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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.
| 451. |
2 mole of X and 1 mole of Y are allowed to react in a 2 litre container.when equilibrium is reached, the following reaction occurs `2X(g)+Y(g) hArr Z(g)` at 300 K. If the moles of Z at equilibrium is 0.5 then what is equilibrium constant `K_C`? |
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Answer» Correct Answer - 4 |
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| 452. |
Calculate the change in pressure ( in atm) when 2 mole of `NO` and `16 g O _(2)` in `6.25` litre originally at `27^(@)C` react to produce the maximum quantity of `NO_(2)` possible according to the equation. `( ` Take `R=(1)/(12)` ltr. Atm`//`mol K `)` `2NO(g)+O_(2)(g) hArr 2NO_(2)(g)`A. 1B. 4C. 5D. 2 |
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Answer» Correct Answer - 4 `{:(,2NO,+,o_(2)(g),rarr,2NO_(2)(g)),("moles",2,,0.5,,0),(,2-2xx0.5,,0,,2xx0.5),(,1,,0,,1),(,n_(f)=1+1,,,,n_(f)=2+0.5),( :. ,Deltan=(2.5-2),=0.5"moles",,,),( :. ,"change in pressure",,,,):} ` `Deltap=(DeltanRT)/(V)=0.5xx(1)/(12)xx(3300)/(8.25)=2atm` |
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| 453. |
The exothermic formation of `C1F_(3)` is represented by the equation `C1_(2(g)) + 3F_(2(g))hArr2C1F_(3(g))` , `Delta H = - 329 kJ` Which of the following will increase the quantity of `C1F_(3)` in an equilibrium mixture of `C1_(2)`, `F_(2)` and `C1F_(3)`A. Increasing the temperatureB. Removing `C1_(2)`C. Increasing the volume of the containerD. Adding `F_(2)` |
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Answer» Correct Answer - D By increasing the amount of `F_(2)` in the reaction the amount of `C1F_(3)` increases. |
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| 454. |
The vapour density of mixture consisting of `NO_2` and `N_2O_4` is 38.3 at `26.7^@C`. Calculate the number of moles of `NO_2` I `100g` of the mixture.A. `0.2`B. `0.4`C. `0.8`D. `1.6` |
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Answer» Correct Answer - B `alpha=(D-d)/(d)` D= Vapour density before dissociation d= vapour density after dissociation `N_(2)O_(4) hArr 2NO_(2)` Vapour density of `N_(2)O_(4)` before dissociation `(D)=(14xx2+16xx4)/(2)=92/2=46` Vapour density after dissociation `(d)=38.3` `:. alpha=(46-38.3)/38.3=0.2` `{:(,N_(2)O_(4),hArr,2NO_(2)),("Initial",1,,0),("At equilibrium",1-alpha,,2alpha):}` Number of moles of `NO_(2)` at eq. `=2alpha=2xx0.2=0.4` |
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| 455. |
The vapour density of a mixture containing `NO_(2)` and `N_(2)O_(4)` is `38.3 at 27^(@)C`. Calculate the mole of `NO_(2)` in `100 g` mixture. |
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Answer» Correct Answer - B::C::D `{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("At equilibrium",(1-x),,2x):}` x (degree of dissociation)=`(D-d)/((n-1)d)` Given, `d=38.3, D=("molecular mass of" N_(2)O_(4))/(2)=92/2=46, n=2` So, `x=(46-38.3)/38.3=0.2` At equilibrium, amount of `N_(2)O_(4)=1-0.2=0.8 "mol"` and amount of `NO_(2)=2xx0.2=0.4 "mol"` Mass of the mixture `=0.8xx92+0.4xx46` `=73.6+18.4=92.0 g` Since, `92 g` of the mixture contains `=0.4 "mol" NO_(2)` So `100g` of the mixture contains `=(0.4xx100)/92` `=0.43 "mol" NO_(2)` |
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| 456. |
The exothermic formation of `C1F_(3)` is represented by the equation `C1_(2(g)) + 3F_(2(g))hArr2C1F_(3(g))` , `Delta H = - 329 kJ` Which of the following will increase the quantity of `C1F_(3)` in an equilibrium mixture of `C1_(2)`, `F_(2)` and `C1F_(3)`A. Increasing the temperatureB. Removing `C1_(2)`C. Increasing the volume of containerD. Adding `F_(2)` |
