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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.
| 251. |
Consider the following equilibrium: `SO_(3) rarr SO_(2)+O_(2)` `8.0 g` of `SO_(3)` are put in a container at `600^(@)C`. The equilibrium pressure and density are `1.8` atm and `1.6 g L^(-1)`, respectively a. Find the value of `K_(p)`. b. Also find the moles of helium that is to be added at equilibrium to double the pressure at constant temperature. |
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Answer» Weight of `SO_(3)` taken `=8.0 g` molecular weight of `SO_(3)=80 g//"mol"` `:.` mole of `SO_(3)=8/80=0.1 "mol"` `{:(SO_(3),hArr,SO_(2),+,1//2O_(2)),(0.1-x,,x,,x//2):}` `PV=nRT` or `PM=dRT` Total moles `=0.1+x//2` `rArr M=(dRT)/(P)=(1.6xx0.0821xx873)/(1.8)=63.6` `n_(1)M_(1)=n_(2)M_(2), n_(1)/n_(2)=M_(2)/M_(1)` `rArr (0.1)/(0.1+x//2)=M_(av)/M_(SO_(3))=63.6/80rArr x=0.05` `:. K_(p)=(((x//2)/(0.1+x//2)P)^(1//2)((xP)/(0.1+x//2)))/(((0.1-x)/(0.1+x//2)P))=0.61` `(P=1.8 "atm")` b. `n_(i)=0.1 x//2` To double pressure by adding inert gas at constant `T` double moles at equilibrium `rArr n_(f)=2(0.1+x//2)` `rArr n_(He)` added `=0.1+x//2=0.1+(0.05//2)=0.125` |
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| 252. |
A container of volume `V L` contains an equilibrium mixture that consists of `2` mol each of gaseous `PCl_(5), Pcl_(3)` and `Cl_(2)` at `3` atm `T K`. Some `Cl_(2)` is added unitl the volume is double keeping `P` and `T` constant. Calculate moles of `Cl_(2)` added and `K_(p)` for `PCl_(5) rarr PCl_(3)+Cl_(2)` |
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Answer» `{:(PCl_(5),hArr,PCl_(3),+,Cl_(2),),(2,,2,,2,P=3.0 "atm"):}` Total moles `=2+2+2=6` `p_(PCl_(5))=2xx3//6=1, p_(PCl_(3))=1` `K_(p)=((1)(1))/(1)=1 "atm"` When volume is double `(i.e., Vrarr 2V)` at constant `P` and `T` total moles brcome `=12` If volume is double, reaction will more backward at constant `P` and `T` `{:(PCl_(5),hArr,PCl_(3),+,Cl_(2),("Let x mole of" Cl_(2) "us added")),(2,,2,,2+x,("Let y mole of" PCl_(5) "is formed")),(2+y,,2-y,,(2+x)-y,):}` `p_(PCl_(5))=((2+y))/(12)xx3, p_(PCl_(3))=((2-y))/(12)xx3` `p_(Cl_(2))=((2+y)-y)/(12)xx3` `:. K_(p)=1=(((2-y)/(12))xx3xx[((2+x)-y)/(12)]xx3)/(((2+y)/(12))xx3) ...(i)` Also `2+y+2-y+(2+x)-y=12 ...(ii)` From equation (i) and (ii), solve, `y=2//3, xrArr 20//3` |
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| 253. |
The plots of `log_(10) K vs. 1//T` leads toa straight line having intercept equal to:A. `DeltaG^(@)`B. `(DeltaG^(@))/(2.303R)`C. `(DeltaS^(@))/(2.303R)`D. `(DeltaH^(@))/(2.303R)` |
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Answer» `DeltaG^(@)=-2.303RT log_(10)K` `log_(10) K=-(DeltaG^(@))/(2.303RT)` `= -(DeltaH^(@)-TDeltaS^(@))/(2.303RT)` `= -(DeltaH^(@))/(2.303RT)+(DeltaS^(@))/(2.303R)` |
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| 254. |
A graph plotted between `log_(10) K_(c)` and `1//T` is straight line with intercept `10` and slpoe equal t0 `0.5.` Calculate : (`i`) pre -exponential factor `A`. (`ii`) heat of reaction at `298 K`. (`iii`) equilibrium constant at `298 K`. (`iv`) equilibrium constant in at `800 k` assuming `DeltaH` remains constant in between `298 K` and `800 K`. |
| Answer» (`i`) `10^(10)`, (`ii`) `-2.303 cal mol^(-1)`, (`iii`) `1.004xx10^(10)`, (`iv`) `1.001xx10^(10)` | |
| 255. |
