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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.
| 751. |
For the reaction: `2NO_(2)(g) hArr 2NO(g)+O_(2)(g)` `K_(c)=1.8xx10^(-6)` at `184^(@)C` `R=0.0831 kJ K^(-1) mol^(-1)` when `K_(p)` and `K_(c)` are compared at `184^(@)C`, it is found thatA. `K_(p)` is greater than `K_(c)`B. `K_(p)` is less than `K_(c)`C. `K_(p)=K_(c)`D. None of the above |
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Answer» Correct Answer - A::B::C As `K_(p)=K_(c )(RT)^(Deltan)` `Deltan=3-2=1` `:. K_(p)=K_(c )(RT)^(Deltan)` or `K_(p) gt K_(c )` |
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| 752. |
Given `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g),K_(1)` `N_(2)(g)+O_(2)(g)hArr2NO(g),K_(2)` `H_(2)(g)+(1)/(2)O_(2)hArrH_(2)O(g),K_(3)` The equilibrium constant for `2NH_(3)(g)+(5)/(2)O_(2)(g)hArr2NO(g)+3H_(2)O(g)` will beA. `K_(1)K_(2)K_(3)`B. `(K_(1)K_(2))/(K_(3))`C. `(K_(1)K_(3)^(2))/(K_(2))`D. `(K_(1)K_(3)^(2))/(K_(3))` |
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Answer» Correct Answer - d |
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| 753. |
At `87^(@)C`, the following equilibrium is established. `H_(2)(g)+S(s) hArrH_(2)S(g), K_(c)=0.08` If `0.3` mole hydrogen and 2 mole sulphur are heated to `87^(@)C` in a `2L` vessel, what will be concentration of `H_(2)S` at equilibrium ?A. `0.11M`B. `0.022M`C. `0.044M`D. `0.08M` |
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Answer» Correct Answer - a |
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| 754. |
Calculate the equilibrium constant for the reaction, `H_(2(g))+CO_(2(g))hArrH_(2)O_((g))+CO_((g))` at `1395 K`, if the equilibrium constants at `1395 K` for the following are: `2H_(2)O_((g))hArr2H_(2)+O_(2(g))` (`K_(1)=2.1xx10^(-13)`) `2CO_(2(g))hArr2CO_((g))+O_(2(g))` (`K_(2)=1.4xx10^(-12)`) |
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Answer» For `2H_(2)O hArr 2H_(2)+O_(2)` `K_(1)=([H_(2)]^(2)[O_(2)])/([H_(2)O]^(2)) …(i)` For `2CO_(2) hArr 2CO+O_(2)` `K_(2)=([CO]^(2)[O_(2)])/([CO_(2)]^(2)) …(ii)` For `CO_(2)+H_(2) hArr H_(2)O+CO` `K=([H_(2)O][CO])/([CO_(2)][H_(2)]) ...(iii)` Thus, dividing equations (ii) by (i), we get `K_(2)/K_(1)=([CO]^(2)[O_(2)])/([CO_(2)]^(2))xx([H_(2)O]^(2))/([H_(2)]^(2)[O_(2)])` `K_(2)/K_(1)=([CO]^(2)[H_(2)O]^(2))/([CO_(2)]^(2)[H_(2)]^(2))=K^(2)` [By equation (iii)] or `K=(K_(2)/K_(1))^(1//2)=((1.4xx10^(-12))/(2.1xx10^(-13)))^(1//2)=2.58` |
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| 755. |
The equilibrium constant `K_(p)` for the reaction `H_(2)(g)+CO_(2)(g)hArrH_(2)(g)+CO(g)`is 4.0 at `1660^(@)C` Inittally 0.80`H_(2) and 0.80 mole CO_(2)` are injecteed into a 5.0 litre flask what is the equilibrium concentraton of `CO_(2)(g)`?A. `0.533`B. `0.0534`C. `0.535`D. none of these |
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Answer» Correct Answer - b |
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| 756. |
Consider equilibrium `H_(2)O(l)hArrH_(2)O(g)`. Choose the correct direction of shifting of equilibrium with relative humidity.A. `R.H.gt1`,rightwardB. `R.H.lt1`, rightwardC. `R.H.gt1`, leftwardD. `R.H.lt1`, leftward |
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Answer» Correct Answer - B::C `R.H.gt1rArr P_(H_(2)OgtV.P.` `rArr QgtKrArr` leftward shift `R.H.lt1rArr P_(H_(2)oltV.P.` `rArr QltKrArr` Rightward shift |
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| 757. |
`N_(2)O_(4)(g) hArr 2NO_(2)(g)` The equilibrium reaction shown is endothermic as written. Which change will increase the amount of `NO_(2)` at equilibrium?A. Adding a catalystB. Decreasing the temperatureC. Increasing the volume of the containerD. Adding an inert gas to increase the pressure |
