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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.
| 101. |
Consider the imaginary equilibrium `4A+5B hArr4X+6Y` The equilibrium constant `K_(c)` has the unitA. `"mole"^(2)litre^(2)`B. `"Litre mole"^(-1)`C. `"Mole litre"^(-1)`D. `"Litre"^(2)"mole"^(2)` |
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Answer» Correct Answer - C Unit of `K_(c)=("unit of concentration")^(Deltan)="(""molelitre"^(-1)")"^(Deltan)` `Deltan=10-9=1` `thereforeK_(c)="mol Litre"^(-1).` |
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| 102. |
`N_(2)+O_(2)hArr2NO-Q` cal In the above reaction which is the essential condition for the higher production of NOA. High temperatureeB. High pressureC. Low temperatureeD. Low pressure |
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Answer» Correct Answer - A `N_(2)+O_(2)hArr2NO,Q` cal The above reaction is endothermic so for higher production of NO, and the temperature should br high. |
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| 103. |
An equeous solution contains `0.10MH_(2)Sand 0.20M HCl.` If the equlibrium constants for the formation of HS from `H_(2)S` is `1.0xx10^(-7)` and that of `S^(2-)` from HS ions is `1.2xx10^(-13)` then the concentration of `S^(2-)` ions in aquesous solution isA. `3xx10^(-20)`B. `6xx10^(-21)`C. `5xx10^(-19)`D. `5xx10^(-8)` |
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Answer» Correct Answer - A `{:(HCl to, H^(+),+, CI^(-)),(,(0.2+x+y),,0.2):}` `{:(H_(2)ShArrH^(+),+, HS^(-):Ka_(1)=10^(-7)),(0.1-x(0.2+x+y),,(x-y)),(HS^(-)hArrH^(+),+,S^(2-):K_(a_(2))=1.2xx10^(-13)):}` |
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| 104. |
One mole of ethyl alcohol was treated with one mole of acetic acid at `25^(@)C.` Two-third of the alcohol change into easter at equlibrium. The equlibrium constant for the reaction will beA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - D `C_(2)H_(2)OH+COOHtoCH_(3)COOC_(2)H_(5)+H_(2)O` `{:("Initial conc.","1 mol","1 mol","0 mole","0 mole"),("At equlibrium", 1-2/3=1/3 "mol",1/3"mol",2/3"mol",2/3"mol"):}` `K_(C)=((2)/(3)xx(2)/(3))/(1/2xx1/3)=4` |
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| 105. |
For the reaction `PCl_(3(g))+Cl_(2(g))hArrPCl_(5(g))` the position of equlibrium can be shifted to the right byA. Increasing the temperatureB. Doubling the volumeC. Additional of `Cl_(2)` at constant volumeD. Additional of equimolar quantities of `PCl_(3) and PCl_(5)` |
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Answer» Correct Answer - C According to Le-chatelier principle when concentration of reactant increases, the equlibrium shift in favour of forward reaction. |
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| 106. |
Change in volume the system does not alter the number of moles in which of the following equilibriumA. `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))` |
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Answer» Correct Answer - A `N_(2(g))+O_(2(g))hArr2NO_((g))," "Deltan=2-2=0` |
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| 107. |
For the reaction `2NO_(2(g))hArr2NO_((g))+O_(2(g))` `K_(c)=1.8xx10^(-6)at 185^(@)C. At 185^(@)C,` the value of `K_(c)` for reaction `NO_((g))+1/2O_(2(g))hArrNO_(2(g))` isA. `0.9xx10^(6)`B. `7.5xx10^(2)`C. `1.95xx10^(-3)`D. `1.95xx10^(3)` |
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Answer» Correct Answer - B Reaction is reversed and halved. `K=sqrt(1//18xx10^(-6))=7.5xx10^(2)` |
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| 108. |