`K_(p)` for the reaction, `N_(2)+3H_(2)hArr2NH_(3)` is `1.6xx10^(-4)atm^(-2)` at `400^(@)C`. What will be `K_(p)` at `500^(@)C`? Heat of reaction in this temperature range is `-25.14 kcal`. |
| Answer» `1.462xx10^(-5) atm^(-2)` , | |
| 256. |
The equilibrium constant for a reaction `A+B hArr C+D` is `1xx10^(-2)` at `298 K` and is `2` at `273 K`. The chemical process resulting in the formation of `C` and `D` isA. exothermicB. endothermicC. unpredictableD. there is no relationship between `K` and `DeltaH` |
| Answer» `K_(c)` increase with temperature and thus reaction is endothermic. | |
| 257. |
The equilibrium constant for a reaction `A+B hArr C+D` is `1.0xx10^(-2)` at `298` and is `2.0` at `373 K`. The chemical process resulting in the foemation of C and D isA. ExothermicB. EndothermicC. UnpredictableD. None |
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Answer» Correct Answer - B Since K increases on increasing temperature, so the reaction will go forward by increasing temperature and hence is endothermic. |
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| 258. |
The value of K for the reaction `O_(3)(g)+OH(g) hArr H(g)+2O_(2)(g)` Changed from `0.096` at `298 K` to `1.4` at `373 K`. Above what temperature will the reaction become thermodynamically spontaneous in the forward direction assuming that `DeltaH^(ɵ)` and `DeltaS^(ɵ)` values for the reaction do not change with change in temperature? Given that `DeltaS_(298)^(ɵ)=10.296 J K^(-1)`. |
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Answer» We have `"log" K_(2)/K_(1)=(DeltaH^(ɵ))/(2.303 R)((T_(2)-T_(1)))/(T_(1)T_(2))` `"log" 1.4/0.096=(DeltaH^(ɵ))/(2.303xx8.314)((373-298)/(373-298))` `DeltaH^(ɵ)=33025 J` Now the temperature above which the forward reaction will be spontaneous is actually the temperature at which the reaction attains equilibrium, that is, when `K=1` or `log K=0` `:. DeltaG^(ɵ)=-2.303RT log K=-2.303RT log 1.0=0` From thermodynamics, we get `DeltaG^(ɵ)=DeltaH^(ɵ)-T DeltaS^(ɵ)` `0=33025-Txx10.296` or `T=320.75 K` |
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| 259. |
A reversible reaction is endothermic in forward direction. Then which of the following is (are) correct?A. In K vs `1//T` will be a straight line with negative slopeB. `d/(dT) ln K gt 0`C. A plot of d ln K against `1//T^(2)` will have positive slopeD. An increase in temperature will shift the reaction in the forward direction. |
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Answer» Correct Answer - A::B::C `K_(eq)=e^(-DeltaH//RT)(A, B, C)` `DeltaH gt 0(f), T uarr`, Reaction proceed in forward direction by absorbimg heat. |
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| 260. |
Solubility of a gas in liquid increases onA. Addition of a catalystB. Decreaseing of pressureC. Increasing of pressureD. Increasing of temperature |
| Answer» Correct Answer - C | |
| 261. |
`N_(2)``O_(4)` is `25%` dissociated at `37^(@)C` and one atmosphere pressure. Calculate (i) `K_(p)` and (ii) the percentage dissociation at 0.1 atm and `37^(@)C` |
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Answer» Correct Answer - (i) `0.266` atm (ii) `63.25%` `N_(2)O_(4)hArr2NO_(2)` `{:(1,0,,,),(1-.25,.50,,,),(.75,.50,,,n_("total")=1.25):}` `P_(N_(2)O_(4))=((P_(NO_(2)))^(2))/((P_(N_(2)O_(4))))=(.50//1.25)^(2)/((.75//1.25))=(.50xx.50)/(1.25xx.75)=(4)/(15)=0.266` At pressure `0.1` atm `N_(2)O_(4)hArr2NO_(2)` `1 0 ` `(1-alpha) 2a` `K_(P)=((1-alpha)/(1+alpha))xx0.1 , P_(NO_(2)=(((2alpha))/(1+alpha))xx0.1` `K_(P)=((2alpha)/(1+alpha)xx0.1)^(2)/(((1-alpha)/(1+alpha))xx0.1), K_(P)=(4alpha^(2)xx0.1)/((1+alpha)(1-alpha))` `0.266=(0.1xx4alpha^(2))/(1-alpha^(2))` `0.665=(1+0.665)alpha^(2). rArr alpha=63.25%` |
