05 KJ `=-11.7xx10^(3)-298xx(-105)=19590 J=19.59 kJ`. |
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| 7. |
For a reaction at 25^(@)C, enthalpy and entropy changes are -11.7xx10^(3)" J "mol^(-1) and -105" J "mol^(-1)K^(-1) respectively. What is the Gibb's free energy? |
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Answer» 15.05 kJ `DeltaG=DeltaH-TDeltaS=-11.7xx10^(3)-298xx(-105)` `=19590J=19.59kJ` |
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| 8. |
For a reaction: A(s) + 2B ^(+) rarr A^(2+) +2 B (s). K_(aq) has been found to be 10^(12) TheE^(o) cell is : |
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Answer» `0.354` VOLT |
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| 9. |
For a reaction Ararr products, the concentration of reactant are C_(0),aC_(0),a^(2)C_(0),a^(3)C_(0)…… after time interval 0, t, 2t…. Where 'a' is constant. Then : |
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Answer» REACTION is of `1^(st)` order and `K=((1)/(1))` LN a |
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| 10. |
For a reaction, AiffB equilibrium constant is 1.66 and k_("forward")=0.166 hr^(-1). Calculate the time (in hour) when concentration of B is 80% of its equilibrium concentration. (Given : In 25=3.20) |
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Answer» |
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| 11. |
For a reaction, Ararr B + C, activation energy is 15 kJ/mole and enthalpy of reaction is +5kJ/mole. The activation energy for the reaction B+Crarr A is |
| Answer» Answer :A | |
| 12. |
For a reaction , AhArrB,(d[A])/(dt)=5xx10^(-4)[B]-4xx10^(-3)[A]M "min"^(-1) Starting with only A at 0.1 M concentration calculate concentration of B after time t=9200sec. [Given:In 2=0.69] |
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Answer» `(0.8)/(9)M` |
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| 13. |
For a reaction : A(g) to nB(g) the rate constat is 6.93xx10^(-4) sec^(-1). the reaction is performed at constant pressure and tomperature of 24.63 atm and 300K starting with 1 mole of pure A. If concentration of B after 2000 sec is 3/(3.25)M then calculate the value of n |
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Answer» |
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| 14. |
For a reaction A(g) hArr B(g) at equilibrium . The partial pressure of B is found to be one fourth ofthe partial pressure of A . The value of DeltaG^(@) of the reaction A rarrB is |
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Answer» RT LN `4` `K_(P) = (P_(6))/(P_(A)) = (P_(A)//4)/(P_(A)) = (1)/(4)` At equilibrium , `DeltaG=O`. `DeltaG^(@) =-RT l nK_(P) =- RTInK_(P) =- RTIn (1)/(4) = RT In 4` |
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| 15. |
For a reaction A(g) iff B(g)Boat equilibrium, the partial pressure of B is found to be one fourth of the partial pressure of A. The value of Delta G^2 for the reaction A toBis |
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Answer» RT In 4 ` Delta G^@ = - RT LN K = - RT ln 1/4 = RT ln 4` |
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| 16. |
For a reaction, activation energy (E_a ) =0 and rate constant, k = 1.5 x 10^4 s^(-1) at 300 K. What is the value of rate constant at 320 K'? |
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Answer» `3.2 X 10^6 s^(-1)` |
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| 17. |
For a reaction , activation energy (E_(a) )= 0 and rate constant (K) = 3.2 xx 10^(6) s^(-1) at 300 K . What is the value of the rate constant at 310 K |
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Answer» `3.2 xx 10^(-12) s^(-1)` |
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| 18. |
For a reaction A to Products, starting with initial concentrations of 5xx10^(-3) M and25xx10^(-4)M, half-lives are found tobe 1.0 and 8.0 hour respectively. If we start with an initial concentration of 1.25xx10^(-3) M,what will be the half-lifeof the reaction ? |
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Answer» SOLUTION :For a REACTION of nth order, `t_(1//2)prop(1)/([A_(0)]^(n-1))` `:.((t_(1//2))_(1))/((t_(1//2))_(2))=([A_(0)]_(2)^(n-1))/([A_(0)]_(1)^(n-1))={([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)` `(1)/(8)=((25xx10^(-4))/(5xx10^(-3)))^(n-1)=((1)/(2))^(n-1)" or "((1)/(2))^(3)=((1)/(2))^(n-1)" or "n-1=3" or "n=4` APPLYING the above formula again forinitial concentration `1.25xx10^(-3)"M and "5xx10^(-3)" M",` `((t_(1//2))_(1))/(1)=((5xx10^(-3))/(1.25xx10^(-3)))^(4-1)=(4)^(3)=64" hours, "i.e.,t_(1//2)=64" hours"` |
