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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 601. |
For the reaction, `A(g)+2B(g) hArr 2C(g)`, the rate constant for forward and the reverse reactions are `1xx10^(-4)` and `2.5xx10^(-2)` respectively. The value of equilibrium constant, K for the reaction would beA. `2xx10^(-4)`B. `3xx10^(-2)`C. `4xx10^(-3)`D. `3xx10^(2)` |
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Answer» Correct Answer - C We know, `K=K_(f)/K_(b)=(1xx10^(-4))/(2.5xx10^(-2))=4xx10^(-3)` |
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| 602. |
When two reactants A and B are mixed to give products C and D, the reaction quotient (Q) at the initial stages of the reactionA. is zeroB. decrease with timeC. independent of timeD. increase with time |
| Answer» At the initial state if only reactats are used `Q=0`, but from initial state to attain equilibrium, it continuously increase with time till it becomes equal to `K_(c)`. | |
| 603. |
When two reactants A and B are mixed to give producys C and D, the reaction quotient (Q) at the initial stages of the reactionA. Is zeroB. Decreases with timeC. Is independent of timeD. Increases with time |
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Answer» Correct Answer - D Reactionquotient is equal to the ratio of the concentration of products to the ratio of the concentrations of the reactants at any stage of the reaction, each concentration term being raised to the power of its stoichiometric coefficient. In the beginning of the reaction, `Q=0`. |
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| 604. |
For the reaction equilibrium, `N_(2)O_(4(g))hArr2NO_(2(g))`, the concentration of `N_(2)O_(4)` and `NO_(2)` at equilibrium are `4.8xx10^(-2)` and `1.2xx10^(-2)` mol/L respectively. The value of `K_(c)` for the reaction is:A. `3xx10^(-3)M`B. `3xx10^(3)M`C. `3.3xx10^(2)M`D. `3xx10^(-1)M` |
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Answer» `K_(c)=([NO_(2)]^(2))/([N_(2)O_(4)])=((1.2xx10^(-2))^(2))/(4.8xx10^(-2))` `=3.0xx10^(-3)M` |
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| 605. |
The rate constant for forward and backward reactions of hydrolysis of ester are `1.1xx10^(-2)` and `1.5xx10^(-3)` per minute respectively. Equilibrium constant for the reaction isA. `4.33`B. `5.33`C. `6.33`D. `7.33` |
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Answer» Correct Answer - D `K_(f) = 1.1xx10^(-2), K_(b) = 1.5xx10^(-3)` `K_(c) = (K_(f))/(K_(b)) = (1.1xx10^(-2))/(1.5xx10^(-3)) = 7.33` |
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| 606. |
When two reactants, A and B are mixed to give products C and D, the reaction quotient Q, at the initial stages of the reaction.A. (A) is zeroB. (B) decrease with timeC. (C) is independent of timeD. (D) increases with time |
| Answer» Correct Answer - D | |
| 607. |
When two reactants, A and B are mixed to give products C and D, the reaction quotient Q, at the initial stages of the reaction.A. is zeroB. decreases with timeC. is independent of timeD. increases with time |
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Answer» Correct Answer - D |
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| 608. |
If the equilibrium constant for the reaction, `N_(2)O_(4)hArr2NO_(2)` is `K=[NO_(2)]^(2)//[N_(2)O_(4)]`, then which of the graphs are correct at constant temperature?A. B. C. D. |