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| 262. |
Which of the following factors will increase the solubility of `NH_(3)` gas in `H_(2)O`? a. Increase in pressure b. Addition of water c. Increase in temperature Decrease in pressure |
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Answer» `NH_(3)(g) hArr NH_(3)(aq)` The solubility increases: (i) By increasing the pressure (ii) By decreasing the temperature (since dissolution of a gas an exothermic process. (iii) By adding solvent |
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| 263. |
Assertion (A) : The value of K for a reaction may increase or decrease with increase in temperature depending upon whether the reaction is exothermic or endothermic. Reason (R) : With increase in temperature, the axtent of reaction increases.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 - B | |
| 264. |
What would happen to a reversible reaction at equilibrium, when a. The temperature is raised, given that its `DeltaH` is positive. b. The temperature is lowered, given that its `DeltaH` is positive. c. The temperature is lowered, given that its `DeltaH` is negative. The prssure is lowered, given that `Deltan` is negative. e. The pressure is incteased, given `Deltan` is negative. |
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Answer» a. More of the products will be formed. b. Less of the products will be formed. c. More of the products will be formed. d. Less of the products will be formed. e. More of the products will be formed. |
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| 265. |
The reaction between `H_(2)` and `CO_(2)` to form `CO` and `H_(2)O` in the gas phase is exothermic. Predict the changes that take place when the system originally at equilibrium is stressed in each of the following wats a. `CO_(2)` is removed. b. `CO` is reamoved. c. The temperature is decreased. d. The pressure of the system is increased. e. The volume of the system is increased. |
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Answer» a. Moves towards reactants side. b. Moves towards products side. c. Moves towards reactants side. d. Remains unchanged. e. Remains unchanged. |
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| 266. |
The relation between `K_(p)` and `K_(c)` is `K_(p)=K_(c)(RT)^(Deltan)` unit of `K_(p)=(atm)^(Deltan)`, unit of `K_(c)=(mol L^(-1))^(Deltan)` `H_(3)ClO_(4)` is a tribasic acid, it undergoes ionisation as `H_(3)ClO_(4) hArr H^(o+)+H_(2)ClO_(4)^(-), K_(1)` `H_(2)ClO_(4)^(-) hArr H^(o+)+HClO_(4)^(2-), K_(2)` `HClO_(4)^(2-) hArr H^(o+)+ClO_(4)^(3-), K_(3)` Then, equilibrium constant for the following reaction will be: `H_(3)ClO_(4) hArr 3H^(o+)+ClO_(4)^(3-)`A. `K_(1)K_(2)K_(3)`B. `((K_(1)K_(3))^(2))/(K_(2))`C. `K_(1)/K_(2)`D. `(K_(1)K_(2))/(K_(3)^(2))` |
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Answer» Correct Answer - A `H_(3)PO_(4) overset(K_(1))(rarr)H^(o+)+H_(2)PO_(4)^(ɵ)` `H_(2)PO_(4)^(ɵ) overset(K_(2))(rarr)H^(o+)+HPO_(4)^(2-)` `HPO_(4)^(2-)overset(K_(3))(rarr)H^(o+)+PO_(4)^(3-)` `:.` For `H_(3)PO_(4) overset(K)(rarr) 3H^(o+)+PO_(4)^(3-)` `K=K_(1)xxK_(2)xxK_(3)` |