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Answer» Correct Answer - B Favourable condition for forward reaction according to Le-Chatelier priencipal are: (a) Decrease in temperature (b) Increase in concentration of reaction of reactant and (c) Increase in pressure. |
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| 457. |
Calculate the degree of dissociation and `K_(P)` for the following reaction. `PCl_(5)(g)iffPCl_(3)(g)+ Cl_(2)(g)``{:(t=0,a,0,0),(t=t,a-x,x,x):}` Since for a mole, `x` moles are dissociated |
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Answer» `therefore"For"1 "mole", (x)/(a)"moles"=alpha` are dissociated `therefore x=a alpha` |
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| 458. |
The vapour density of a mixture containing `NO_(2) " and "N_(2)O_(4)` is `38.3 " at " 33_(@)C` calculate the no. of moles of `NO_(2)` if `100kg "of" N_(2)O_(4)` were taken initially. |
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Answer» `N_(2)O_(4)(g)iff2NO_(2)(g)` `M_(mix)=2xx38.3=76.6` `N_(2)O_(4)iff2NO_(2)` `{:(t=0,a,0),(t=t,a-aalpha,2aalpha):}` no. of moles of `NO_(2)=`"no. of moles of" N`O_(2)=2_(aalpha)=(2xx100xx0.2)/(92)=0.435` |
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| 459. |
In which of the following reaction, the yield of the products does not increase by increase in thepressure?A. `N_(2)(g)+O_(2)(g) = 2NO(g)`B. `2SO_(2)(g)+O_(2)(g)=2SO_(3)(g)`C. `N_(2)(g)+3H_(2)(g)=2NH_(3)(g)`D. `PCl_(3)(g)+Cl_(2)(g)=PCl_(5)(g)` |
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Answer» Correct Answer - A In which `Deltan=0` i. `Deltan=2-(2+2)=0` ii. `Deltan=2-(2+1)=-1` iii. `Deltan=2-(1+3)=-2` iv. `Deltan=1-(1+1)=-1` |
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| 460. |
Reaction in which yield of product will increase with increase in pressure isA. `H_(2(g)) + I_(2(g))hArr2HI_((g))`B. `H_(2)O_((g)) + CO_((g))hArrCO_(2(g)) + H_(2(g))`C. `H_(2)O_((g)) + C_((s))hArrCO_((g)) + H_(2(g))`D. `CO_((g)) + 3H_(2(g))hArrCH_(4(g)) + H_(2)O_((g))` |
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Answer» Correct Answer - D In reaction `CO + 3H_(2)hArrCH_(4) + H_(2)O` Volume is decreasing in forward direction so no increasing pressure the yield of product will increase. |
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| 461. |
In which of the following gaseous equilibrium an increase in pressure will increase the yield of the productsA. `2HI hArrH_(2) + I_(2)`B. `2SO_(2) + O_(2)hArr2SO_(3)`C. `H_(2) + Br_(2)hArr 2HBr`D. `H_(2)O + COhArrH_(2) + CO_(2)` |
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Answer» Correct Answer - A Increaase in pressure causes the equilibrium to shift in that direction in which no. of moles (volume) is less. |
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| 462. |
Given a system in equilibrium, an increase in concentration of the products is always produced by arise in temperature, when the reaction is:A. gas phaseB. spontaneousC. exothermicD. endothermic |
| Answer» An endothermic reaction is favoured with rise in temperature. | |
| 463. |
The reaction quotient `(Q)` predicts:A. the direction of equilibrium to be attainedB. the ratio of activities at equilibrium i.e., `K_(c)`C. the ratio of activities at any-timeD. all of these |
| Answer» Correct Answer - These are facts. | |
| 464. |
