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 solution of dibasic acid the molarity (M) and normality (N) are related as |
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Answer» N=M/2 |
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| 2. |
For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here x_L and x_M represent mole fraction of L and M, respectively, in the solution. The correct statement(s) applicable to this system is(are) |
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Answer» The POINT Z represents VAPOUR pressure of pure liquid M and Raoult's LAW is obeyed from `x_L` = 0 to `x_L` = 1 |
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| 3. |
For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown is the following figure. Here x_(L) and x_(M) represent mole fractions of L and M, respectively, in the solution. The correct statement (s) applicable to this system is (are) |
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Answer» The point Z represents vapour pressure of pure LIQUID M and Raoult's law is obeyed from `x_(L) = 0` to `x_(L) = 1` |
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| 5. |
For a second order reaction, t_(75%) = x t _(50%) find the value of x |
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Answer» |
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| 6. |
For a second order reaction, 2" A"toProducts, a plot of log t_(1//2) vs log a (where a is the initial concentration) will give an intercept equal to which of the following ? |
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Answer» `1//k` `k= (1)/(t)[(1)/([A])-(1)/([A]_(0))]` When `t = t_(1//2), [A] = ([A]_(0))/(2)` Hence, `t_(1//2) = (1)/(k)[(2)/([A]_(0))-(1)/([A]_(0))] = (1)/(k[A]_(0))` `therefore log t_(1//2) = - log k - log [A]_(0) =- log k - log a` Plot of log `t_(1//2)` vs log a will be linear with intercept = log k.
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| 7. |
For a second order reaction rate at a particular time is x . If the initial concentration is tripled , the rate will become : |
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Answer» Solution :( C ) Rate = `k[A]^(2) = X ` If conc . Is tripled i.e A = [3A] Rate = `k[3A]^(2) = k.9[A]=x ` `:. (x)/(x)=9` i.e BECOMES nine times (9x) |
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| 8. |
For a second order reaction, 2Ararr Products, a plot of logt_(1//2) vs log a (where a is initial concentration) will give an intercept equal to which one of the following? |
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Answer» `(1)/(K)` |
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| 9. |
For a sample of perfect gas when its pressure is changed isothermally from p_(i) to p_(f), the entropy change is given by |
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Answer» `DeltaS=RTln((p_(i))/(p_(F)))` In isothermal PROCESS `T_(i)=T_(f)` `DeltaS_("sys")=nRln.(P_(i))/(P_(f))` |
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| 10. |
For a sample of an ideal gas at given temperature (T), speed distribution curve is given as follows. Then the speeds corresponding to point A, B and C are respectively known as: |
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Answer» Most probable, AVERAGE and root mean SQUARE |
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| 11. |
For a reversible reaction : X_((g))+3Y_((g))iff2Z_((g)), DeltaH=-40kJ the standard entropies of X, Y and Z are 60, 40 and 50 JK^(-1) mol^(-1) respectively. The temperature at which the above reaction attains equilibrium is about |
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Answer» Solution : `X_((g))+3Y_((g))iff2Z_((g))` `DeltaS^(@)=2S^(@)(Z)-{S^(@)(X)+3S^(@)(Y)}` `=2xx50-{60+3xx40}=100-180 "= - 80 J K"^(-1) mol^(-1)` `DeltaH^(@)=-40 kJ=-40,000 J,"" DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` At equilibrium, `DeltaG^(@)=0 ""therefore DeltaH^(@)=TDeltaS^(@)` `or, T = (DeltaH^(@))/(DeltaS^(@))=(40000)/(80)=500 K`. |
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| 12. |
For a reversible reaction : X_((g))+3Y_((g))hArr 2Z_((g)), Delta H=-40 kJ, the standard entropies of X,Y and Z are 60, 40 and 50 JK^(-1)mol^(-1) respectively. The temperature at which the above reaction attains equilibrium is about |
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Answer» 400 K `Delta S^(@)=2S^(@)(Z)-{S^(@)(X)+3S^(@)(Y)}` `=2xx50-{60+3xx40}=100-180` `=-80J K^(-1)MOL^(-1)`. Given `Delta H^(@)=-40 kJ =-40,000 J` `Delta G^(@)=Delta H^(@)-T Delta S^(@)` At equilibrium, `Delta G^(@)=0` `therefore Delta H^(@)=T Delta S^(@)` or,`T=(Delta H^(@))/(Delta S^(@))=(40000)/(80)=500 K`. |
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| 13. |
For a reversible reaction the rate constant for the forward reaction is 2.38xx10^(-4) and for the backward reaction is 8.15xx10^(-5) The k_c of the reaction is: |
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Answer» 0.342 |
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| 14. |
For a reversible reaction if the concentration of the reactants are doubled, the equilibrium constant will be : |
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Answer» HALVED |
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| 15. |
For a reversible reaction C hArr D, heat of reaction at constant volume is -33.0 kJ mol^(-1), calculate: (i) The equilibrium constant at 300 K. (ii) If E_(f) and E_(b) are energy of activation for forward and backward reactions respectively, calculate E_(f) and E_(b) at 300 K. Given that E_(f) : E_(b)=20:31 Assumw pre exponential factor same for forward and backward reaction. |
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| 16. |