For the reaction `2NO_(2(g))hArr2NO_((g))+O_(2(g))` `(K_(c)=1.8xx10^(-6)at 184^(@)C)` `(R=0.0831kJ//(mol.K)`A. `K_(p)` is greater then `K_(c)`B. `K_(p)` is less than `K_(c)`C. `K_(p)=K_(e)`D. Wheter `K_(p)` is greater than, less than or equal to `K_(c)` depnds upon the total gas pressure |
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Answer» Correct Answer - A `K_(p)=K_(c)(RT)^(Deltan)` `Deltan=3-2=1,k_(p)gtk_(c).` |
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| 109. |
The equlibrium constant is `6.0xx10^(-4)` for the `N_(2)+O_(2)hArr2NO` reaction. If the concentration of nitrogen is `0.10` mol/L and concentration of oxygen is 0.20 mol/L at equilibrium. Then the concentration of nitric oxide at equlibrium isA. `10.9xx10^(-3)mol//L`B. `1.09xx10^(-3)mol//L`C. `10.9xx10^(-5)mol//L`D. `1.09xx10^(-5)mol//L` |
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Answer» Correct Answer - B `K_(c)=([NO]^(2))/([N_(2)][O_(2)])` `6xx10^(-4)=([NO]^(2))/(0.10xx0.20` `[NO]=sqrt(6.0xx10^(-4)xx0.10xx0.20)` `=1.09xx10^(-3)mol//L` |
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| 110. |
For a reaction if `k_(p)gtk_(c)` the forward reaction is favoured byA. Low pressureB. hifh pressureC. High temperatureD. Low temperature |
| Answer» Correct Answer - A | |
| 111. |
For the reaction, `PCl_(5)(g)to PCl_(3)(g)+Cl_(2)(g),` The forward reaction at constant temperature is favoured byA. Introducing an inert gas at constant volumeB. Introducting chlorine gas at constant volumeC. Increasing the volume of the containerD. Introducing `PCl_(5)` at constant volume |
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Answer» Correct Answer - C::D Introduction of an inert gas at constant pressure causes the equlibrium to shift in a direction in which number of moles increases. The forward reaction is furter accelerated by increase in the quantity of substrate, i.e., `PCl_(5)` and by the increase of space, i.e., volume of container. |
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| 112. |
For the reaction equlibrium `N_(2)O_(4)hArr2NO_(2(g))`the concentration of `N_(2)O_(4) and NO_(2)` at equlibrium are `4.8xx10^(-2)and 1.2xx10^(-2)"mol lite"^(-1)` respectively. The value of `K_(c)` for the reaction isA. `3.3xx10^(2)"mol litre"^(-1)`B. `3xx10^(-1)"mol litre"^(-1)`C. `3xx10^(-3)"mol litre"^(-1)`D. `3xx10^(3)"mol litre"^(-1)` |
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Answer» Correct Answer - C `K=([NO_(2)]^(2))/([N_(2)O_(4)])=([1.2xx10^(-2)])/([4.8xx10^(-2)])=0.3xx10^(-2)=3xx10^(-3)` |
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| 113. |
In which of the following equlibrium system is the rate of the backward reaction favoured by increase of pressureA. `PCl_(5)hArrPCl_(3)+Cl_(2)`B. `2SO_(2)hArr2SO_(3)`C. `N_(2)+3H_(2)hArr2NH_(3)`D. `N_(2)+O_(2)hArr2NO` |
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Answer» Correct Answer - A The rate of backward reactin favoured by increase of pressure in the reaction as `Deltan` is positive `PCl_(4)hArrPCl_(3)+Cl_(2)` |
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| 114. |
At constant teperature, the equlimbeijm constnt `(K_(p))` for the decomposition reaction `N_(2)O_(4)hArr2NO_(2)` is expressed by `K_(p)=(4x^(2)P)(1-x^(2)),` where P = pressure, x= ectent of decomposition. Which one of the following statements is trueA. `K_(p)` increases with increase of PB. `K_(p)` increase with increase of xC. `K_(p)` increase with decrease of xD. `K_(p)` remains constant with change in P and x |
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Answer» Correct Answer - D With change of pressurr, will change in such a way that `K_(p)` remains a constant. |
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| 115. |