A saturated solution of iodine in water contain `0.330g I_(2) per litre`. More than this `I_(2)` can be dissolved in `KI` solution because of the following equilibrium. `I_(2(g))+I^(-)hArrI_(3)^(-)` A `0.100 M KI` solution actually dissolves `12.5 g "iodine per litre"`, most of which is converted to `I_(3)^(-)`. Assuming that the concentration of `I_(2)` in all saturated solution is the same, calculate the equilibrium constant for the above reaction. What is the effect of adding water to a clear saturated solution of `I_(2)` in the `KI` solution? |
| Answer» `707 litre mol^(-1)` , | |
| 465. |
Calculate `K_(c )` for the reaction `KI + I_(2) hArr KI_(3)`. Given that initial weight of `KI` is `1.326 g` weight of `KI_(3)` is `0.105 g` and number of moles of free `I_(2)` is `0.0025` at equilibrium the volume of solution is `1-L`.A. `0.032`B. `0.024`C. `0.064`D. 0.012 |
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Answer» Correct Answer - A |
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| 466. |
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant `K_(c)` is:A. `DeltaG^(@)=RTIn K`B. `DeltaG^(@)=-RTIn K`C. `DeltaG=RTIn K`D. `DeltaG=-RTIn K` |
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Answer» Correct Answer - b |
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| 467. |
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant `K_(c)` is:A. `-DeltaG^(@)=RT "in" K`B. `DeltaG=RT "In" K`C. `-DeltaG=RT "In" K`D. `DeltaG^(@)=RT "In" K` |
| Answer» Correct Answer - A | |
| 468. |
For the reaction , `O_2(g)+2F_2(g)to2OF_2(g),K_p=4.1` If `P_(O_2)(g)=0.116` atm and `P_(F_2)(g)=0.0461` atm at equilibrium , what is the pressure of `OF_2(g)`?A. 0.101 atmB. 0.032 atmC. 0.760 atmD. 166 atm |
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Answer» Correct Answer - B |
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| 469. |
In Q.No .5, if the mixture of gases was allowed to come to quilibrium .The volume of the reaction vessel was then rapidly increased by a factor of two .As a result of the change the reaction quotient `(Q_(c))` would:A. increase because of the pressure decreaseB. decrease because of the pressure decreaseC. remain the same because the equilibrium constant is indendent of volumeD. increase because the reaction is endothermioc |
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Answer» Correct Answer - a |
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| 470. |
Pure `PCl_(5) ` is introduced into an evacaated chamber and to equilibrium at `247^(@)C` and 2.0 atm .The equilibrium gases mixure contains 40% choririne by volume . Calculate `K_(p)` at `247^(@)C` for the reaction `PCl_(5)(g)hArrPCl_(3)(G)+Cl_(2)(g)`A. `0.625` atmB. `4 atm`C. `1.6 atm`D. none of these |
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Answer» Correct Answer - c |
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| 471. |
For the reaction Calculate `K_(P)`at900K where the equilibrium straem -hydrogen mixture was 45% `H_(2)` by volume :A. `1.49`B. `1.22`C. `0.67`D. none of these |
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Answer» Correct Answer - a |
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| 472. |
Consider the reaction `SO_(2)Cl_(2) hArr SO_(2)(g)+Cl_(2)(g)` at `375^(@)C`, the value of equilibrium constant for the reaction is `0.0032`. It was observed that the concentration of the three species is `0.050 mol L^(-1)` each at a certain instant. Discuss what will happen in the reaction vessel? |