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Answer» Concentration of `NO_(2)` is in square. See `[NO_(2)]^(2)=K[N_(2)O_(4)]` |
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| 609. |
A reaction mixture containing `H_(2), N_(2)` and `NH_(3)` has partial pressures `2` atm, `1` atm, and `3` at,. Respectively, at `725 K`. If the value of `K_(p)` for the reaction, `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` is `4.28xx10^(-5) atm^(-2)` at `725 K`, in which direction the net reaction will go?A. ForwardB. BackwardC. No net reactionD. Direction of reaction cannot be predicted. |
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Answer» `Q_(p)=((p_(NH_(3)))^(3))/(p_(N_(2))xx(P_(H_(2)))^(3))=((3)^(2))/((1)(2)^(3))=9/8 "atm"^(-2)=1.125 "atm"^(-2)` Since the value of `Q_(p)` is larger then `K_(p) (4.28xx10^(-5) "atm"^(-2))`, it indicates net reaction will proceed in the backward direction. |
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| 610. |
The partial pressures of `N_(2)O_(4) "and" NO_(2) "at" 40^(@)C` for the following equilibrium `N_(2)O_(4)(g)hArr2NO_(2)(g)` are `0.1 "atm and" 0.3` atm respectively. Find `K_(P)` for the reaction. |
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Answer» Correct Answer - `0.9"atm"` `P_(N_2O)_(4)=0.1 "atm" P_(NO_(2))=0.3` atm. reaction is `N_(2)O_(4)(g)hArr2NO_(2)(g)` `K_(P)=((P_NO_(2)))/((P_(N_(2)_O_(4))))=((P_(NO_(2)))^(2))/((P_(N_(2)_O_2)))=(0.3)^(2)/((0.1))=0.9` atm |
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| 611. |
Calculate partial pressure of B at equilibrium in the following equilibrium `A(s) iff B(g) +2C(g), " " K_(P)=32 atm^(3)`. |
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Answer» Correct Answer - 2 |
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| 612. |
The homogeneous reversible reaction, `C_(2)H_(5)OH+COOHhArrCH_(3)COOC_(2)H_(5)+H_(2)O` is studied at various initial concentrations of the reactants at constant temperature. Calculate `K` in each case. `{:(,"Moles of acid", "Moles of alcohol","Moles of ester"),(,"Per litre (initial)","Per litre(initial)","Per litre at equilibrium"),((i),1,1,0.637),((ii),1,4,0.93):}` |
| Answer» Correct Answer - `K=4, K=4.02` | |
| 613. |
For a gasous reaction, `2A+B hArr 2C`, the partial pressures of A,B and C at equilibrium are 0.3 atm, 0.4 atm and 0.6 atm respectively. The value of `K_P` for the reaction would be :A. `10 "atm"^(-1)`B. `1/10 "atm"^(-1)`C. `0.2 "atm"^(-1)`D. `5 "atm"^(-1)` |
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Answer» Correct Answer - A |
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| 614. |
In the dissociation of `PCl_(5)` as `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` If the degree of dissociation is `alpha` at equilibrium pressure P, then the equilibrium constant for the reaction isA. `K_(p)=alpha^(2)/(1+alpha^(2)P)`B. `K_(p)=(alpha^(2)P^(2))/(1-alpha^(2))`C. `K_(p)=P^(2)/(1-alpha^(2))`D. `K_(p)=(alpha^(2)P)/(1-alpha^(2))` |
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Answer» Correct Answer - D `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial",1,,0,,0),("Final",1-alpha,,alpha,,alpha),("Partial pressure"=,(1-alpha)/(1+alpha).P,,alpha/(1+alpha).P,,alpha/(1+alpha).P):}` Total "mole" `=1-alpha+alpha+alpha=1+alpha` `K_(p)=(alpha/(1+alpha)P.alpha/(1+alpha).P)/((1-alpha)/(1+alpha)P)=(alpha^(2)P)/(1-alpha^(2))` |