In the equilibrium, `2AhArrB+C,` the equilibrium concentrations of `A.B and C at 300K` are `3xx10^(-4)M, 1xx10^(-4)M and 4.5xx10^(-4)M` respectively. Thte vlue of `K_(c)` for the above equilibrium at 300 K isA. `0.5`B. `0.05`C. `5.0`D. `1.5` |
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Answer» Correct Answer - A For the equlibrium, `2AhArrB+C` `K_(c)=([B][C])/([A]^(2))` `where [B]=1xx10^(-4)M,[C]=4.5xx10^(-4)M and [A]=3xx10^(-4)M` Then `k_(c)=((1xx10^(-4))(4.5xx10^(-4)))/((3xx10^(-4))^(2))=(4.5)/(9)=0.5` |
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| 116. |
In a chemical equilibrium, the rate constant of the backward reaction is `7.5xx10^(-4)` and the equilibrium constant is 1.5. So the rate constant of the forward reaction isA. `5xx10^(-4)`B. `2xx10^(-3)`C. `1.125xx10^(-3)`D. `9.0xx10^(-4)` |
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Answer» Correct Answer - C `K_(c)=(K_(f))/(K_(b))` `K_(f)=K_(c)xxK_(b)=1.5xx7.5xx10^(-4)=1.125xx10^(-3)` |
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| 117. |
`DeltaG^(@)(HI,g)-=+1.7kJ.` What is the equilibrium constnt at `25^(@)C"for"2 HI_((g))hArrH_(2(g))+I_(2(g))`A. `24.0`B. `3.9`C. `2.0`D. `0.5` |
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Answer» Correct Answer - D `DeltaG^(@)=-2.303xx8.314xxx10^(-3)xx298logK_(p)` `1.7=-2.303xx8.314xx10^(-3)xx298xxlog K_(p)` `K_(p)=0.5` |
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| 118. |
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 is `CH_(3)OOC_(2)H_(5)+H_(2)OhArrCH_(3)COOH+C_(2)H_(5)OH`A. `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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| 119. |
`H_(2)+I_(2)hArr 2HI` In the above equlibrium system if the conectration of the reactants at `25^(@)C` is increased, the value of `K_(c)` willA. IncreaseB. DecreaseC. Remains the sameD. Depends on the nature of the reactants |
| Answer» Correct Answer - C | |
| 120. |
If `alpha` is the reaction of HI dissociation at equilibrium n the reaction, `2HI hArrH_(2)+I_(2)` then starting with 2 moles of reactants and products at equlibrium is |
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Answer» Correct Answer - 2 `2HIhArrH_(2)+I_(2)` `{:("Initial mole",2,0,0),("At eq.",2(1-alpha),alpha,alpha):}` Total mole `=2(1-alpha)+alpha+alpha=2` |
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| 121. |
When two reactants, A and B are mixed to given products C and D, the ractin 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 At initial stage of reactin, concentration of each product will increase and hence Q will increase. |
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| 122. |
In chemical equilibrium, the vlue of `Deltan` (number of molecules of products-number of molecules of reactants) is negative, then the relationship between `K_(p)and K_(c)` will beA. `K_(p)-K_(c)=0`B. `K_(p)=K_(c)(RT)^(+Deltan)`C. `K_(p)=K_(c)(RT)^(-Deltan)`D. `K_(p)=(1)/(K_(c))` |
| Answer» Correct Answer - B | |
| 123. |
Consider the following reactions in which all the reactants and the products are in gaseous state None `2PQhArrP_(2)+Q_(2)," "K_(1)=2.5xx10^(5)` `PQ+1//2R_(2)hArrPQR," "K_(2)=5xx10^(-3)` The value of `K_(1)` for the equlibrium `1//2P_(2)+1//2Q_(2)+1//2R_(2)hArrPQR, is`A. `2.5xx10^(-3)`B. `2.5xx10^(3)`C. `1.0xx10^(-5)`D. `5xx10^(3)` |
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Answer» Correct Answer - C Given, `2PQhArrP_(2)+Q_(2)` `K_(1)=([P_(2)][Q_(2)])/([PQ]^(2))=2.5xx10^(5)` `PQ+1//2R_(2)hArrPQR.` `K_(2)=([PQR])/([PQ][R_(2)]^(1//2))=5xx10^(-3)` Required equlibrium is: `1//2P_(2)+1//2Q_(2)+1//2R_(2)hArrPQR,` `K_(3)=([PQR])/([P_(2)]^(1//2)[Q_(2)]^(1//2)[R_(2)]^(1//2))` `K_(3)=K_(2)xxsqrt((1)/(K_(1)))` `=5xx10^(-3)xxsqrt((1)/(2.5xx10^(5)))=(5xx10^(-3))/(0.5xx10^(3))=1xx10^(-5)` |