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Answer» Correct Answer - C |
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| 758. |
In the reaction `C(s)+CO_(2)(g) hArr 2CO(g)`, the equilibrium pressure is `12` atm. If `50%` of `CO_(2)` reacts, calculate `K_(p)`.A. 12 atmB. 16 atmC. 20 atmD. 24 atm |
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Answer» Correct Answer - B |
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| 759. |
In the reaction `C(s)+CO_(2)(g) hArr 2CO(g)`, the equilibrium pressure is `12` atm. If `50%` of `CO_(2)` reacts, calculate `K_(p)`. |
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Answer» Correct Answer - A `{:(,C(s),+,CO_(2),hArr,2CO(g)),("Gaseous moles",-,,1,,0),("before dissociation",,,,,),("Gaseous moles",-,,(1-50/100),,(2xx50)/100),("after dissociation",-,,0.5,,1):}` Total moles `=1.5` and `Deltan=1` Total pressure given at equilibrium `=12 "atm"` `K_(p)=((n_(CO))^(2))/((n_(CO_(2))))xx[P/(Sigman)]^(Sigma_(n)) ((1)^(2))/0.5xx(12/1.5)^(1)` `K_(p)=12/(1.5xx0.5)=16 "atm"` |
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| 760. |
If reaction `A+BhArrC+D`, taken place in `5` liter close vessel, the rate constant of forward reaction is nine times of rate of backward reaction. If initially one mole of each reactant present in the container, then find the correct option//is.A. `([C])/([B])=(3)/(1)`B. `"log" K_(P)="log"K_(C)`C. `[D]_(eq)=15xx10^(-2) "mole" L-1`D. `K_(eq)=9` |
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Answer» Correct Answer - A::B::C::D `{:(A,+,B,hArr,C,+,D,),((1-x)/(5),,(1-x)/(5),,(x)/(5),,(x)/(5),t=eq):}` `K_(F)=9K_(b) K_(C)=(K_(F))/(K_(b))=(9)/(1)` `K_(C)=(5//5)^(2)/(((1-x)/(5))^(2))=(9)/(1) (x)/(1-x)=3` `x=3-3x` `4x=3` `x=(3)/(4)=0.75` `[A]=(0.25)/(5xx100)=5xx10^(-2) "mole" L^(-1) [B]=[A] C=(0.75)/(500)=15xx10-2 "mol" L^(-1)` |
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| 761. |
For the reaction `N_(2)O(g) hArr 2NO_(2)(g). DeltaH=57.49kJ //` mole, the vapour density of equilibrium mixture `………………………..` with increase of temperature.A. increasesB. decreasesC. Remain sameD. can not be predicted |
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Answer» Correct Answer - 2 `DeltaH=+ve rArr` endothermic reaction go forward`" "rArr" "`mole increses molar mass decreases. |
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| 762. |
`C(s)+CO_(2)(g) hArr 2CO(g)`. If this system is at equilibrium, which change(s) will alter the value of `K_(P)`? (P) Raising the temperature (Q) Adding solid C (R) Decreasing the pressureA. P onlyB. Q onlyC. P and Q onlyD. Q and R only |
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Answer» Correct Answer - A |
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| 763. |
For the reaction, `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)" "DeltaH^(@)lt0` Which change(s) will increase the fraction of `SO_(3)(g)` in the equilibrium mixture? (P) Increasing the pressure (Q) Increasing the temperature (R) Adding a catalystA. P onlyB. R onlyC. P and R onlyD. P, Q and R |
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Answer» Correct Answer - A |
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| 764. |
For a gaseous phase reaction `A+2B hArr AB_(2), K_(c)=0.3475 L^(2) "mole"^(-2)` at `200^(@)C`. When `2` moles of B are mixed with one "mole" of A, what total pressure is required to convert `60%` of A in `AB_(2)`? |
| Answer» Correct Answer - `181.5 atm`, | |