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Answer» In this equation, concentration of three species i.e., `SO_(2)Cl_(2)(g), SO_(2)(g)`, and `Cl_(2)(g)` each is given, but it is not mentioned that whether the system is at equilibrium or not. So first check it. Find reaction coefficient for given equation. `Q=([SO_(2)][Cl_(2)])/([SO_(2)Cl_(2)])=((0.05)(0.05))/((0.05))=0.05` `rArr Q ne K_(eq)`, so system is not at equilibrium state. As `Q gt K_(eq)`, the concentration must adjust till `Q=K_(eq)` for equilibrium. This can happen only if reaction shifts backwards, and products recombine to give back reacyants. Hence, in the vessel, the system move backward so that it can achieve equilibrium state. |
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| 473. |
A system at equilibrium is described by the equation of fixed temperature `T`. `SO_(2)Cl_(2)(g)hArr SO_(2)(g)+Cl_(2)(g)` What effect will an increases in the total pressure caused by a decrease in volume have on the equilibrium ?A. Concentration of `SO_(2)Cl_(2)(g)` increasesB. Concentration of `SO_(2)(g)` increasesC. Concentration of `Cl_(2)(g)` increasesD. Concentration of all gases increases |
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Answer» Correct Answer - D Concentration of all gases increases and equilibrium shifts toward less no. of moles but new equilibrium concentration of every gas would be higher than earlier. |
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| 474. |
For the reaction at `300K` `A(g) hArr V(g)+S(g)` `Delta_(r) H^(@)=-30kJ //mol`, `Delta_(r)S^(@)=-0.1Kj.k^(-1).mol^(-1)` What is the value of equilibrium constant ? |
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Answer» Correct Answer - 2 `Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@)` `=-30+300xx0.1=0` `Delta_(r)G^(@)=-2.303RTlog K` `K=1` |
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| 475. |
Given the following reaction at equilibrium `N_(2)(g) + 3H_(2)(g)hArr2NH_(3)(g)`. Some inert gas at constant pressure is added to the system. Predict which of the following facts:A. more `NH_(3)`(g) is producedB. less `NH_(3)` (g) is producedC. no effect on the equilibriumD. `K_(p)` of the reaction is decreeased |
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Answer» Correct Answer - B On adding inert gas at constant pressure, effect on equilibrium will be similar to as if volume of container has been increased. |
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| 476. |
The equilibrium constant K for the reaction `2HI(g) hArr H_(2)(g)+I_(2)(g)` at room temperature is `2.85` and that at `698 K` is `1.4 xx10^(-2)`. This impliesA. `HI` is exothermic compoundB. `HI` is very stable at room temperatureC. `HI` is relatively less stable than `H_(2)` and `I_(2)` at room temperatureD. `HI` is resonance stablised |
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Answer» Correct Answer - C With increase in temperature, `K` value decreases, which means that at high temperature the reaction proceeds in backward direction or proceeds forward at room temperature. In another words, at room temperature, HI dissociates or `HI` is less stable than `H_(2)` and `I_(2)`. |
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| 477. |
Under what pressure conditions `CuSO_(4).5H_(2)O` be efforescent at `25^(@)C`. How good a drying agent is `CuSO_(4).3H_(2)O` at the same temperature? Given `CuSO_(4).5H_(2)O(s) hArr CuSO_(4).3H_(2)O(s)+2H_(2)O(v)` `K_(p)=1.086xx10^(-4) atm^(2)` at `25^(@)C`. Vapour pressure of water at `25^(@)C` is `23.8` mm of Hg. |