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| 124. |
In which of the following system, dobbling the volume of the container cause a shify to the rightA. `H_(2(g))+Cl_(2(g))=2HCl_((g))`B. `2CO_((g))+O_(2(g))=CO_(2(g))`C. `N_(2(g))+3H_(2(g))=2NH_(3(g))`D. `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))` |
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Answer» Correct Answer - D Increase in volume. i.e., decrease in pressure shifts the equlibrium in the direction in which number of moles increases `(Deltan` positive) |
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| 125. |
If the concentration of `OH^(-)` ions in the reaction `Fe(OH)_(3) hArr Fe^(3+)(aq)+3OH^(-)(aq)` is decreased by `1/4` times, then equlibrium concentration of `Fe^(3+)` will increase byA. 64 timesB. 4 timesC. 8 timesD. 16 times |
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Answer» Correct Answer - A `K_(c)=[Fe^(3+)][OH^(-)]^(3)` |
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| 126. |
20.0 kg of `N_(2(g))and 3.0kg of H_(2(g))` are mixed to produce `NH_(3(g)).` The amount of `NH_(3(g))` formed isA. 17 kgB. 34 kgC. 20 kgD. 3 kg |
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Answer» Correct Answer - A We know that `N_(2)+3H_(2)hArr2NH_(3)` `{:(28g,6g,34g),(14g,3g,17g):}` Here given `H_(2)` is 3 kg and `N_(2)` is 20 kg but 3 kg of `H_(2)` can only react with 14 g of `N_(2)` and thus the obrained `NH_(3)` will be of 17 kg. |
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| 127. |
At room temperaturee, for the reaction `NH_(4)SH_((s))hArrNH_(3(s))+H_(2)S_((g))`A. `K_(p)=K_(c)`B. `K_(p)gtK_(c)`C. `K_(p)ltK_(c)`D. `K_(p)and K_(c)` do not relate |
| Answer» Correct Answer - B | |
| 128. |
`NH_(4)^(+)` is kept in group zero becauseA. Its salts are highly solublr in waterB. The `K_(sp)` of salts of ammonium is highC. `K_(sp)` of salts of `NH_(4)^(+)` are lowD. Ammonium salts are insoluble in water |
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Answer» Correct Answer - A::B (a)and (b) are correct because `NH_(4)^(+)` salts are ionic and can from H-bonds with water, therefore they are highly solublr in water. Higher the solubility, higher will be `K_(sp).` `therefore` © and (d) are not correct. |
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| 129. |
In third group of qualitive analysis, `NH_(4)Cl` is added before `NH_(4)OH` so thatA. `OH^(-)` concentration decreasesB. only group III radical get precipitated as hydroxide but others with high solubility product do notC. `K_(sp)` of group III hydroxide is highD. Group III radicals get precipitated as chlorides |
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Answer» Correct Answer - A::B `NH_(4)OH` is a weak electrolyte and dissociates partially as `NH_(4)OHhArrNH_(4)^(+)+OH^(-).` On adding `NH_(4)Cl,` the dissociation of `NH_(4)OH` is suppressed such that `[OH^(-)]` is very less and only less soluble hydroxides are precipitated. Hence choices (a) and (b) are correct while (c) and (d) are incorrect. |
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| 130. |
For the reaction `PCl_(3(g))+Cl_(2(g))hArrPCl_(5(g))at 250^(@)C,` then value of `K_(c)` is 26, then the value of `K_(p)` the same temperaturee will beA. `0.61`B. `0.57`C. `0.83`D. `0.46` |
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Answer» Correct Answer - A `K_(p)=K_(c)(RT)^(Deltan)=26(0.0821xx523)^(-1)=0.61.` `Deltan_(g)=1-2=-1` |
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| 131. |