| 765. |
For the reversible reaction,`A+BhArrC`, the specific reaction rates for forward and reverse reactions are `1.25xx10^(3) "and" 2.75xx10^(4)` respectively. The equilibrium constant for the reaction is:A. `45.45`B. `0.022`C. `2.20`D. `0.4545` |
| Answer» Correct Answer - A | |
| 766. |
Which of the following change will shift the reaction in forward direction? `I_(2)(g) hArr 2I(g), DeltaH^(ɵ)=+150 kJ`A. Increase in concentration of `I_(2)`B. Decrease in concenrtation of `I_(2)`C. Increase in temperatureD. Increase in total pressure |
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Answer» Correct Answer - C `Delta H` is positive so it will shift toward the product by increase in temperature. |
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| 767. |
For which of the following reactions at equilibrium at constant temperature, doubling the volume will cause a shift to the right?A. `N_(2)O_(4)(g) hArr 2NO_(2)(g)`B. `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)`C. `2CO(g)+O_(2)(g) hArr 2CO_(2)(g)`D. `N_(2)(g)+O_(2)(g) hArr 2NO(g)` |
| Answer» Correct Answer - A::B | |
| 768. |
Write the expression for the equilibrium constant `K_(c )` for each of the following reactions: a. `2NOCl(g) hArr 2NO(g)+Cl_(2)(g)` b. `2Cu(NO_(3))_(2)(s) hArr 2CuO(s)+4NO_(2)(g)+O_(2)(g)` c. `CH_(3)COOC_(2)H_(5)(aq)+H_(2)O(1) hArr CH_(3)COOH(aq)+C_(2)H_(5)OH(aq)` d. `Fe^(3+)(aq)+3OH^(Θ)(aq) hArr Fe(OH)_(3)(s)` e. `I_(2)(s)+5F_(2) hArr 2IF_(5)` |
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Answer» (i) `K_(C)=([NO(g)]^(2)[Cl_(2)(g)])/([NOCl(g)]^(2))` (ii) `K_(C)=[NO_(2)(g)]^(4)[O_(2)(g)]` `K_(C)==([CH_(3)COOH(aq)][C_(2)H_(5)OH(aq)])/([CH_(3)COOC_(2)H_(5)(aq)][H_(2)O(l)])` (iv) ` K_(C)=(1)/([Fe^(3+)(aq)[OH_(-)(aq)]_(3)])` (v) `K_(C)=(1)/([F_(2)(g)]^(5))` |
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| 769. |
`2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)" "DeltaHlt0` Which change(s) will increase the quantity of `SO_(3)(g)` at equilibrium? (P) Increasing the temperature. (Q) Reducing the volume of the container. (R) Adding He to increase the pressure keeping volumeA. P onlyB. Q onlyC. P and Q onlyD. Q and R only |
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Answer» Correct Answer - B |
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| 770. |
Consider the system at equilibrium: `2SO_(2)(g) + O_(2)(g) hArr 2SO_(3)(g)` for which `DeltaHlt0`. Which change(S) will increase the yield of `SO_(3)(g)`? (P) Increasing the temperature (Q) Increasing the volume of the containerA. P onlyB. Q onlyC. Both P and QD. Neither P nor Q |
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Answer» Correct Answer - D |
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| 771. |
The equilibrium constant for the reaction `N_(2)(g)+O_(2)(g) hArr 2NO(g)` at temperature T is `4xx10^(-4)`. The value of `K_(c)` for the reaction `NO(g) hArr 1/2 N_(2)(g)+1/2 O_(2)(g)` at the same temperature isA. `4xx10^(-4)`B. 50C. `2.5xx10^(2)`D. `0.02` |
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Answer» Correct Answer - b |
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| 772. |
For the equilibrium `AB(g) hArr A(g)+B(g)`. `K_(p)` is equal to four times the total pressure. Calculate the number moles of A formed if one mol of AB is taken initially. |