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| 615. |
The degree of dissociation is `0.4` at `400 K` and `1.0` atm for the gaseous reaction `PCl_(5) hArr PCl_(3)+Cl_(2)` assuming ideal behaviour of all gases, calculate the density of equilibrium mixture at `400 K` and `1.0` atm (relative atomic mass of P is `31.0` and of Cl is `35.5`). |
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Answer» `{:(,PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),("Initial",1,,0,,0),("At equilibrium",(1-0.4),,0.4,,0.4):}` Total moles at equilibrium is `1-0.4+0.4+0.4=1.4` `PV=nRT` or `V=(nRT)/(P)=(1.4xx0.0821xx400)/(1)=45.976 L` Density =`("Mass")/("Volume")` `=208.5/45.976(("No change in mass"),("during reaction"))=4.54 g L^(-1)` |
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| 616. |
The value of `log_(10)K` for a reaction `A hArr B` is (Given: `Delta_(f)H_(298K)^(Theta) =- 54.07 kJ mol^(-1)`, `Delta_(r)S_(298K)^(Theta) =10 JK^(=1) mol^(-1)`, and `R = 8.314 J K^(-1) mol^(-1)`A. `5`B. `10`C. `95`D. `100` |
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Answer» `DeltaG^(@)=-2.303RT log K` or `DeltaH^(@)-TDeltaS^(@)=-2.303Rt logK` `[-54.07xx10^(3)-298xx10]=-2.303xx8.314xx298 log K` `:. Log K=(57050)/(5705)=10` |
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| 617. |
The value of `log_(10)K` for a reaction `A hArr B` is (Given: `Delta_(f)H_(298K)^(Theta) =- 54.07 kJ mol^(-1)`, `Delta_(r)S_(298K)^(Theta) =10 JK^(=1) mol^(-1)`, and `R = 8.314 J K^(-1) mol^(-1)`A. 5B. 10C. 95D. 100 |
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Answer» Correct Answer - B |
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| 618. |
At `273K` and `1` atm ,a moles of `N_(2)O_(4)` decomposes to `NO_(2)` according to equation `N_(2)O_(4)(g) hArr 2NO_(2)(g)`. To what extent has the decomposition proceeded when the original volume is `25%` less than that of exisiting volume? |
| Answer» Correct Answer - `33%` , | |
| 619. |
An equilibrium mixture at `300 K` contains `N_(2)O_(4)` and `NO_(2)` at `0.28` and `1.1 atm`, respectively. If the volume of container is doubles, calculate the new equilibrium pressure of two gases. |
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Answer» `{:(,N_(2)O_(4),hArr,2NO_(2)),("Pressure at equilibrium",0.28,,1.1):}` `:. K_(p)=((p_(NO_(2)))/p_(N_(2)O_(4)))^(2)=((1.1)^(2))/(0.28)=4.32 "atm"` Now if the volume of container is doubled, i.e., pressure decreases and will become half, the reaction will proceed in the direction where the reaction shows an increase in moles i.e., decomposition of `N_(2)O_(4)` is favoured. `{:(,,N_(2)O_(4),hArr,2NO_(2)),("New pressure at equilibrium",,(0.28/2-P),,(1.1/2+2P)):}` where reactant `N_(2)O_(4)` equivalent to pressure `P` is used up in doing so. Again `K_(p)=([(1.1//2)+2P]^(2))/([(0.28//2)-P])=([0.55+2P]^(2))/([.14-P])=4.32` `P=0.045` `:.` Now `P_(N_(2)O_(4))=0.14-0.045=0.095 "atm"` `P_(NO_(2))` at new equilibrium=`0.55+2xx0.045` `:. P_(NO_(2))=0.64 "atm"` |
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| 620. |
The oxidation of `SO_(2)` by `O_(2)` to `SO_(3)` is an exothermic reaction . The yield of `SO_(2)` will be maximum ifA. temperature is increased and pressure is kept constantB. temperature is decreased and pressure is increasedC. both temperature and pressure are increasedD. both temperature and pressure are decreased |