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| 19. |
For a reaction a, A toProducts ,the unitsof rateconstantare givenas L mol^(-1) s^(-1) . Writethe rate expression . |
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Answer» SOLUTION :Units of RATE constant of the reaction `=("units of rate")/("(concentration)"^(n))` `"L mol"^(-1)s^(-1)=("mol L"^(-1)s^(-1))/("(mol L"^(-1)")"^(n))` Therefore the value of n is equal to 2. Rate EQUATION for the reaction is given as, `"rate = K "[A]^(2)`. |
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| 20. |
For a reaction A to C + D, initial concentration of A is 0.010 M. After 100s, the concentration of A is 0.01 M. The rate constant has numerical value 9.0. The reaction is of: |
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Answer» ZERO ORDER |
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| 21. |
For a reaction :A to B toProducts , the rate of the reaction at various concentrations are given below : The rate law for the above reaction is |
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Answer» A) `R = k [A]^(2) [B]` Rate = `k [A]^(alpha) [B]^(BETA)` From expt. No. 1 `2 = k[0.2]^(alpha) [0.2]^(beta) "" … (1)` From expt. No. 2 `4 = k [0.2]^(alpha) [0.4]^(beta) "" …. (2)` dividing eq. (1) and (2) we get `(2)/(4) = (k[0.2]^(alpha) [0.2]^(beta))/(k[0.2]^(alpha) [0.4]^(beta)) or (1)/(2) = [(1)/(2)]^(beta) implies beta = 1` From expt. No. 3 `36 k = [0.6]^(alpha) [0.4]^(beta) "" ... (3)` dividing eqn. (2) and (3) we get `(4)/(36) = (k[0.2]^(alpha) [0.4]^(beta))/(k[0.6]^(alpha) [0.4]^(beta))or (1)/(9) = [(1)/(3)]^(alpha) implies alpha= 2 ` `implies` Rate law for the given REACTION is `r = k [A]^(2) [B]^(1)` |
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| 22. |
For a reaction A to B + C, the initial concentration of A was reduced from 2M to 1 M in one hour and from 1 M to 0.25 M in two hours. What is the order of reaction? |
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Answer» |
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| 23. |
For a reaction A to B, the rate of reaction quadrupled when the concentration of A is doubled . The rate expressions of the reaction is r = K (A)^(n) , when the value of n is |
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Answer» 1 |
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| 24. |
For a reaction Ato2B, as time proceeds |
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Answer» [A] decreases, but [B] INCREASES |
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| 25. |
For a reaction A (s) + B^(2+) rarr B(s) + A ^(2+) ,at 25^(@) C E^(o) "" _(cell) = 0.2955 V.Hence, K_(eq) will be : |
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Answer» 10 |
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| 26. |
For a reaction A rarr B rarr C _(t1//2) for A & B are 4 and 2 minutes respectively. How much time would be required for the B to reach maximum concentration. |
| Answer» SOLUTION :`t=4` MIN | |
| 27. |
For a reaction : A rarr Products it is found that the rate of reaction increases by a factor of 6.25 when the concentration of A is increased by a factor of 2.5 . The order of the reaction with respect to A is : |
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Answer» `2.5 ` |
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| 28. |
For a reaction, A hArr P, the plots of [A[ and [P] with time at temperature T_(1) and T_2 are given ahead: . If T_(1) gt T_(1), the correct statement(s) is (are): (Assume DeltaH^(@) and DeltaS^(@) are independent of temperature and ratio of ln K at T_(1) to ln K at T_(2) as greater than (T_(2))/(T_(1)). Here, H,S,G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively. |
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Answer» `DeltaH^(@) LT 0, DeltaS^(@) lt0` `("ln "K_(1))/("ln "K_(2)) gt (T_(2))/(T_(1))""(Here" "T_(2)gtT_(1))` `T_(1)" ln "K_(1) gt T_(2)lnK_(2)` `RT_(1)lnK_(1)gtRT_(2)lnK_(2)`(It SHOWS that `DeltaG^(@)lt0)` `(-DeltaH^(@)-T_(1)DeltaS^(@))gt - (DeltaG^(@)-T_(2)DeltaS^(@))` `-DeltaH^(@)+T_(1)DeltaS^(@) gt -DeltaH^(@)+T_(2)DeltaS^(@)` Above expression will be correct ONY when `(DeltaH^(@) lt 0)` `T_(1)DeltaS^(@) gt T_(2)DeltaS^(@)` Given that `T_(2) gt T_(1)` `thereforeDeltaS^(@) lt0` |