An amount of solid `NGH_(4)HS` is placed ina flask already containg ammonia gas at a certain temperaturee and 0.50 atm. Pressue. Ammounium hydrogen sulphide decomposes to yield `H_(3)and H_(2)S` gases in the flask. When the decomposition reaction reaches equlibrium, the total pressure in the flask rises to `0.84` atm. The equilibrium constant for `NH_(4)HS` decompositions at this temperaturee isA. `0.30`B. `0.18`C. `0.17`D. `1.11` |
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Answer» Correct Answer - D `{:(NH_(4)HShArrNH_(3(g))+H_(2)S_((g))),(" ""a ""0.5atm"),(""a-x " "0.5+x " "x):}` Total pressure `=0.5+2x=0.84 i.e., x=0.17` `K_(p)=P_(NH_(3)).P_(H_(2)S)=(0.67).(0.17)=0.1139` |
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| 132. |
Thermal decomposition of gaseous `X_(2)`to gaseous X at 298 K takes place accoeding to the following equation: `X_(2_(g)hArr2X(g)` is positive. At the start of the reaction, there is one mole of `X_(2)` and no X. As the reaction proceeds, the number of moles of X formed is given by `beta.` The `beta_("equulibrium")` is the number of moles of X formed at equlibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: The standard reaction Gibbs energy, `Delta_(r)G^(@),` of this reaction `R=0.08L barK^(-1)mol^(-1)`) The equlibrium constatn `K_(p)` for this reaction at 298 K, in terms of `beta_("equlibrium,")` isA. `(8beta_("equlibrium")^(2))/(2-beta_("equlibrium"))`B. `(8beta_("equlibrium")^(2))/(4-beta_("equlibrium")^(2))`C. `(4beta_("equlibrium")^(2))/(2-beta_("equlibrium"))`D. `(4beta_("equlibrium")^(2))/(4-beta_("equlibrium")^(2))` |
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Answer» Correct Answer - B `underset(1)(X_(2))(g)hArr2XC(g)` `-(beta_(e))/(2)" "beta_(e)` Total number of moles at equlibrium `implies1-(beta_(e))/(2)+beta_(e)implies1+(beta_(e))/(2)` `K_(P)=((P_(x))^(2))/(P_(x_(2)))` `=((((beta_(e)xx2)/(1+(beta_(e))/(2)))^(2))/((1-(beta_(e))/(2))xx2))/(1+(beta_(e))/(2))=(2beta_(e)^(2))/(1-(beta_(e^(2)))/(4))impliesK_(p)=(8beta_(e)^(2))/(4-beta_(e)^(2))` |
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| 133. |
If the equilibrium constant of the reaction `2HIhArrH_(2)+I_(2)is 0.25,` then the equlibrium constnat of the reaction `H_(2)+I_(2)hArr2HI` would beA. `1.0`B. `2.0`C. `3.0`D. `4.0` |
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Answer» Correct Answer - D `K_(1)` for reaction `2HIhArr H_(2)+I_(2)is 0.25 K_(2)` for reaction `H_(2)+I_(2)hArr2HI "will be"K_(2)=(1)/(K_(1))=(1)/(0.25)=4` Because `II^(nd)` reaction is reverse of `I^(st).` |
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| 134. |
For the following reaction in gaseous phase `CO+1/2O_(2)toCO_(2),K_(p)//K_(c)is`A. `(RT)^(1//2)`B. `(RT)^(-1//2)`C. (RT)D. `(RT)^(-1)` |
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Answer» Correct Answer - B `K_(p)=K_(c)[RT]^(Deltan_(g))` `Deltan_(g)=1-1.5=-0.5` `K_(p)=K_(c)[RT]^(1//2)therefore(K_(p))/(K_(c))=[RT]^(-1//2)` |
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| 135. |
In an equilibrium reaction for which `DeltaG=0,` the equlibrium constant K = |
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Answer» Correct Answer - B `If DeltaG^(@)=0` `logK_(p)=0" "(becauselog1=0)` `K_(p)=1.` |
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| 136. |
Calculate `DeltaG` for conversion of oxtgen to ozone `3//2O_(2(g))toO_(3(g))at 298K, if K_(p)` for this coversion is `2.47xx10^(-29)`A. `163kJ mol^(-1)`B. `2.4xx10^(2)kJmol^(-1)`C. `1.63kJ mol^(-1)`D. `2.38xx10^(6)kJ mol^(-1)` |
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Answer» Correct Answer - A As we know that, `DeltaG^(@)=-2.303RT logK_(p)` Therefore, `DeltaG^(@)=-2.303xx(8.314)xx(298)(log2.47xx10^(-29))` `DeltaG^(@)=16,300J mol^(-1)=163KJmol^(-1).` |
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| 137. |