| Answer» Correct Answer - B | |
| 621. |
In which of the following reaction is `K_(p)gtK_(c)`A. `H_(2)+l_(2)rarr2HI`B. `N_(2)+3H_(2)rarr2NH_(3)`C. `2SO_(3)rarr2SO_(2)+O_(2)`D. `PCl_(3)+Cl_(2)rarrPCl_(5)` |
| Answer» Correct Answer - C | |
| 622. |
When `0.15` mol of CO taken in a `2.5 L` flask is maintained at `750 K` along with a catalyst, the following reaction takes place `CO(g)+2H_(2)(g) hArr CH_(3)OH(g)` Hydrogen is introduced until the total pressure of the system is `8.5` atm at equilibrium and `0.08` mol of methanol is formed. Calculate a. `K_(p)` and `K_(c)` b. The final pressure, if the same amount of CO and `H_(2)` as before are used, but with no catalyst so that the reaction does not take place. |
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Answer» (`i`) Given, `{:(,,CO_((g)),+,2H_(2(g)),hArr,CH_(3)OH_((g))),(,"Mole before reaction",0.15,,a,,0),(,"Mole after reaction",(0.15-x),,(a-2x),,x):}` and `x=0.08` Total mole at equilibrium `=0.15-x+a-2x+x=0.15+a-2x` `=0.15+a-0.16=a-0.01` Also total mole at equilibrium are obtained by, `n=(PV)/(RT)` `n=(8.5xx2.5)/(0.0821xx750) :. P=8.5 atm, V=2.5 litre, T=750 K` at eqm. `:. n=0.345` `:. a-0.01=0.345` `:. a=0.355` At equilibrium, Mole of `CO=0.15-0.08=0.07` Mole of `H_(2)=0.355-0.16=0.195` Mole of `CH_(3)OH=0.08` `:. K_(c)=([CH_(3)OH])/([H_(2)]^(2)[CO])=(0.08//2.5)/(((0.07)/(2.5))((0.195)/(2.5))^(2))` `=187.85 mol^(-2)litre^(2)` Also `K_(p)=K_(c)(RT)^(Deltan)=187.85xx(0.0821xx750)^(2)` `=0.05 atm^(-2)` (`ii`) If reaction does not take place, then Mole of `CO=0.15` Mole of `H_(2)=0.355` `:. `Total mole `=0.505` `:. Pxx2.5=0.505xx0.0821xx750` `:. P=12.438 atm` |
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| 623. |
At 1000 K, the value of `K_(p)` for the reaction: `A(g) + 2B(g)hArr3C(g) + D(g)` is `0.05` atmosphere. The value of `K_(c)` in terms of R would be:A. 20000 RB. `0.02 R`C. `5xx10^(-5) R`D. `5xx10^(-5)xxR^(-1)` |
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Answer» Correct Answer - D `K_(p) = K_(c)(RT)^(Delta n), Delta n = 4 - 3 = 1` `0.05 = K_(c)Rxx1000` `K_(c) = 5xx10^(-5)xxR^(-1)` |
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| 624. |
In a reversible reaction, if the concentration of reactants are doubles, the equilibrium constant K will:A. Also be doubledB. Be halvedC. Become one-fourthD. Remain the same |
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Answer» Correct Answer - D `K_(c)` is a characteristic constant for the given reaction. |
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| 625. |
For the reaction `CO(g)+2H_(2)(g)hArrCH_(3)OH(g)` Hydrogen gas is introduced into a five-litre flask at `327^(@)C`, containing `0.2` mol of `CO(g)` and a catalyst, untill the pressure is `4.92 atm`. At this point, `0.1` mol of `CH_(3)OH(g)` is formed. Calculate the equilibrium constants `K_(p)` and `K_(c )`. |
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Answer» `{:(,,CO_((g)),+,2H_(2(g)),hArr,CH_(3)OH_((g))),(,"Initial mole",0.2,,a,,0),(,"Mole at equilibrium",(0.2-0.1),,(a-0.2),,0.1):}` Now total mole at equilibrium `=0.1+a-0.2+0.1=a` Also Mole `(n)=(PV)/(RT)=(4.92xx5)/(0.0821xx600)=0.499` `a=0.499` `:. [CH_(3)OH]=(0.1)/(5)`, `[CO]=(0.2-0.1)/(5)=(0.1)/(5)`, `[H_(2)]=(0.499-0.20)/(5)=(0.299)/(5)` `:. K_(c)=([CH_(3)OH])/([CO][H_(2)]^(2))=(0.1//5)/(0.1//5xx(0.299//5)^(2))` `=279.64 litre^(2)mol^(-2)` `K_(p)=K_(c)(RT)^(Deltan)` `=279.64xx(0.0821xx600)^(-2)` `=0.115 atm^(-2)` |