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Answer» An efflorescent salt is one that loses `H_(2)O` to atmosphere. For the reaction `CuSO_(4).5H_(2)O CuSO_(4).3H_(2)O(s)+H_(2)O(v)` `K_(P)== 1.086xx10^(-4)` `:. =1.042xx10^(-2) "atm" =7.92 mm` Given at `25^(@)C, (i.e., 23.8)` which is greater than `7.92` mm and thus, reaction will proceed in the backward direction, i.e., `CuSO_(4).3H_(2)O+2H_(2)O(v) CuSO_(4).5H_(2)O(s)` Thus, `CuSO_(4).5H_(2)O` will not act as efflorescent but on the contrary `CuSO_(4).3H_(2)O` will absorb moisture from the atmosphere under given conditions. |
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| 478. |
From the data given below derive which of the following reactant is most effective drying agent at `0^(@)C`. Given `P_(H_(2)O)^(@)=4.58 mm` at `0^(@)C`. i. `SrCl_(2).6H_(2)O(s) hArr SrCl_(2).2H_(2)O(s)+4H_(1)O(g), K_(p)=6.9xx10^(-12) atm^(4)` ii. `Na_(2)SO_(4).10H_(2)O(s) hArr Na_(2)SO_(4)(s)+10H_(2)O(g), K_(p)=4.08xx10^(-25) atm^(10)` iii. `Na_(2)HPO_(4). 12H_(2)O(s) hArr Na_(2)HPO_(4). 7H_(2)O(s)+5H_(2)O(g), K_(p)=5.525xx10^(-13) atm^(5)` |
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Answer» For (i) `K_(p)=(P_(H_(2)O))^(4)` `:. P_(H_(2)O)=root(4)(sqrt(K_(p)))=root(4)(sqrt(6.9xx10^(-12)))=1.62xx10^(-3) "atm"` For (ii) `K_(p)=(P_(H_(2)O))^(10)` `P_(H_(2)O)=root(10)(sqrt(K_(P)))=root(10)(sqrt(4.08xx10^(-25)))=3.64xx10^(-3) "atm"` For (iii) `K_(p)=(P_(H_(2)O))^(5)` `:. P_(H_(2)O)=root(5)(sqrt(K_(P)))=root(5)(sqrt(5.25xx10^(-13)))=3.5xx10^(-3) "atm"` All these reactant can be used as drying agent since their `P_(H_(2)O)` is lesser than `V.P.` of `H_(2)O` in atmosphere, i.e., `4.58 mm` or `6.02xx10^(-3) "atm"`. The efficiency order is `SrCl_(2).6H_(2)O gt Na_(2)HPO_(4). 12H_(2)O gt Na_(2)SO_(4). 10 H_(2)O` |
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| 479. |
One "mole" of `N_(2)O_(4)(g)` at `100 K` is kept in a closed container at `1.0` atm pressure. It is heated to `400 K`, where `30%` by mass of `N_(2)O_(4)(g)` decomposes to `NO_(2)(g)`. The resultant pressure will beA. `4.2`B. `5.2`C. `3.2`D. `6.2` |
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Answer» b. Since volume is constant. Therefore on increasing the temperature four times, Pressure becomes four times `:. P=4 "atm"` Decomposition of `N_(2)O_(4) 30%` by man `alpha 30%` by mole `:. a=30/100=0.3` `{:(,N_(2)O_(4),hArr,2NO_(2)),("Initial",1,,0),("at. equilibrium",1-0.3,,2xx0.3),(,=0.7,,=0.6),("Total moles"=0.7+0.6=1.3,,,):}` `("Initial mole")/("moles at equilibrium")=(("Initial pressure after"),("chabge of temperature"))/("Equilibrium pressure")` `1/1.3=4/PrArr P=5.2 "atm"` Alternate method: `alpha=(T_(1)P_(2)-T_(2)P_(1))/(T_(2)P_(1))` `0.3=(100xxP_(2)-400xx1)/(400xx1)` `rArr P_(2)=5.2 "atm"` |
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| 480. |
The equilibrium constant for the reaction, `A+B hArr C+D` is `2.85` at room temperature and `1.4xx10^(-2)` at `698 K`. This shows that the forward reaction isA. ExothermicB. EndothermicC. UnpredictableD. There is no relationship between `DeltaH` and `K`. |
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Answer» Correct Answer - A Because equilibrium constant decreases on increasing temperature |
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| 481. |
The decomposition of `N_(2)O_(4)` to `NO_(2)` is carried out at `280^(@)C` in chloroform. When equilibrium is reached, `0.2` mol of `N_(2)O_(4)` and `2xx10^(-3)` mol of `NO_(2)` are present in a `2L` solution. The equilibrium constant for the reaction `N_(2)O_(4) hArr 2NO_(2)` isA. `1xx10^(-2)`B. `2xx10^(-3)`C. `1xx10^(-5)`D. `2xx10^(-5)` |
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Answer» Correct Answer - C `{:(,N_(2)O_(4),hArr,2NO_(2)),("Initial",1,,0),("Final",1-x,,2x),(,1-x=0.2 "mol",,2x=2xx10^(-3) "mol"):}` `[N_(2)O_(4)]=0.2/2=0.1 M [NO_(2)]=(2xx10^(-3))/(2)=10^(-3) M` `K=([NO_(2)]^(2))/([N_(2)O_(4)])=((10^(-3))^(2))/0.1=10^(-5)` |