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| 267. |
The relation between `K_(p)` and `K_(c)` is `K_(p)=K_(c)(RT)^(Deltan)` unit of `K_(p)=(atm)^(Deltan)`, unit of `K_(c)=(mol L^(-1))^(Deltan)` Consider the following reactions: i. `CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g), K_(1)` ii. `CH_(4)(g)+H_(2)O(g) hArr CO(g)+3H_(2)(g), K_(2)` iii. `CH_(4)(g)+2H_(2)O(g) hArr CO_(2)(g)+4H_(2)(g), K_(3)` Which of the following is correct?A. `K_(3)=K_(1)//K_(2)`B. `K_(3)=K_(1)^(2)//K_(2)^(3)`C. `K_(3)=K_(1)xxK_(2)`D. `K_(3)=K_(1)sqrt(K_(2))` |
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Answer» Correct Answer - C `K_(3)=K_(1)xxK_(2)` |
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| 268. |
The relation between `K_(p)` and `K_(c)` is `K_(p)=K_(c)(RT)^(Deltan)` unit of `K_(p)=(atm)^(Deltan)`, unit of `K_(c)=(mol L^(-1))^(Deltan)` Given: `2NO(g)+O_(2)(g) hArr 2NO_(2)(g), K_(1)` `2NO_(2)(g) hArr N_(2)O_(4)(g), K_(2)` `2NO(g)+O_(2)(g) hArr N_(2)O_(4)(g), K_(3)` Which of the following relations is correct?A. `K_(3)=K_(1)//K_(2)`B. `K_(3)=K_(1)xxK_(2)`C. `K_(3)=K_(1)+K_(2)`D. `K_(3)=K_(1)//sqrt(K_(2))` |
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Answer» Correct Answer - B `K_(3)=K_(1)xxK_(2)` |
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| 269. |
The relation between `K_(p)` and `K_(c)` is `K_(p)=K_(c)(RT)^(Deltan)` unit of `K_(p)=(atm)^(Deltan)`, unit of `K_(c)=(mol L^(-1))^(Deltan)` Consider the two reaction: `XeF_(6)(g)+H_(2)O(g) hArr XeO_(3)F_(4)(g)+2HF(g), K_(1)` `XeO_(4)(g)+XeF_(6)(g) hArr XeOF_(4)(g)+XeO_(3)F_(2)(g), K_(2)` Then the equilibrium constant for the following reaction `XeO_(4)(g)+2HF(g) hArr XeO_(3)F_(2)(g)` is given by:A. `K_(1)//K_(2)^(2)`B. `(K_(1)//K_(2))^(1//2)`C. `K_(1)^(2)//K_(2)^(3)`D. `K_(2)//K_(1)` |
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Answer» Correct Answer - D `K=K_(2)/K_(1)` |
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| 270. |
The equilibrium constant has no unit if `Deltan`= …………. |
| Answer» Correct Answer - `Deltan=0`, zero | |
| 271. |
At equilibrium , the amount of each constituent of reaction mixture becomes ………… |
| Answer» Correct Answer - Constant | |
| 272. |
If `Ag^(+)+NH_(3)hArr[Ag(NH_(3))]^(+)`, `K_(1)=3.5xx10^(-3)` and `[Ag(NH_(3))]^(+)+NH_(3)hArr[Ag(NH_(3))_(2)]^(+)`, `K_(2)=1.74xx10^(-3)`. The formation constant of `[Ag(NH_(3))_(2)]^(+)` is :A. `6.08xx10^(-6)`B. `6.8xx10^(-6)`C. `1.6xx10^(3)`D. `1.088xx10^(7)` |
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Answer» `K_(1)=([Ag(NH_(3))]^(+))/([Ag^(+)][NH_(3)])=3.5xx10^(-3)`...(`i`) `K_(2)=([Ag(NH_(3))_(2)]^(+))/([Ag(NH_(3))]^(+)[NH_(3)])=1.74xx10^(-3)`....(`ii`) `K_(3)=K_(1)xxK_(2)=([Ag(NH_(3))_(2)]^(+))/([Ag^(+)][NH_(3)]^(2))` ....(`iii`) `K_(3)` for `Ag^(+)+2NH_(3)to[Ag(NH_(3))_(2)]^(+)` By eqs. (`i`), (`ii`) and (`iii`), `K_(3)=K_(1)xxK_(2)` `=3.5xx10^(-3)xx1.74xx10^(-3)` `=6.08xx10^(-6)` |
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| 273. |
`N_(2)O_(4)` is `25%` dissociated at `37^(@)C` and `1 atm`. Calculate (i) `K_(p)` and (ii) the percentage dissociation at `0.1` atm and `37^(@)C`. |