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| 626. |
Consider the following reaction `N_(2)O_(4)(g)hArr2NO_(2)(g)DeltaH=58.6KJ` What will be the effect of the following changes on the concentration of `N_(2)O_(4)` at equilibrium? (i) Increasing the pressure (ii) Increasing the temperature (iii) Increasing the volume (iv) Adding more `NO_(2)(g)` to the system without changing temperature and pressure (v) Adding catalyst. |
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Answer» (i) Backward (ii) Forward (iii) Forward (iv) Backward (v) No change |
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| 627. |
Consider the following equilibrium in a closed container: `N_(2)O_(4)(g)hArr2NO_(2)(g)` At a fixed temperature, the volume of the reaction container is halved. For this change which of the following statements holds true regarding the equilibrium constant `(K_(p))` and degree of dissociation `(alpha)`?A. neigher `K_(p)` no `alpha` changesB. Both `K_(p)` and `alpha` changeC. `K_(p)` changes but `alpha` does not changeD. `K_(p)` does not chamge but `alpha` changes |
| Answer» Correct Answer - B::C::D | |
| 628. |
`N_(2)O_(4)` dissociates as `N_(2)O_(4)hArr2NO_(2)`. At `55^(@)C` and one atmospehere, `%` decomposition of `N_(2)O_(4)` is `50.3%`. At what `P` and same temperature, the equilibrium mixture will have the ratio of `N_(2)O_(4):NO_(2)` as `1:8` ? |
| Answer» Correct Answer - `0.19 atm`, | |
| 629. |
Nitrogen combines with oxygen to form nitric oxide, `N_(2(g))+O_(2(g))hArr2NO_((g))`, (`DeltaH=90 kJ mol^(-1)`) The decomposition of `NO` is favoured by:A. decrease of temperatureB. increase of temperatureC. increase in the concentration of nitric oxideD. decrease of pressure |
| Answer» Correct Answer - (a,c) | |
| 630. |
`N_(2)O_(4)` dissociates as `N_(2)O_(4)(g)hArr2NO_(2)(g)` At `40^(@)C` and one atmosphere `%` decomposition of `N_(2)O_(4)` is `50.3%`. At what pressure and same temperature, the equilibrium mixture has the ratio of `N_(2)O_(4): NO_(2)` as `1:8`? |
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Answer» Case I `N_(2)O_(4)(g)hArr2NO_(2)(g)` `{:("At Eq",,(1-x),,2x,),(,,"total moles" =1+x,,):}` `p_(N_(2)O_(4))=((1-x))/((1+x))xxP, p_(NO_(2))=(2x)/((1+x))xxP` `K_(p)=(((2x)/(1+x).P)^(2))/(((1-x)/(1+x).P))=(4x^(2)P)/((1-x^(2)))` The `%` dissociation of `N_(2)O_(4)=50.3%` (given) Hence, degree of dissociation `=50.3/100=0.503` and `P=1` `:. K_(P)=(4xx(0.503)^(2)xx1)/([1-(0.503)^(2)])` `rArr K_(p)=1.3548` atm Case II `N_(2)O_(4)hArr2NO_(2)` `{:((1-x),(2x)):}` Given, `((1-x))/(2x)=1/8` `x=0.8` Let the new pressure be `P` atm. `K_(p)=(4x^(2)P)/((1-x^(2)))=(4xx0.8xx0.8xxP)/((1+0.8)(1-0.8))=1.3548` `P=0.19 "atm"` |
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| 631. |
0.1 mole of `N_(2O_(4)(g)` was sealed in a tude under one atmospheric conditions at `25^(@)C` Calculate the number of moles of `NO_(2)(g)` preesent , if the equilibrium `N_(2)O_(4)(g)hArr2NO_(2)(g)(K_(P)=0.14)` is reached after some time :A. `1.8xx10^(2)`B. `2.8xx10^(2)`C. `0.034`D. `2.8xx10^(-2)` |
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Answer» Correct Answer - c |
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| 632. |