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| 482. |
For the reaction `A(g) +3B(g) hArr 2C(g) `at `27^(@)C`, 2 moles of `A`, 4 moles of B and 6 moles of C are present in 2 litre vessel. If `K_(c)` for the reaction is 1.2, the reaction will proceed in `:`A. forward directionB. backward directionC. neither directionD. none of these |
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Answer» Correct Answer - 1 `Q_(c)=(((6)/(2))^(2))/(((2)/(2))((4)/(8))^(3))=(9)/(8)` `Q_(c) lt K_(c)` so reaction will proceed in forward direction. |
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| 483. |
At a certain temperature, `K_(c)` for `SO_(2)(g)+NO_(2)(g) hArr SO_(3)(g)+NO(g)` is `16`. If we take one mole of each of all the equilibrium concentration of NO and `NO_(2)`?A. `1.6 mol L^(-1)`B. `0.8 mol L^(-1)`C. `0.4 mol L^(-1)`D. `0.6 mol L^(-1)` |
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Answer» Correct Answer - C `{:(SO_(2)(g),+,NO_(2)(g),hArr,SO_(3)(g),+,NO(g),),(1,,1,,1,,1,"Initial conc"),(1-x,,1-x,,1+x,,1+x,"At Eq."):}` `K=([SO_(3)][NO])/([SO_(2)][NO_(2)])=((1+x)(1+x))/((1-x)(1-x))` `16=((1+x)^(2))/((1-x)^(2))rArr((1+x))/((1-x))=4` or `x=0.6` `[NO_(2)]=1-x=1-0.6=0.4 "mol" L^(-1)` |
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| 484. |
On cooling of following system at equilibrium `CO_(2)(s)hArrCO_(2)(g)`A. There is no effect on the equilibrium stateB. More gas is formedC. More gas solidifiesD. None of above |
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Answer» Correct Answer - C `T datt`, then `V darr`, then `P uarr` , equilibrium shifts is such direction so as decrease pressure, i.e., backward direction. |
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| 485. |
Partial pressure of `O_(2)` in the reaction `2Ag_(2)O(s) hArr 4Ag(s)+O_(2)(g)` isA. `K_(p)`B. `sqrt(K_(p))`C. `root(3)(K_(p))`D. `(K_(p))^(2)` |
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Answer» Correct Answer - A::B::C `2Ag_(2)O(s) hArr 4Ag(s)+O_(2)(g)` `:. K_(p)=P_(O_(2))` |
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| 486. |
For which reaction at `298K`, the value of `(K_(p))/(K_(c))` is maximum and minimum respectively:A. `N_(2)O_(4)hArr2No_(2)`B. `2SO_(2)+O_(2)hArr2SO_(3)`C. `X+YhArr4Z`D. `A+3BhArr7C` |
| Answer» Correct Answer - B::D | |
| 487. |
For the following equilibrium, `2SO_(2)(g) +O_(2)(g) hArr 2SO_(3)(g)` the total equilibrium pressure is `P_(1)`. If volume of the system is reduced to `1//2` of this initial volume then equilibrium is restablished. The new equilibrium total pressure will be `:`A. `2P_(1)`B. `3P_(1)`C. `3.5P_(1)`D. less than `2P_(1)` |
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Answer» Correct Answer - 4 `{:(,2SO_(2)(g),+,O_(2)(g),hArr,2SO_(3)(g)),("at Eq",x,,y,,z),("on reducing vol",2x,,2y,,2z),("new Eq.",2x-2a,,2y-a,,2z-2a),("total pressure",=2(x+y+z)-a,=2P_(1)-a,,,lt2P_(1)):}` |
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| 488. |
If `CI_(2)` HCI and `O_2` are mixed in such a manner that the partial pressure of each is 2 atm and the mixture (V.P.=0.5 atm). What would be the approximate partial pressure of `CI_2` when equilibrium is attained at temperature (T)? If your answer in scientific notation is `xx x10^(-y)` ,write the value of y. [Given:` 2H_2O(g)+2CI_2(g)hArr4HCI(g)+O_2(g),K_(p)=12xx10^(8)`atm) |
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Answer» Correct Answer - 3 |
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| 489. |