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Answer» `{:(,N_(2)O_(4),hArr,2NO_(2)),("Initial",1,,0),("At equilibrium",(1-a),,2a):}` Total moles`=1+a` `P_(N_(2)O_(4))=((1-a))/((1+a))p` `P_(NO_(2))=(2a)/((1+a))p` `K_(p)=([NO_(2)]^(2))/([N_(2)O_(4)])=([(2a)/((1+a))p]^(2))/([((1-a))/((1+a))p])=(4a^(2)p)/(1-a^(2))` Putting `a=0.25`, we get `K_(p)=0.267` Calculating the percentage of dissociation at a pressure of `0.1` atm, we get `0.267=(4a^(2)xx0.1)/(1-a^(2))` or `a~~0.63` Percentage dissociation `=63%` |
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| 274. |
For the reaction `N_(2)(g)+O_(2)(g) hArr 2NO(g)`, the equilibrium constant is `K_(1)`. The equilibrium constant is `K_(2)` for the reaction `2NO(g)+O_(2) hArr 2NO_(2)(g)` What is `K` for the reaction `NO_(2)(g) hArr 1/2 N_(2)(g)+O_(2)(g)`?A. `(1)/((K_(1)K_(2)))`B. `(1)/((2K_(1)K_(2)))`C. `(1)/((4K_(1)K_(2)))`D. `(1/((K_(1)K_(2))))^(1//2)` |
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Answer» Correct Answer - D `N_(2) + O_(2)hArr2NO,K_(1) = ([NO]^(2))/([N_(2)] [O_(2)])` …….(1) `2NO + O_(2)hArr2NO_(2),K_(2) = ([NO_(2)]^(2))/([NO]^(2)[O_(2)])` …..(2) Aim `NO_(2) rarr (1)/(2) N_(2) + O_(2), K_(c) = ([NO_(2)]^(1//2)[O_(2)])/([NO]^(2))` ....(3) To get aim reverse eq. `(2)xx(1)/(2) +` eqn. `(1)xx(1)/(2)` By observation `K_(c) = sqrt((1)/(K_(1) K_(2)))` |
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| 275. |
`2` mole of `PCI_(5)` were heated in a `5` litre vessel. It dissociated. `80%` at equilibrium find out the value of equilibrium constatn. Report your answer as `K_(C)xx50`. |
| Answer» Correct Answer - 64 | |
| 276. |
2 moles of `PCI_(5)` was heated in a closed vessel of 2 litre capacity. At equilibrium, `40%` of `PCI_(5)` is dissociated it `PCI_(3)` and `CI_(2)`. The value of equilibrium constant isA. `0.266`B. `0.53`C. `2.66`D. `5.3` |
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Answer» Correct Answer - A `underset2(PCl_(5))harrunderset0(PCl_(3))+underset0(Cl_(2))` `(2xx60)/(100)` `(2xx40)/(100)` `(2xx40)/(100)` Volume of container = 2 L `K_(c) = ((2xx40)/(100xx2)xx(2xx40)/(100xx2))/((2xx60)/(100xx2)) = 0.266` |
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| 277. |
The ratio of the rate of diffusion of a sample of `N_(2)O_(4)` partially dissociated in to `NO_(2)` to pure hydrogen was found to be `1:5` . Calculate the degree of dissociation of `N_(2)O_(4)`.A. `0.84`B. `0.54`C. `0.42`D. `0.64` |
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Answer» Correct Answer - 1 `(M_(mix))/(M_(H_(2)))=25` `M_(mixture)=50` `{:(N_(2)O_(4)(g),hArr,2NO_(2)(g),,),(a,,,,),(a,,2aalpha,,):}` `axx92=50xxa(1+alpha)` `alpha=(42)/(50)=0.84` |
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| 278. |
At `200^(@)C PCl_(5)` dissociates as follows : `PCl_(5)(g0hArrPCl_(3)(g)+Cl_(2)(g)` It was found that the equilibrium vapours are 62 times as heavy as hydreogen .The degree of dissociation of `PCl_(5)` at `200^(@)C` is nearly :A. `10%`B. `42%`C. `50%`D. `68%` |
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Answer» Correct Answer - d |
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| 279. |
The equilibrium constants of the following areA. `K_1K_(3)^(3)//K_2`B. `K_2K_(3)^(3)//K_1`C. `K_2K_(3)//K_1`D. `K_2^(3)K_(3)//K_1` |
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Answer» Correct Answer - B Given, `N_2+3H_2hArr2NH_3,K_1` .....(1) `N_2+O_2hArr2NO,K_2` .....(ii) `H_2+(1)/(2)O_2toH_2O,K_2` .......(iii) To calculate , `2NH_3+(5)/(2)O_2overset(K)hArr2NO+3H_2O` `K=?` .....(iv) On reversing the equaiton (1) and multiplying the equation (iii) by 3, we get `2NH_3hArrN_2+3H_2,(1)/(K_1)` ....(v) `3H_2+(3)/(2)O_2to 3H_2O,K_(3)^(3)` ......(vi) Now, add equation (ii),(v) and (vi), we get the resultant equation (iv). `2NH_3+(5)/(2)O_2overset(K)hArr2NO+3H_2O` `K=(K_2K^2)/(K_1)` |