`K_(p)` for the reaction `N_(2)+3H_(2) hArr 2NH_(3)` at `400^(@)C` is `1.64xx10^(-4)`. Find `K_(c)`. Also find `DeltaG^(ɵ)` using `K_(p)` and `K_(c)` values and interprest the difference. |
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Answer» Correct Answer - A::C::D `N_(2)+3H_(2) hArr 2NH_(3)` `Deltan=Sigman` products - `Sigman` reactions `=2-4=-2` `K_(p)=K_(c)(RT)^(Deltan)` `1.64xx10^(-4)=K_(c)(0.0821xx673)^(-2)` `:. K_(c)=0.5006` Now since `DeltaG^(ɵ)=-2.303 RT log K` If `K=K_(p)` then `DeltaG^(ɵ)=-2.303xx2xx673 log (1.64xx10^(-4))` `=+11.733 kcal` If `K=K_(c)` then `DeltaG^(ɵ)=-2.303xx2xx673 log (0.5006)=+931 cal` where `DeltaG^(ɵ)` is the free energy change when all the reactants and products are in the standard state. For `DeltaG^(ɵ)=-2.303 RT log K_(p)`, standard state means the partial pressure of each as `1` atm. When `DeltaG^(ɵ)=-2.303 RT log K_(C)` standard state means that concentration of each is `1` mole `L^(-1)`. So the value are different. |
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| 633. |
`K_(p)` for the reaction `N_(2)+3H_(2) hArr 2NH_(3)` at `400^(@)C` is `1.64xx10^(-4)`. Find `K_(c)`. Also find `DeltaG^(ɵ)` using `K_(p)` and `K_(c)` values and interpret the difference. |
| Answer» `K_(c)=0.5006 mol^(-2) litre^(+2)`, `+11.733kcal`, `+931 cal` , | |
| 634. |
`N_(3) +3H_(2) hArr 2NH_(3)` Starting with one mole of nitrogen and 3 moles of hydrogen, at equiliibrium `50%` of each had reacted. If the equilibrium pressure is `P`, the partial pressure of hydrogen at equilibrium would beA. `P/2`B. `P/3`C. `P/4`D. `P/6` |
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Answer» Correct Answer - A |
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| 635. |
The equilibrium constant `K_(p)` for the reaction, `N_(2)+3H_(2) hArr 2NH_(3)` is `1.64xx10^(-4)` at `400^(@)C` and `0.144xx10^(-4)` at `500^(@)C`. Calculate the mean heat of formation of `1` mol of `NH_(3)` from its elements in this temperature range. |
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Answer» We have `log_(10) K_(2)/K_(1)=(DeltaH)/(2.303R)(1/T_(1)-1/T_(2))` `"log" 0.144/1.64=(DeltaH)/(2.303xx1.987xx10^(-3))(1/673-1/773)` `DeltaH=-25.14 kcal for 2 "mol" =-12.57 kcal "mol"^(-1)` |
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| 636. |
At `450^(@)C` the equilibrium constant `K_(p)` for the reaction `N_(2)+3H_(2) hArr 2NH_(3)` was found to be `1.6xx10^(-5)` at a pressure of `200` atm. If `N_(2)` and `H_(2)` are taken in `1:3` ratio. What is `%` of `NH_(3)` formed at this temperature? |
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Answer» Correct Answer - A::B::C `{:(,N_(2),+,3H_(2),hArr,2NH_(3)),("moles before eq",1,,3,,0),("moles after eq",(1-x),,3(1-x),,2x):}` `:. K_(p)=((2x^(2))(4-2x)^(2))/((1-x)[3(1-x)]^(3)xxP^(2))` `:. K_(p)=((2x^(2))(4-2x)^(2))/((1-x)[3(1-x)]^(3)xxP^(2))` `K_(p)=(16xx x^(2) xx(2-x)^(2))/(27(1-x)^(4)xxP^(2))` or, `1.6xx10^(-5)=(16x^(2)xx(2-x)^(2))/(27(1-x)^(4)xx(200)^(2))` `(x^(2)(2-x)^(2))/((1-x)^(4))=(1.6xx10^(-5)xx27xx(200)^(2))/(16)` `=(16xx10^(-6)xx27xx(200)^(2))/16` `:. (x(2-x))/((1-x)^(2))=200xx10^(-3)xxsqrt(27)=1.039` `:. x=0.301` `:.` mole of `NH_(3)` formed `=2xx0.301=0.602` Total "moles" at equilibrium `=4-2xx0.301=3.398` `:. %` of `NH_(3)` at equilibrium `=0.602/3.398xx100=17.76%` |
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| 637. |