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Answer» Correct Answer - A::B Let the total equilibrium pressure be=P atm Given `K_(p)=4P` Let the start be made with `1` mol of AB(g) and the degree if dissociation be x. `AB(g) hArr A(g)+B(g)` `{:("at equilibrium,",1-x,,x,,x):}` Total "moles" at equilibrium `=1-x+x+x=1+x` Thus, `p_(A)`=Partial pressure of A =`x/(1+x)P` `p_(B)` =Partial pressure of B `=x/(1+x).P` `p_(AB)`=Partial pressure of AB `=(1-x)/(1+x).P` Applying the law of mass action `K_(p)=(p_(A)xxp_(B))/p_(AB)=((x/(1+x).P)(x/(1+x).P))/(((1-x)/(1+x).P))` So `4P=x^(2)/(1-x^(2)).P` or `4-4x^(2)=x^(2)` or `5x^(2)=4` or `x=2/sqrt(5)` Hence, number of "moles" of A formed `=2//sqrt(5)` times initial moles of AB taken. |
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| 773. |
Which of the following reaction will shift in forward direction. When the respective change is made at equilibriumA. `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` increase in pressure at eq.B. `H_(2)O(s)hArrH_(2)O(l)` addition of inert gas at constant volumeC. `PCI_(5)(g)hArrPCI_(3)(g)+CI_(2)(g)` addition of inert gas at constant pressureD. `H_(2)+I_(2)hArr2Hi` increase in temperature |
| Answer» Correct Answer - A::B::C | |
| 774. |
Increased pressure shifts the equilibrium of the reaction: `N_(2)(g)+3H_(2)(g)harr2NH_(3)(g)` so as toA. form more ammonia gasB. produce more `N_(2)(g)"and" H_(2)(g)`C. Keep the conversion to ammonia unalteredD. produce more `H_(2)(g)`. |
| Answer» Correct Answer - A | |
| 775. |
According to Le Chateier principle, when an equilibrium is subjected to any external change, the equilibrium/reaction shifts to comensate the effect of the change. This principle helps in shifting the reaction towards appropriate diections so as to increase % yield of any reaction. Which of the following changes cannot cause an incresase in extent of dissociation of `CH_(3)COOH` in its aqueous solution as per the reaction? If is given that conversion of graphite to diamond in an endothermic reaction and the conversion `C_("graphite")hArrC_("diamond")` attains equilibria at `1.5xx10^(9)` Pa at 300K then comment at what pressure equilibria can be attained at 500 K?A. `P gt 1.5xx10^(9)Pa`B. `P lt 1.5xx10^(9)Pa`C. `P = 1.5xx10^(9)Pa`D. At ant pressure |
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Answer» Correct Answer - D |
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| 776. |
According to Le Chateier principle, when an equilibrium is subjected to any external change, the equilibrium/reaction shifts to comensate the effect of the change. This principle helps in shifting the reaction towards appropriate diections so as to increase % yield of any reaction. Which of the following changes cannot cause an incresase in extent of dissociation of `CH_(3)COOH` in its aqueous solution as per the reaction? A,B,C and D are in equilibrium in a 2 litre container at 400 K and their moles are respectively 4,5,8 and 6. If the reaction involved is `3A(g)+2B(g)hArrC(g)+5D(g)` then calculate equilibrium concentration of C when volume is increased to 10 litre.A. 4MB. 0.8MC. 5MD. none of these |
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Answer» Correct Answer - D |
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| 777. |
According to Le Chateier principle, when an equilibrium is subjected to any external change, the equilibrium/reaction shifts to comensate the effect of the change. This principle helps in shifting the reaction towards appropriate diections so as to increase % yield of any reaction. Which of the following changes cannot cause an incresase in extent of dissociation of `CH_(3)COOH` in its aqueous solution as per the reaction? `CH_(3)COOH(aq)+H_(2)O(l)hArrCH_(3)COO^(-)(aq)+H^(+)(aq)`A. Addition of water into the solutionB. Addition of NaOH into the solutionC. Addition of HCI into the solutionD. Remove of `CH_(3)COO^(-)` from solution. |