For a reversible reaction following graph is obtained between log_(10)K_(C) and (1)/(T) hence DeltaH for the reaction is:- |
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Answer» `-19.14` J/mole |
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| 17. |
For a reversible reaction ALeftrightarrowBwhich one of the following statements is wrong from the given energy profile diagram ? |
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Answer» Activation energy of forward REACTION is greater thanbackward reaction where, `E_(a)` = activation energy of forward reaction ` E_(a)` = activation energy of baclward reaction the above profile DIAGRAM shows that `E_(a) gt E' _(a)` the potential energy of the product is greater than that of the reactant , so the reaction is endothemic . ` E_(a)= E'_(a) + DELTAE` ` E_(t)= E_(a) or E_(t) gt E'_(a)` |
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| 18. |
For a reversible reaction A underset(k_(2)=2 xx 10^(2)s^(-1))overset(k_(1)=4xx10^(2)s^(-1))hArr B initial concentration of A is 21 mol L^(-1) . Select the correct statement: |
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Answer» Equilibrium CONCENTRATION of A is 14 MOL `L^(-1)` |
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| 19. |
For a reversible reaction, A hArr B, which one of the following statements is wrong from given energy profile diagram ? |
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Answer» Activation energy of forward REACTION is greater than that of backward reaction. |
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| 20. |
For a reversible first-order reaction, A underset(K_(2))overset(K_(1))(hArr)B, K_(f)=10^(-2) s^(-1) and B_(eq.)/A_(eq.)=4, If A_(0)=0.01 ML^(-1) and B_(0)=0, what will be concentration of B after 30 sec? |
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| 21. |
For a reversible chemical reaction where the forward process is exothermic which of the following statements is correct |
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Answer» The backward REACTION has HIGHER activation ENERGY than the forward reaction |
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| 22. |
For a reversible chemical reaction where the forward process is exothermic. Which of the following statements is correct? |
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Answer» The backward REACTION has HIGHER activation energy than the forward reaction |
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| 23. |
For a real gas y at T=60 k, select the only correct option (a/(bR)=270k) |
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Answer» `because T lt T_c IMPLIES` on applying pressure GAS will liquify |
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| 24. |
For a redox reaction to proceed in a cell, the e.m.f. must be: |
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Answer» Positive |
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| 25. |
For a redox reaction to proceed spontaneously in a given direction , the emf should |
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Answer» be ZERO |
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| 26. |
For a redox reaction to proceed spontaneously in a given direction, the emf should: |
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Answer» be zero |
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| 27. |
For a real gas the P -V curve was experimentally plotted and it hed the following appearance. With respect to liquifaction. Choose the correct statement. |
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Answer» at `T = 500K, P = 40` atm, the STATE will be liquid (a) at `T = 500K, P = 40 atm` corresponds to 'a' SUBSTANCE - ga (b) at `T= 300 K, P = 50 atm` corresponds to 'b' substance -liquid (c ) at `T lt 300 K, P gt 20 atm` corresponds to 'c' substance -liquid (d) at `T lt 500 K, P gt 50 atm` corresponds to 'd' substance -liquid So, Answer (D) |
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| 28. |
For a real gas x at its critical temperature the correct option is - |
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Answer» (II)(iv)(P) & liquid & VAPOUR phase become indistinguishable `:.` Their pysical properties are same `because Z lt 1` , attractive TENDENCIES dominate. |
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| 29. |
For a real gas the P-V curve was experimentally plotted and it had the following appearance. With respect to liquefaction, choose the correct statement. |
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Answer» at `T = 500 K, P = 40` ATM, the STATE will be liquid |
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| 30. |
For a real gas critical volume (Vc) is '2x' times the actual volume occupied by gas molecules, 'x' value is |
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| 31. |
For a reactions , A+B rarr product , therate law 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 "0.5` | |
| 32. |
For a real gas, deviations from ideal gas behaviour are maximum at : |
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Answer» `-10^@C` and 5.0 ATM |
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| 33. |
For a reactionAtoB with activation energy E_(a) and rate constant k=Ae^(-Ea//RT). The rate of the reaction (Rate =k[A]) increases by increasing the temperature because |
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Answer» ACTIVATION ENERGY decreases with increase in TEMPERATURE |
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| 34. |
For a reaction , X+Y rarr Product , quadrupling [x] . Increases the rate by afactor of 8 . Quadrupling both [x] and [y] , increases the rate by a factor of 16. Find the order of the reaction with respect to x and y . What is the overall order of the reaction. |