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| 280. |
Each question contains STATEMENT-1 (Assertion) and STATEMENT-2( Reason). Examine the statements carefully and mark the correct answer according to the instruction given below: STATEMENT-1: For the reaction `A(g)iff B(g)+C(g)`, `K_(p)`=1atm. If we start with equal moles of all gases at 9 tm of initial pressure, then at equilibrium partial pressure of A increases. STATEMENT-2: Reaction quotient `Q_(p)gtK_(p)` hence equilibrium shifts in backward direction.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - A |
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| 281. |
Each question contains STATEMENT-1 (Assertion) and STATEMENT-2( Reason). Examine the statements carefully and mark the correct answer according to the instruction given below: STATEMENT-1:The gas phase reaction `PCl_(3)(g)+Cl_(2)(g) iffPCl_(5)(g)` shifts to the right on increasing pressure. STATEMENT-2: When pressure increase, equilibrium shifts towards more number of moles.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - C |
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| 282. |
In which of the following equilibrium, the value of `K_(p)` is less than `K_(c)`A. `H_(2) +I_(2)hArr2HI`B. `N_(2) + 3H_(2)hArr2NH_(3)`C. `N_(2) + O_(2)hArr 2NO`D. `CO + H_(2)OhArrCO_(2) + H_(2)` |
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Answer» Correct Answer - B `K_(p) = K_(c)(RT)^(Delta n)` , `Delta n = 2 - (2 + 1) = - 1` , i.e., negative, `K_(p) lt K_(c)` . |
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| 283. |
In which of the following case `K_(p)` is less than `K_(c)`A. `H_(2) + C1_(2)hArr2HC1`B. `2SO_(2) + O_(2)hArr2SO_(3)`C. `N_(2) + O_(2)hArr2NO`D. `PC1_(5)hArrPC1_(3) + C1_(2)` |
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Answer» Correct Answer - B ` 2SO_(2) + O_(2) hArr 2SO_(3) ` for this reaction `Delta n = -1` , `K_(c) gt K_(p)` |
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| 284. |
For which of the following `K_(p)` is less than `K_(c)` ?A. `N_(2)O_(4)hArr2NO_(2)`B. `N_(2)+3H_(2)hArr2NH_(3)`C. `H_(2)+I_(2)hArr2HI`D. `CO+H_(2)OhArrCO_(2)+H_(2)` |
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Answer» Correct Answer - B `N_(2) + 3H_(2)hArr2NH_(3)` `K_(p) = K_(c) (RT)^(Delta n)` `Delta n = P_("mole") - R_("mole")` `Delta n = (2 - 4) = - 2` `K_(p) = K_(c) (RT)^(-2)` `K_(p) ltK_(c)` |
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| 285. |
For the following reaction through stages I, II and III `A overset(I)(rarr)B overset(II)(rarr)C overset(III)(rarr)D` quantity of the product formed (x) varies with temperature (T) as given. Select the correct statement. A. Stages I and III are endothermic but II is exothermic.B. Stages I and III are exothermic but II is endothermicC. Stages II and III are exothermic but I is endothermicD. Stage I is exothermic but stages II and III are endothermic. |