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Answer» Correct Answer - C |
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| 778. |
In the reaction `C(s) + CO_(2)(g)hArr2CO(g)` the following amounts of sbstance were formed in `0.2` litre flask `CO_(2) = 0.06` mole. The equilibrium constant isA. `0.208`B. `4.10`C. `0.30`D. `0.416` |
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Answer» Correct Answer - A `K = ([CO]^(2))/([CO_(2)]) = (((0.05)/(0.2))^(2))/(((0.06)/(0.2))` `K = 0.208` |
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| 779. |
In which of the following reactions, the formation of product is favoured by decrease in temperature ? (1)`N_(2)(g)+O_(2)(g)hArr2NO(g), DeltaH^(@)=181`(2)`2CO_(2)(g)hArr2CO(g)+O_(2)(g), DeltaH^(@)=566` (3)`H_(2)(g)+I_(2)hArr2HI(g), DeltaH^(@)=-9.4`(4)`H_(2)(g)+F_(2)(g)hArr2HF(g),DeltaH^(@)=-541`A. `1,2`B. 2 onlyC. 1,2,3D. 3,4 |
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Answer» Correct Answer - d |
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| 780. |
`2Cl(s)+O_(2)(g)hArr2CO(g)` `DeltaHlt0`. If the reaction is at equilibrium with excess `C(s)` remaining, what change will increase the quantity of `CO(g)` for the reaction at equilibrium ? (P) Adding C(s) (Q) Increasing the temperature (R) Increasing the pressure.A. P onlyB. R onlyC. P,Q and RD. None of these |
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Answer» Correct Answer - D |
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| 781. |
`N_(2)O_(2) hArr 2NO, K_(1),` `(1/2)N_(2)+(1/2)O_(2) hArr NO, K_(2),` `2NO hArr N_(2)+O_(2),K_(3),` `NO hArr (1/2)N_(2)+(1/2)O_(2),K_(4)` Correct relaton(s) between `K_(1), K_(2), K_(3)` and `K_(4)` is/areA. `K_(1)xxK_(3)=1`B. `sqrt(K_(1))xxK_(4)=1`C. `sqrt(K_(3))xxK_(2)=1`D. None |
| Answer» Correct Answer - A::B::C | |
| 782. |
Which of the following will not affect the value of equilibrium constant of a reaction?A. Change in temperatureB. Addtion of catalystC. Change in concentration of the reactantsD. Change in pressure |
| Answer» Correct Answer - B::C::D | |
| 783. |
Assertion (A) : The value of K increases with increase in temperature in case of endothermic reaction Reason (R) : The increase in temperature shifts the equilibrium in the backward direction in case of exothermic reaction.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. |
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Answer» Correct Answer - D In endothermic reaction, on increasing the temperature, reaction shifts to forward direction. |
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| 784. |
Consider the following reactions .In which case the formation of product is favoured by decreasein pressure? (1)`CO_(2)(g)+C(s)hArr2CO(g),DeltaH^(@)=+172.5Kj` (2)`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) hArr2NH_(3)(g),DeltaH^(@)=-91.8KJ` (3) `N_(2)(g)+O_(2)(g)hArr2NO(g),DeltaH^(@)=181KJ` (4) `2H_(2)O(g)hArr2H_(2)(g)+O_(2)(g),DeltaH^(@)=484.6KJ`A. `2,3`B. `3,4`C. `2,4`D. `1,4` |
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Answer» Correct Answer - d |
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| 785. |
In which case does the reaction go farthest to completion: `K=1, K=10^(-10)`, and why? |
| Answer» The ratio `(["Product"])/(["Reactant"])` is maximum when `K=10K^(10)`, and thus, the reaction goes farthest to completion when `K=10^(10)`. | |
| 786. |