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Answer» Solution :`{:(,X+yrarr,"Product(Z)"),(,x+yrarr,z),("condition"-1implies,4x+y,8),("condition"-2implies,4x+4yrarr,16):}` `z=k[x]^m[y]^n""(1)` `8Z=k[4x]^m[y]^n""(2)` `16z=k[4x]^m[4y]^n""...(3)` Dividing Eq (2) by Eq (1) we get `(8z)/z=(k[4x]^m[y]^n)/(k[x]^m[y]^n)` `8=4^mrArr2^3=(2^2)^mimplies2^3=2^(2m),2m=3,m=3//2` 1.5 ORDER with respect to x . Dividing Eq (3) by Eq (1) we get , `(16z)/z = (k[4x]^m[4y]^n)/(k[x]^m[y]^n)` `16=4^m.4^n` `16=4^2.4^n` `16/16=4^n` `1 = 4n ` `:. n = 0 ` [Zero order with respect to y ] OVERALL order of the reaction , `k[x]^m[y]^n` `k[x]^(1.5)[y]^0` Order `=(1.5+0)=1.5` |
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| 35. |
For a reaction(1)/(2)A to 2B rate of disappearance of A is related to rate of apperarance of B by the expression |
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Answer» `(-d[A])/(DT)=(1)/(2)(d[B])/(dt)` |
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| 36. |
For a reaction X + Y to Z, rate prop [X]. What is (i) molecularity and (ii) order of reaction ? |
| Answer» Answer :A | |
| 37. |
For a reaction, X+Y rarr Product , quadrupling [x], increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with suspect to x and y. what is the overall order of the reaction? |
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Answer» Solution :`""x+y RARR "Product (z)"` `"" x+y rarr z` Condition - 1 `rArr 4x+y rarr 8` Condition - 2 `rArrr 4x+4y rarr 16` `""z=k[x]^(m)[y]^(n)"………………….(1)"` `""8Z=k[4x]^(m)[y]^(n)"......................(2)"` `""16z=k[4x]^(m)[4y]^(n)".....................(3)"` Dividing EQ (2) by Eq (1) we get, `(8z)/z=(k[4x]^(m)[y]^(n))/(k[x]^(m)[y]^(n))` `8=4^(m) rArr 2^(3)=(2^(2))^(m) rArr 2^(3)=2^(2m), 2m=3, m=3//2` 1.5 order with respect to x. Dividing Eq (3) by Eq (1) we get, `""(16z)/z=(k[4x]^(m)[4y]^(n))/(k[x]^(m)[y]^(n))` `""16=4^(m).4^(n)` `""16=4^(2).4^(n)` `""16/16=4^(n)` `""1=4n` `""THEREFORE n=0` [Zero order with respect to y] Overall order of the reaction, `""k[x]^(m)[y]^(n)` `""k[x]^(1.5]y^(0)` Order `=(1.5+0)=1.5` |
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| 38. |
For a reaction, X + Y rarr Z there is no entropy charge. Enthaply change for the reaction is 100J mol^(-1). Delta G is |
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Answer» `- 100 J MOL^(-1)` `= 100 J mol^(-1)` |
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| 39. |
For a reaction X rarr Y the rate of the reaction has been found to be third order with respect to X . What happens when the concentration of X is doubled : |
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Answer» rate becomes DOUBLE when X becomes double rate becoems `2^(3)` = 8 times |
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| 40. |
For a reaction to occur spontaneously |
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Answer» `DELTA S` MUST be NEGATIVE |
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| 41. |
For a reaction, the threshold energy is equal to |
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Answer» Activation energy + initial POTENTIAL energy of reactants |
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| 43. |
For a reaction ,the value of slope of a plot in Kto (1)/(T)=………. |
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Answer» `-(E_(a))/(2.303)` |
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| 44. |
For a reaction to be spontaneous at all temperatures |
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Answer» `Delta G` and `Delta H` should be NEGATIVE `Delta G=Delta H-T Delta S` For a REACTION to be spontaneus, `Delta G` must be negative.If `Delta G` and `Delta H` are negative, then reaction will be SPONTANEOUS at all temperatures. |
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| 45. |
For a reaction, the rate constant is 2.34 s^(-1). The half-life period for the reaction is |
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Answer» 0.30 s |
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| 46. |
For a reaction ,the rate of reaction was found to increase about .1.8 times when the temperature was increased by 10^@C . The increase in rate is due to : |
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Answer» Increase in number of active molecules |
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| 47. |
For a reaction, the rate of reaction was found to increase about 1.8 times when the temperature was increased by 10^(@)C. The increase in rate is not due to |
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Answer» INCREASE in NUMBER of active MOLECULES |
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| 48. |
For a reaction the free energy change, DeltaG=-RT ln K_(p)+RTlnQ_(p) where K_(P) = equilibrium constant, Q_(P)= reaction quotient. For the reaction to be in equilibrium state |
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Answer» `(Q_(p))/(K_(p))gt1` |
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| 49. |
For a reaction the graph of the rate of the reaction against molar concentration of the reactant is as shown : What is the order of the reaction ? |
| Answer» SOLUTION :ZERO ORDER. | |
| 50. |
For a reaction the graph drawn between half life period and reciprocal of initial concentration(1/a) gives a straight line with positive slope. The magnitude of the slope is 500 and iniital concentration of the reactant is 2M. If the initial rate of the reaction is represented as x xx10^(-3) units, then the value of x is _____________ |
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