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| 29. |
For a reaction : A rarr B If k_(1) and k_-1 are the rate constants for the forward and the backard reactions respectively then equilibrium constant of the reaction is given as : |
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Answer» `K = (k_(1))/(k_(-1))` |
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| 30. |
For a reaction a graph plotted between log(dx/dt)andlog(a-x) along y and x axes respectively shows a straight line with a positive slope of 45^(@). The order of the reaction is |
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Answer» 0 |
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| 31. |
For a reaction A+Bto products ,the rate law is ---- Rate =k[A][B]^((3)/(2)).Can the reaction be an elementary reaction?Explain. |
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Answer» SOLUTION :The reaction is not ELEMENTARY because ORDER of the reaction is fractional `1+(3)/(2)=(5)/(2)` Fractional order reaction can.t be elementary. |
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| 32. |
For a reaction A+Bto" Products, the rate law is : Rate "=k[A][B]^(3//2). Can the reaction be anelementary reaction ? Explain. |
| Answer» Solution :No, the given reaction cannot be an elementary reaction. This is because the elementary step of the SLOWEST step INVOLVES 1 molecule of A and `3//2` MOLECULES of B, i.e., FRACTIONAL. | |
| 33. |
For a reaction, A+Bto Product, the rate law is given by : r=k[A]^(1//2)[B]^(2) Whatis the order of the reaction ? |
| Answer» SOLUTION :ORDER of REACTION `=(1)/(2)+2=2(1)/(2)" or "2.5.` | |
| 34. |
For a reaction A+BtoProducts, rate law is -(d[A])/(dt)=k[A]_(0). The concentration of A left after time t when t=(1)/(k) is |
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Answer» `([A]_(0))/(e)` When `t = (1)/(k), [A] = [A]_(0)e""^(-k(1)/(2))[A]_(0)e^(-1)=([A]_(0))/(e)` |
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| 35. |
For a reaction A+Bto productss the rateof reactions is increased by 8 times when the concentration af A and B are doubled. The rate of reaction in doubled when the concentration of A alone is doubled. Then |
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Answer» Order of reaction with respect to A is 1 |
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| 36. |
For a reaction A + B to Product, the rate law is given by r = k[A]^(1//2) [B]^2. What is the order of the reaction ? |
| Answer» SOLUTION :ORDER of REACTION = 1/2+2=2.5 | |
| 37. |
For a reaction, A+B to Product, the rate is given by, r=k[A]^(1//2)[B]^(2). What is the order of the reaction? |
| Answer» SOLUTION :ORDER of the REACTION `=(1)/(2) +2=2(1)/(2) or 2.5`. | |
| 38. |
For a reactionA + B to C + D if the concentration of A is doubled without altering the concentration of B , the rate gets doubled . If the concentration of B is increased by nine times without altering the concentration of A , the rate gets tripled . The order of the reaction is |
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Answer» 2 r = K `[A]^(alpha) [B]^(beta) ""` … (i) 2r = k `[2A]^(alpha) [B]^(beta) "" … (ii)` 3R = k `[A]^(alpha) [9B]^(beta) ""` … (iii) Dividing eqn. (ii) by eqn. (i) `3 = 9^(beta)` or `3 = 3^(2 beta)` or , `2 beta = 1` or ` beta = 1//2` `therefore` Order of the REACTION = ` 1+ (1)/(2) = (3)/(2)`. |
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| 39. |
For a reaction : A+B rarr Products the rate of the reaction of various concentrations are given below :{:("Expt No.",[A], [B]," ""Rate (mol dm"^(-3)s^(-1)),(1,0.2,0.2,""2),(2,0.2,0.4,""4),(3,0.6,0.4,""36):} The rate law for the above reaction is : |
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Answer» `r= k[A]^(2)[B]` Therefore order with respect to [A] =2 Rate =`k[A]^(2) [B]^(1)` |
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| 40. |
For a reaction A + B rarr Products, the rate of the reaction was doubled when the concentration of A was doubled. When the concentration of A and B were doubled, the rate was again doubled, the order of the reaction with respect to A and B are: |