For the reaction `4NO_(2)(g) +O_(2)(g)hArr 2N_(2)O_(5)(g)`, which of the following facts holds good ?A. `K_(p)=K_(c)`B. `K_(p)gt K_(c)`C. `K_(p)ltK_(c)`D. `K_(p)` and `K_(c)` cannot unless pressure of the system is provided. |
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Answer» Correct Answer - C `Deltan lt 0 (i.e. -4-1+2=-3)` so`K_(p) lt K_(c)` as, `K_(p)=K_(c)(RT)^(Deltan)` `Deltanlt0(-4-1+2=-3)` |
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| 638. |
When the system `2HI(g) hArr H_(2)(g)+I_(2)(g)` is at equilibrium, inert gas is introduced. Dissociation of `HI` is …………. |
| Answer» Correct Answer - No change | |
| 639. |
The equilibrium constant for, `2H_(2)S(g)hArr2H_(2)(g)+S_(2)(g) "is" 0.0118 "at" 1300 K` while the heat of dissociation is `597.4KJ`. The standard equilibrium constant of the reaction at `1200K` is:A. `1.180xx10^(-4)`B. `11.80`C. `118.0`D. cannot be calculated from given data |
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Answer» Correct Answer - A `log(K_(2))/(K_(1))=(DeltaH^(@))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]` `log(K_(2))/(0.0118)=(597.4xx10^(3))/(2.303xx8.314)[(1)/(1300)-(1)/(1200)]` `log K_(2)=-2+log(0.0118)=-3.928 rArr K_(2)=1.18xx10^(-4)` Therefore. (A) option is correct. |
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| 640. |
Assertion (A) : For `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`, the equilibrium constant is `K`. The for `1/2 N_(2)(g)+3/2H_(2)(g) hArr NH_(3)(g)`, the equilibrium constant will be `sqrt(K)`. Reason (R) : If concentrations are changed to half, the equilibrium constants will be halved.A. If both (A) and (R) are correct, and (R) is the correct explanation for (A)B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A)C. If (A) is correct, but (R) is incorrectD. If (A) is incorrect, but (R) is correct. |
| Answer» Correct Answer - C | |
| 641. |
An equilibrium mixture for the reaction `2H_(2)S(g) hArr 2H_(2)(g) + S_(2)(g)` had 1 mole of `H_(2)S, 0.2` mole of `H_(2)` and 0.8 mole of `S_(2)` in a 2 litre flask. The value of `K_(c)` in mol `L^(-1)` isA. (A) `0.08`B. (B) `0.016`C. (C) `0.004`D. (D) `0.160` |
| Answer» Correct Answer - B | |
| 642. |
For the equilibrium, `SrCl_(2).6H_(2)O(s) hArr SrCl_(2).2H_(2)O(s)+4H_(2)O(g)` the equilibrium constant `K_(P)=16xx10^(-12)"atm"^(4)` at `1^(@)C`.If one litre of air saturated with water vapour at `1^(@)C` is exposed to a large quantity of `SrCl_(2).2H_(2)O(s)`, what weight of water vapour will be absorbed? Saturated vapour pressure of water at `1^(@)C=7.6` torr.A. `6.4` mgB. `3.25` mgC. `2.3` mgD. `8.5` mg |
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Answer» Correct Answer - A |
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| 643. |
`K_(p)` for `3//2H_(2)+1//2N_(2) hArr NH_(3)` are `0.0266` and `0.0129 atm^(-1)`, respectively, at `350^(@)C` and `400^(@)C`. Calculate the heat of formation of `NH_(3)`. |
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Answer» Correct Answer - A::B::C::D `2.303 "log"K_(p_(2))/K_(p_(1))=(DeltaH)/(R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `2.303"log"0.0129/0.0266=(DeltaH)/(2)[(673-623)/(673xx623)]` `:. DeltaH=12140 cal=-12140 cal` |
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| 644. |
Equilibrium constant `K_(p)` for `H_(2)S(g) hArr 2H_(2)(g)+S_(2)(g)` is `0.0118` atm at `1065^(@)C` and heat of dissociation is `42.4` Kcal. Find equilibrium constant at `1132^(@)C`. |