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Answer» Correct Answer - A Stage I (A to B): The quantity of `B` is more than that of `A`. This means formation of `B` is favoured with increase in temperature `T`. Thus, stage I must be endothermic. Stage II (B to C): The quenty of `C` is less than that of `B`. Thus, the formation of `C` is less favoured with an increase in temperature. Hence, stage II must be exothermic. Stage III (C to D): Following the above statements stage III must be endothermic. |
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| 286. |
A 0.20 M solution of methanoic acid has degree of ionization of 0.032. Its dissociation constant would beA. `2.1xx10-2`B. `2.1xx10-4`C. `1.1xx10-6`D. `1.6xx10-8` |
| Answer» Correct Answer - A::B | |
| 287. |
Which among the following reactions will be favoured at low pressure?A. `N_(2)(g)+O_(2)(g) hArr 2NO(g)`B. `H_(2)(g)+I_(2)(g) hArr 2HI(g)`C. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`D. `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` |
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Answer» Correct Answer - C On lowering the pressure, equilibrium favour the direction of higher volume. |
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| 288. |
Which measurable property becomes constant in water`hArrwater vapour ` equilibrium at constant temperature. |
| Answer» Correct Answer - Vapour pressure. | |
| 289. |
Which of the following reaction will be favoured at low pressure?A. `H_(2)(g)+I_(2)(g) hArr 2HI(g)`B. `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`C. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`D. `N_(2)(g)+O_(2)(g) hArr 2NO(g)` |
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Answer» Correct Answer - C Number of moles are less on reactant side, hence on lowering the pressure, reaction will proceed in forward direction. |
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| 290. |
At equilibrium, the value of equilibrium constant K isA. `1`B. `2`C. `3`D. `0` |
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Answer» Correct Answer - A At equilibrium `DeltaG=0` `DeltaG=-nRT ln K` `0=- nRT ln K` or `K=1` |
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| 291. |
At equilibrium stage, the rate of forward reaction is ……….. To the rate of backward reaction. |
| Answer» Correct Answer - Equal | |
| 292. |
`COCl_(2)` gas decomposes as: `COCl_(2)(g) hArr CO(g)+Cl_(2)(g)` If one mole of He gas is added in the vessel at equilibrium at constant pressure thenA. `[COCl_(2)]` increases.B. "moles" of `CO` will increases.C. The reaction goes in forward goes in forward direction.D. `K_(c )=1` |
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Answer» Correct Answer - B::C `COCl_(2)(g) hArr CO(g)+Cl_(2)(g), Deltan_(g)=2-1=1` When an inert gas is added at constant `P`, the reaction shifts where `Deltan_(g) gt 0` (forward direction). Hence, (b), (c ) are correct. |
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| 293. |
1.50 moles each of hydrogen and iodine were placed in a sealed 10 litre container maintained at 717 K. At equilibrium `1.25` moles each of hydrogen and iodine were left behind. The equilibrium constant, `K_(c)` for the reaction , `H_(2)(g)+I_(2)(g) hArr 2Hl(g)` at 717 K isA. `0.4`B. `0.16`C. 25D. 50 |
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Answer» Correct Answer - 2 `{:(,H_(2)(g),+,l_(2)(g),hArr,2Hl(g)),(t=0,1.5,,1.5,,0),(t-t_(eq),1.5-x,,1.5-x,,2x):}` We know,` 1.5-x=1.25`,or`" "x=25` `K_(c)=((.5)^(2))/((1.25)^(2))=.16` |
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| 294. |