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Answer» 1,1 |
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| 41. |
For a reaction A +B rarr C, it is observed that half life of A (when B is taken in very large amount) is independent of concentration of A and the graph of (1)/([B]) us time is given below when [A] = [B]. What is order of reaction with respect to B ? |
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Answer» 4 |
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| 42. |
For a reaction , A + B rarr product, the rate law is given by r = k[A]^((1)/(2))[B]^(2). What is the order of the reaction ? |
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Answer» SOLUTION :Order of REACTION `= (1)/(2) + 2` `= (1+4)/(2) = (5)/(2)` `=2.5` |
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| 43. |
For a reactionA+B rarr Products, it is observed that doubling the concentration of B causes the reaction rate to increase four times, but doubling the concentration of A has no effect on the rate of reaction. The rate equation is threfore |
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Answer» `RATE=K[A]^2` |
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| 44. |
For a reaction A + B rarr C, the rate of the reaction is denoted (-dA)/(dt) or (-dB)/(dt) or (+dC)/(dt). State the significance of plus and minus sign. |
| Answer» Solution :Minus sign i.e. `(-dA)/(DT)` or `(-DB)/(dt)` indicates decreases in the concentration of REACTANTS whereas+ signindicrease in the concentration of PRODUCTS with time i.e. `(+dC)/(dt)`. | |
| 45. |
For a reaction A+B rarr C+D if the concentration of A is doubled without altering the concentration of B, the rate gets doubled. If the concentration of B is increased by nine times without altering the concentration of A, the rate gets tripled. The order of the reaction is |
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Answer» 2 `r=K[A]^(alpha)[B]^(beta) ""`….(i) `2r=K[2A]^(alpha)[B]^(beta) ""`….(ii) `3r=K[A]^(alpha)[9B]^(beta) ""`…..(iii) Dividing eqn.(ii)by eqn.(i) `2=2^(alpha)`or,`alpha =1` Dividing eqn.(iii)by eqn.(i) `3=9^(beta)`or,`3=3^(2beta)`or,`2beta =1`or,`beta = 1//2` `therefore`ORDER of the REACTION `=1+(1)/(2)=(3)/(2)`. |
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| 46. |
For a reaction 2H_(2)O_(2) underset("alkaline medium") overset(I^(-)) to 2H_(2)O+O_(2) (i) H_(2)O_(2)+I^(-) to H_(2)O +IO^(-) (slow) (ii) H_(2)O_(2)+IO^(-1) to H_(2)O +I^(-) + O_(2) (fast) (a) Write rate law for the reaction. (b) Write the overall order of reaction. (c) Out of steps (1) and (2), which one is rate determining step? |
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Answer» Solution :(i) RATE `=k[H_(2)O_(2)][l]` (ii) order = 2 (iii) STEP 1 |
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| 47. |
For a reaction 2SO_(2) + O_(2) ltimplies2SO_(3), rate of disappearance of O_(2) is 2xx10^(-4) mol L^(-1). The rate of appearance of SO_(3) is: |
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Answer» `2xx10^(-4)mol L^(-1)s^(-1)` Rate of reaction: `-1//2 (d[SO_(2)])/(dt) = (-d[O_(2)])/(dt) = 1/2(d[SO_(3)])/(dt)` `2 xx 10^(-4) mol L^(-1)=1/2 (d[SO_(2)])/(dt)` `(d[SO_(3)])/(dt) = 2 xx 2 xx 10^(-4) mol L^(-1)s^(-1)` `=4xx10^(-4)mol L^(-1)s^(-1)` |
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| 48. |
For a reaction A+2BrarrC, the amount of C formed by starting the reaction with 5 moles of A and 8 moles of B is |
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Answer» Solution :`A+2BrarrC` 1 mole of A reacts with 2 moles of B `therefore"5 moles of A will REACT with 10 moles of B."` But we have only 8 moles of B. HENCE, B is the limiting reactant 2 moles of B form 1 mole of C `therefore"8 mole of B will form C = 4 moles."` |
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| 49. |
For a reaction A + 2BtoC , the amount of C formed by starting the reaction with 5 moles of A and 8 moles of B is |
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Answer» 5 moles 8 moles of B require 4 MOLE of A . THEREFORE LIMITING reagent is B . 2 moles of B give 1 mole of C |
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