Variation of `log_(10)K` with `(1)/(T)` is shown by the following graph in which straight line is at `45^(@)` , hence `Delta H^(@)` is: A. `+4.606` calB. `-4.606` calC. 2 calD. `-2` cal |
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Answer» Correct Answer - B Slope of plot `= (-Delta H^(@))/(2.303 R) = 1` `implies Delta H^(@) = - 2.303xx2 = - 4.606` cal |
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| 490. |
`C(s)hArr2A(g)+B(s)` If the dissociation of `C(s)` is `alpha` and `d` is the density of the gaseous mixture in the container. Initially container have only `C(s)` and the reaction is carried at constant temperature and pressure.A. B. C. D. |
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Answer» Correct Answer - D As in the gaseous mixture only `A` will be present so the molecular weight of the gaseous mixture will be `M_(A)` because `PM_(A)=dRT "where" R,P,T` are const so `dpropMA` and hence it does not depend on `alpha` so `d` will remain constant. option (d) is correct. |
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| 491. |
When pressure is applied to the equilibrium system, `IcehArrWater`. Which of the following phenomenon will happen?A. More water will formB. More ice will formC. There will be no a effect equilibriumD. Water will decompose into `H_(2)` and `O_(2)` |
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Answer» Correct Answer - A `underset("more volume")("Ice")hArrunderset("less volume")(Water)` On increasing pressure, equilibrium shifts forward. |
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| 492. |
Variation of `log_(10)K` with `(1)/(T)` is shown by the following graph in which straight line is at `45^(@)` , hence `Delta H^(@)` is: A. `+4.606cal`B. `-4.606cal`C. `2cal`D. `-2cal` |
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Answer» Correct Answer - B `K=Ae^(-DeltaH//RT)` `logK=logA-(DeltaH)/(2.303RT)`. `log K=logA-(DeltaH)/(2.303R)xx(1)/(T)`. `log K=[-(DeltaH)/(2.303R)]xx(1)/(T)+logA`. (-DeltaH)/(2.303R)=1`. ltbgt `DeltaH=-2.303R=-4.606cal`. |
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| 493. |
The value of `K_(p)` for the reaction at `27^(@)C` `Br_(2)(l)+CI_(2)(g)hArr2BrCI(g)` is `1atm`. At equilibrium in a closed container partial pressure of `BrCI` gas `0.1atm` and at this temperature the vapour pressure of `Br_(2)(l)` is also `0.1`atm. Then what will be minimum moles of `Br_(2)(l)` to be added to `1` mole of `CI_(2)`, initially, to get above equilibrium stiuation,A. `(10)/(6) mol es`B. `(5)/(6)mol es`C. `(15)/(6)mol es`D. `2mol es` |
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Answer» Correct Answer - C `Br_(2)(l)+CI_(2)(g)hArr2BrCI(g)` `t=0 1 0` `(1-x) 2x` `K_(p)=((P_(BrCl)^(2)))/(P_(Cl_(2)))=1 "so", P_(Cl_(2))=(P_(BrCl))^(2))=0.01atm` then at equilibrium, `(n_(BrCl))/(n_(Cl_(2))=(0.1)/(0.01)=10=(2x)/(1-x)` So, `10-10x=2x "or" x=(10)/(12)=(5)/(6)mol es` Mole of `Br_(2)(l)` required for maintaining vapour pressure of `01atm` `=2xx(5)/(6)mol es=(10)/(6)mol es="moles of" BrCl(g)`. Moles required for taking part in reaction`="moles of" Cl_(2)` used up`=(5)/(6)mol es`. Hence total moles required`=(5)/(6)+(10)/(6)=(15)/(6)mol es`. |
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| 494. |
What is the effect of temperature and pressure on the yields of products? a. `N_(2)(s)+3H_(2)(g) hArr 2NH_(3)+x cal` b. `N_(2)(g)+O_(2)(g) hArr 2NO(g)-y cal` c. `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)+46.9 kcal` d. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)-15.0 kcal` |