In line kilns, the following reaction, `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` proceeds to completion because ofA. High temperatureB. `CO_(2)` escapesC. Low temperature and low pressureD. molecular mass of `CaO` is less than that of `CaCO_(3)` |
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Answer» Correct Answer - B Liberation of `CO_(2)(g)` shifts the reaction on forward direction. |
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| 295. |
For the reaction: `A + B + Q hArr C + D`, if the temperature is increased, then concentration of the products willA. IncreaseB. DecreaseC. Remain sameD. Become zero |
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Answer» Correct Answer - A `{:(A,+,B,+,Q,harr,C,+,D),(,,2,,,,,2,):}` The reaction is endothermic so no increasing temperature concentration of product will increase. |
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| 296. |
`K_(p)` and `K_(c)` are inter related as `K_(p)=K_(c)(RT)^(Deltan)` Answer the following questions: Which of the following have same units `K_(p)` ?A. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`B. `AB_(2)(g) hArr AB(g)+B(g)`C. `NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g)`D. `2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g)` |
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Answer» Correct Answer - A::B `K_(p)` for reaction `(a)=(P_((PCl_(3)))xxP_(Cl_(2)))/(P_(PCl_(5)))="atm"` `K_(p)` for reaction `(b)=(P_((AB))xxP_((B)))/(P_((AB_(2))))="atm"` `:. a, b` |
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| 297. |
`K_(p)` and `K_(c)` are inter related as `K_(p)=K_(c)(RT)^(Deltan)` Answer the following questions: The unit of equilibrium constant for `H_(2)(g)+I_(2)(g) hArr 2HI(g)`A. mol `L^(-2)`B. `mol^(2) L^(-2)`C. `L mol^(-2)`D. None of these |
| Answer» Correct Answer - D | |
| 298. |
`K_(p)` and `K_(c)` are inter related as `K_(p)=K_(c)(RT)^(Deltan)` Answer the following questions: In which of the following equilibria `K_(p)` is less than `K_(c)`?A. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`B. `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`C. `H_(2)(g)+Cl_(2)(g) hArr 2HCl(g)`D. `2H_(2)O(l) hArr 2H_(2)(g)+O_(2)(g)` |
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Answer» Correct Answer - B For reaction (b) `K_(p)=K_(c)(RT)^(Deltan)=K_(c)(RT)^(-2)` `:. K_(c)=K_(p)(RT)^(2)` `:. K_(p) lt K_(c)` |
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| 299. |
Which reaction is not affected by change in pressure ?A. `H_(2) + I_(2)hArr2HI`B. `2C + O_(2)hArr 2CO`C. `N_(2) + 3H_(2)hArr2NH_(3)`D. `PC1_(5)hArrPC1_(3) + C1_(2)` |
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Answer» Correct Answer - A `H_(2) + I_(2)hArr 2HI implies Delta n = 2 - 2 = 0` |
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| 300. |
For the reaction `2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g)-x kcal`, which is correct?A. degree of dissociation will increase on decreasing temperatureB. on decreasing the volume of container degree of dissociation will increase.C. `K_(c)` will decrease on increasing temperature.D. on adding inert gas at constant pressure the amount of ammonia will decrease. |
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Answer» Correct Answer - 4 As reaction is endothermic So on increasing temperature reaction gas forward. |
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