Pure ammonia is placed in a vessel at a temperature where its dissociation constant `(alpha)` is appreciable. At equilibrium,A. concentration of ammonia does not change with pressure.B. its degree of dissociation, `a` does not change with pressureC. `K_(p)` does not change significantly with pressure.D. concentration of hydrogen is less than that of nitrogen. |
| Answer» Correct Answer - C | |
| 787. |
Pure ammonia is placed in a vessel at a temperature where its dissociation constant `(alpha)` is appreciable. At equilibrium,A. `K_p` does not change significantly with pressureB. `alpha` does not change with pressureC. concentration of `NH_3` does not change with pressure .D. concentration of hydrogen is less than that of nitrogen |
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Answer» Correct Answer - A |
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| 788. |
Pure ammonia is placed in a vessel at a temperature where its dissociation constant `(alpha)` is appreciable. At equilibrium,A. `K_(p)` does not change significantly with pressureB. `alpha` does not change with pressureC. Concrntration of `NH_(3)` dose not change with pressureD. Concentration of `H_(2)` is less than that of `N_(2)` |
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Answer» Correct Answer - A `K_(p)` is constant and dose not change with pressure. |
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| 789. |
Pure ammonia is placed in a vessel at a temperature where its dissociation constant `(alpha)` is appreciable. At equilibrium,A. `K_(p)` does not change with pressureB. `alpha` does not change with pressureC. `[NH_(3)]` does not change with pressureD. `[H_(2)]lt[N_(2)]` |
| Answer» `K_(c)` or `K_(p)` is characterstic value for a given reaction and is independent of `P`. It however changes with temperature. | |
| 790. |
Pure ammonia is placed in a vessel at a temperature where its dissociation constant `(alpha)` is appreciable. At equilibrium,A. `K_(p)` does not change significantly with pressureB. `alpha` does not change with pressureC. concentration of `NH_(3)` does not cahange with pressureD. concentration of hydrogen is less than that of nitrogen |
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Answer» Correct Answer - A `K_(p)` depends only on temperature so `alpha` will change on changing pressure and `P_(H_(2)) gt P_(N_(2))` . |
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| 791. |
Which of the following will not affect the value of equilibrium constant of a reaction?A. Change in the concentration of the reactantsB. Change in temperatureC. Change in pressureD. Addition of catalyst |
| Answer» Correct Answer - A::C::D | |
| 792. |
(1) Consider the following cases- The nature of flow of energy in case `(I)` is same as that in- (A) `II` , (B) `III` , (C ) `II and III` , (D) None |
| Answer» (D) None, Because in `II` and `III`, the flow of energy or matter is taking place only in one direction.While in equilibrium state, the flow of energy takes place in both directions equally. Thus `(I)` is a dynamic equilibrium while states in `II` and `III` are called steady state (static equilibrium). | |
| 793. |
Pure ammonia is placed in a vessel at a temperature where its dissociation constant `(alpha)` is appreciable. At equilibrium,A. `K_(p)` does not change significantly with pressureB. `alpha` does not change with pressureC. The concentration of `NH_(3)` does not change with pressure.D. The concentration of hydrogen is less than that of nitrogen. |