The equilibrium constant oa reaction is 300. If the volume of the reaction flaskis tripled, the equilibrium constatn will beA. 100B. 900C. 600D. 300 |
| Answer» Correct Answer - D | |
| 138. |
Assertion: For a gaseouss reaction, `xA+yBhArr1C+mD,K_(p)=K_(C).` Reason: Concentration of gaseous reactant is taken to be unity.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertin and reason are true but reason is not the correct explanation of the assertion.C. If assertin is true but reason is false.D. If the assertion and reason both are false. |
| Answer» Correct Answer - D | |
| 139. |
The reaction `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` is in equlibrium. Now the reaction mixture is compressed to half the volumeA. More of ammonia will be formedB. Ammonia will dissociate back into `N_(2)and H_(2)`C. There will be no effect on equlibriumD. Equlibrium constant of the reaction will change |
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Answer» Correct Answer - A `K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3)).` What volume is halved, concentrations are doubled. To keep `K_(c)` constatn, increase of `[NH_(3)]` should be more. |
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| 140. |
On addition of an inert gas at constant volume to the reaction `N_(2)+3H_(2)hArr2NH_(3)` at equlibriumA. The reaction remains unaffectedB. Forward reacton is favoredC. The reaction haltsD. Backward reaction is favoured |
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Answer» Correct Answer - A Addition of an inert gas of constant volume condition to an equlibrium has no effect. |
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| 141. |
In which one of the following gaseous equlibria `K_(p)` is less then `K_(c)`A. `N_(2)O_(4)hArr2NO_(2)`B. `2HlhArrH_(2)+I_(2)`C. `2SO_(2)+O_(2)hArr2SO_(3)`D. `N_(2)+O_(2)hArr2NO` |
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Answer» Correct Answer - C `K_(p)=K_(c)(RT)^(Deltan)` `Deltan=-"for reaction"2SO_(2)+O_(2)hArr2SO_(3)` So for this reation `K_(p)` is less than `K_(c).` |
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| 142. |
The equlibrium constant for the reaction `N_(2)+3H_(2)hArr2NH_(3)` is K, then the equlibrium constant for the equlibrium `NH_(3) 1/2N_(2)+3/2H_(2)` isA. `1//K`B. `1//K^(2)`C. `sqrtK`D. `(1)/(sqrtK)` |
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Answer» Correct Answer - D `K_(1)"for" N_(2)+3H_(2)hArr2NH_(3)` `K_(2)"for" NH_(3)hArr1/2N_(2)++3/2H_(2)` `K_(1)xxK_(2)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))xx([N_(2)]^(1//2)[H_(2)]^(3//2))/([NH_(3)])` `K_(1)xxK_(2)=(1)/(K_(2)),K_(2)=(1)/(sqrt(K_(1)))` |
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| 143. |
The reaction quotient (Q) for the reaction `N_(2(g))+3H_(2(g))hArr2NH_(3(g))` is given by `Q=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3)).` The reaction will proceed from right to left isA. `Q=0`B. `Q=K_(c)`C. `QltK_(c)`D. `QgtK_(c)` |
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Answer» Correct Answer - D If `QgtK_(c)` reaction will proceed right to left to decrease concentration of product. |
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| 144. |
A chemicla reaction is catalyzed by a catalyst X. Hence XA. Reduces enthalpy of the reactionB. Decreases rate constant of the reactionC. Does not affect equlibrium constant of reactionD. Of the following which change will shift the reaction towards the product |
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Answer» Correct Answer - D Catalyst does not affect equlibrium constant. |
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