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Answer» a. Greater yield of `NH_(3)` at low temperature and high pressure. b. Geater yield of `NO` at high temperature. Pressure has no effect. c. Greater yield of `SO_(3)` at low temperature and high pressure. d. Greater yield of `PCl_(3)` an `Cl_(2)` at high temperature and low pressure. |
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| 495. |
When pressure is applied to the equilibrium system, `IcehArrWater`. Which of the following phenomenon will happen?A. More ice will be formedB. Water will evaporateC. More water will be formedD. Equilibrium will not be formed |
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Answer» `V_(Ice)gtV_(Water)`, Thus increase in pressure will show forward reaction. |
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| 496. |
The rate of a stoichiometric reaction between a solid and gas in a container may be increased by increasing all of the following factors EXCEPT the :A. pressure of the gas.B. temperature of the gas.C. volume of the container.D. surface area of the solid. |
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Answer» Correct Answer - C |
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| 497. |
`DeltaG^(@)` for `(1)/(2)N_(2)+(3)/(2)H_(2)hArrNH_(3)` is `-16.5 kJmol^(-1)`. Find out `K_(p)` for the reaction. Also report `K_(p)` and `DeltaG^(@)` for : `N_(2)+3H_(2)hArr2NH_(3)` at `25^(@)C`. |
| Answer» `779.41 atm^(-1)`, `6.07xx10^(5) atm^(-2)`, `32.998 kJ` , | |
| 498. |
Write a stoichiometric equation for the reaction between `A_(2)` and `C` whose mechanism is given below. Determine the value of equilibrium constant for the first step. Write a rate law equation for the over all reaction in terms of its initial reactants. (`i`) `A_(2)overset(K_(1))underset(K_(2))hArr2A` `K_(1)=10^(10)s^(-1)` and `K_(2)=10^(10)M^(-1)s^(-1)` (`ii`) `A+CtoAC` `K=10^(-4)M^(-1)S^(-1)` |
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Answer» It is apparent from both the steps that step (`ii`) is slowest and thus `Rate=K[A][C]` ….(`1`) However overall rate constant `K` can be obtained in terms of `A_(2)` as follows, `eq.(`i`)+2xxeq.(`ii`)`, `A_(2)+2Cto2AC` Also for step (`i`), `K_(c)=(K_(1))/(K_(2))=([A]^(2))/([A_(2)])=(10^(10))/(10^(10))=1` or `[A_(2)]=[A]^(2)` or `[A]=[A_(2)]^(1//2)` Thus by eq. (`1`), `Rate=K[C][A_(2)]^(1//2)` `=K[C][A_(2)]^(1//2)` |
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| 499. |
`DeltaG^(@)` for `(1)/(2)N_(2)+(3)/(2)H_(2)hArrNH_(3)` is `-16.5 kJ mol^(-1)` at `25^(@)C`. Find out `K_(p)` for the reaction. Also report `K_(p)` and `DeltaG^(@)` for: `N_(2)+3H_(2)hArr2NH_(3)` at `25^(@)C` |
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Answer» `-DeltaG^(@)=2.303RTlogK_(p)` `-(-16.5xx10^(3))=2.303xx8.314xx298 logK_(p)` `logK_(p)=(16500)/(2.303xx8.314xx298)` `:. K_(p)=779.41 atm^(-1)` Also , `K_(P_(1))` for `N_(2)+3H_(2)hArr2NH_(3)` `K_(p_(1))=(K_(p))^(2)=(779.41)^(2)` `K_(p_(1))=6.07xx10^(5)atm^(-2)` Also, `-DeltaG_(1)^(@)=2.303xx8.314xx298 log (6.07xx10^(5))J` `=32.998 kJ` `:. DeltaG_(1)^(@)=-32.998 kJ mol^(-1)` |
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| 500. |
Consider the following equations for cell reaction: `A+BhArrC+D` ….(`1`) and `2A+BhArr2C+2D` …(`2`) How are `E_(cell)^(@)` for equations (`1`) and (`2`) are related? What is relation between two equilibrium constants? |
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Answer» `E_(cell)^(@)` for eq. (`1`) and (`2`) is same as it does not depend upon stoichiometry of chemical equation. Also `K_(1)=([C][D])/([A][B])`, `K_(2)=([C]^(2)[D]^(2))/([A]^(2)[B]^(2))` `:. K_(1)=sqrt(K_(2))` |
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