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Answer» Correct Answer - A The equilibrium constant does not change at all with changes in concentrations, pressure, presence of catalyst, etc. It changes only with changes in temperature of the system. For endothermic reaction, the value of K increases with increase in temperature and vice versa. For exothermic reaction, the value of K decreases with increase in temperature and vice versa. |
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| 794. |
For the reaction `NOBr (g)iffNO(g)+(1)/(2)Br_(2)(g)` `K_(P)=0.15 atm` at `90^(@)C`. If `NOBr, NO` and `Br_(2)` are mixed at this temperature having partial pressures `0.5 atm`, and `0.2 atm` respectively, will `Br_(2)` be consumed or formed? |
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Answer» `Q_(P)=([P_(Br_(2))]^(1//2)[P_(NO)])/([P_(NOBr]])=([0.2]^(1//2)[0.4])/([0.50])=0.36` `K_(P)=0.15` `therefore Q_(P)gtK_(P)` Hence, reaction will shift in backward direction `therefore Br_(2)` will be consumed |
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| 795. |
For the reaction `H_(2)(g)+I_(2)(g) hArr 2HI(g)` The equilibrium constant `K_(p)` changes withA. Total pressureB. CatalystC. The amounts of `H_(2)` and `I_(2)` presentD. Temperature |
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Answer» Correct Answer - D The equilibrium constant does no change at all with changes in concentrations, volume, pressure, presence of catalyst, etc. It changes only with changes in temperature of the system. For endothermic reaction, the value of K increases with increase in temperature and vice versa. For exothermic reaction, the value of K decreases with increase in temperature and vice versa. |
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| 796. |
In which of the following equilibrium, change in the volume of the system does not alter the number of moles?A. `N_(2(g))+O_(2(g))hArr2NO_((g))`B. `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`C. `N_(2(g))+3H_(2(g))hArr2NH_(3(g))`D. `SO_(2)Cl_(2(g))hArrSO_(2(g))+Cl_(2(g))` |
| Answer» `Deltan=0` and thus no change in equilibrium concentrations. | |
| 797. |
In which of the following equilibrium, change in the volume of the system does not alter the number of moles?A. `N_(2(g)) + O_(2(g))hArr2NO_((g))`B. `PC1_(5(g))hArrPC1_(3(g)) + C1_(2(g))`C. `N_(2(g)) + 3H_(2(g))hArr2NH_(3(g))`D. `SO_(2)C1_(2(g))hArrSO_(2(g)) + C1_(2(g))` |
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Answer» Correct Answer - A `Delta n =0` and thus no changbe in equilibrium concentrations. |
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| 798. |
In which of following reactions, increase in the volume at constant temperature does not affect the number of moles of at equilibrium?A. `2NH_(3)hArrN_(2)+N_(2)+3H_(2)`B. `C(g)+(1//2)O_(2)(g)hArrCO(g)`C. `H_(2)(g)+O_(2)(g)hArrH_(2)O_(2)(g)`D. None of these |
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Answer» Correct Answer - D If in the reaction the ratio of number of moles of reactants to products in same `i.e., 1:1`, then change in volume will not alter the number of moles. |
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| 799. |
Write expression for `K_(p) "and"K_(c)` for the reaction `CaCO_(3)(S)harrCaO(S)+CO_(2)(g)`. |
| Answer» `K_(c)=([CO_(2)(g)])/(1)`. Taking active masses of solids as unity, `K_(c)=[CO_(2)(g)]`. Similarly, `K_(P)=P_(CO_(2))`. | |
| 800. |
statement-1 : A catalyst does not influence the values of equilibrium constant. statement-2 : Catalyst influence the rate of both forward and backward reactions equality.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statment-2is True ,Statement-2is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A |
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