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Answer» Correct Answer - A::B `2.303 "log"K_(p_(2))/K_(p_(1))=(DeltaH)/(R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `2.303"log"K_(p_(2))/K_(p_(1))=(42.4xx10^(3))/(2)[(1405-1338)/(1405xx1338)]` `:. K_(p_(2))/K_(p_(1))=2.129` `:. K_(p_(2))=2.129xx0.0118=0.025 "atm"` |
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| 645. |
Calculate `K_(c)` for the reaction: `2H_(2)(g)+S_(2)(g) hArr 2H_(2)S(g)` if `1.58` mol `H_(2)S, 1.27` mol `H_(2)` and `2.78xx10^(-6)` mol of `S_(2)` are in equilibrium in a flask of capacity `180 L` at `750^(@)C`. |
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Answer» Correct Answer - A::B::C `{:(,2H_(2)(g),+,S_(2)(g),hArr,2H_(2)S(g)),("moles at",1.27,,2.78xx10^(-6),,1.58),("equilibrium",,,,,):}` Volume of container `=180 L` `:. K_(C)=([H_(2)S]^(2))/([H_(2)]^(2)[S_(2)])=((1.58/180)^(2))/((1.27/180)^(2)((2.78xx10^(-6))/180))` `=((1.58)^(2)xx180)/((1.27)^(2)xx(2.78xx10^(-6)))` `K_(C)=1.002xx10^(8)L "mol"^(-1)` |
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| 646. |
From the following data i. `H_(2)(g)+CO_(2)(g) hArr H_(2)O(g)+CO(g), K_(2000 K)=4.40` ii. `2H_(2)O(g) hArr 2H_(2)(g)+O_(2)(g), K_(2000 K)^(I)=5.31xx10^(-10)` iii. `2CO(g)+O_(2)(g) hArr 2CO_(2)(g), K_(1000 K)=2.24xx10^(22)` Show whether reaction (iii) is exothermic or endothermic. |
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Answer» Correct Answer - reaction is exothermic. Equation (iii)`=-[2xx(i)+(ii)]` `therefore K_(20000_(iii)=(1)/(K_(1)^(2)K_(2))=(1)/((4.4)^(2)xx5.31xxxx10^(-10)))=9.7xx10^(7)` `because TuparrowKdownarrow rArr` reaction is exothermic. |
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| 647. |
For which reaction does `K_p=K_c`?A. `2C(s)+O_2(g)to2CO(g)`B. `N_2(g)+3H_2(g)to2NH_3(g)`C. `2H_2(g)+O_2(g)to2H_2O(g)`D. `H_2(g)+I_2(g)to2HI(g)` |
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Answer» Correct Answer - D |
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| 648. |
Consider the following hypothetical equilibrium `2B(g)hArrB_(2)(g)` If `d` is observed vapour density and `D` is theoretical vapour density, then degree of association `(alpha)` will beA. `alpha=((D-d)/(d))`B. `alpha=(2D-d)/(D)`C. `alpha=2-(2D)/(d)`D. `alpha=(2D)/(D-d)` |
| Answer» Correct Answer - C | |
| 649. |
Which statement is correct about a system at equilibrium ?A. The forward and reverse reactions occur at identical ratesB. The concentrations of reacants must equal the concentrations of the products.C. The concentrations of reactants and products can be changed by adding a catalystD. The concentrations of reactants and products are not affected by a change in terperature. |
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Answer» Correct Answer - A |
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| 650. |
An example of a reversible reaction isA. (A) `Pb(NO_(3))_(2)(aq)+2Na!(aq)iffPbl_(2)(S)+2NaNO_(2)(aq)`B. (B) `AgNO_(3)(aq)+HC!(aq)iffAgCl(S)+HNO_(3)(aq)`C. (C) `2Na(S)+H_(2)O(l)iff2NaOH(aq)+H_(2)(g)`D. (D) `KNO_(3)(aq)+NaCl(aq)iffKCl(aq)+NaNO_(3)(aq)` |
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Answer» Correct Answer - D Precipitation reactions, acid base reactions and reactions in which gases are liberated and are taking place in open container will be irreversible reactions. |
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