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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.
| 51. |
In an exothermic reaction, high yield of the product is obtained atA. High temperatureB. Low temperatureC. Low concentrationD. None of these |
| Answer» Correct Answer - B | |
| 52. |
For a system in equilbrium `DeltaG=0` under conditions of constantA. temperaturee and pressureB. Temperatue and volumeC. Energy and volumeD. Pressure and volume |
| Answer» Correct Answer - A | |
| 53. |
For which of the following reactions does the equlibrium constant depend on the units of concentrationA. `NO"_((g))hArr1/2N_(2(g))+1/2O_(2(g))`B. `ZN_((s))+CU_((aq))^(2+)hArrCu_((s))+Zn_((aq))^(2+)`C. `C_(2)H_(5)OH_((1))+CH_(3)COOH_((1))hArrCH_(3)COOC_(2)H_(5(1))+H_(2)O_((1))` (Reaction carried in an inert solvent)D. `COCl_(2(g))hArrCO_((g))+Cl_(2(g))` |
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Answer» Correct Answer - D `Deltan=1` for this change So the equlibrium constant depends on the unit of concentration. |
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| 54. |
For a reversible chemical reaction where the forward process is exothermic which of the following statements is correctA. The backward reaction has higher activation energy than the forward reactionB. The backward and the forward Processes have the same activation energyC. The backward reaction has lower activation energyD. No activation energy is required at all since energy is liberated in the process |
| Answer» Correct Answer - A | |
| 55. |
Which of the following is reversible processA. Melting of ice at `10^(@)C`B. Mixing of two gases by diffusionC. Evaporation of water at `100^(@)C` and 1 atm pressureD. None of the above |
| Answer» Correct Answer - A | |
| 56. |
A liquid is in equlibrium with its vapour at its boiling point. On the average, the molecules in the two phases have equalA. Inter-molecular forcesB. Potential energyC. Total energyD. Kinetic energy |
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Answer» Correct Answer - C Vapours aond liquid are at the same temperatue. |
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| 57. |
The compound insolubel in acetic acid isA. Calcium oxideB. Calcium carbonateC. Calcium oxalateD. Calcium hydroxide |
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Answer» Correct Answer - C `CaO,CaCO_(3)and Ca(OH)_(2)"dissolve in"CH_(3)COOH "to from" (CH_(3)COO)_(2)Ca. But CaC_(2)O_(4)` does not dissolve as `CH_(3)COO^(-)` is a stronger conjugate base than `C_(2)O_(4)^(2-).` |
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| 58. |
The equilibrium constant in a reversible reaction at a given temperatureeA. Depends on the initial concentration of the reactantsB. Depends on the concentration of the products at equilibriumC. Does not depend on the initial concentrationsD. It is not characteristic of the reaction |
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Answer» Correct Answer - C Equlibrium constant is independent of original concentration of reactant. |
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| 59. |
Pure ammonia is placed in a vessel at temperature where its dissociation constant `(alpha)` is appreciable. At equlibriumA. `K_(p)` does not change significantly with pressureB. `alpha` does not change with pressureC. Concentration of `NH_(3)` does 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 does not change with pressure. |
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| 60. |
An example of a reversible reaction isA. `Pb(NO_(3))_(2)(aq)+2Na(aq)=PnI_(2)(s)+2NaNO_(3)(aq)`B. `AgNO_(3)(aq)+HCl(aq)=Ag(Cl(s0+NaNO_(3)(aq)`C. `2Na(s)+H_(2)O(1)=2NaOH(aq)+H_(2)(g)`D. `KNO_(3)+(aq)+NaCl(aq)=KCl(aq)+NaNO_(3)(aq)` |
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Answer» Correct Answer - D As all the reactants and products are present in aqueous from in (d) so it is reversible reaction. In other either solid or gas is generated which is insoluble or valatile and hance makes the reaction unidirectional. |
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| 61. |
If the value of an equilibrium constant for a particular reaction is `1.6xx10^(12),` then at equilibrium the system will containA. Mostly reactantsB. Mostly productsC. Similar amounts of reactants and productsD. All reactants |
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Answer» Correct Answer - B `K=1.6xx10^(12)=[("Product")/("Reactan t")]` k have high value of products have very high conc than reactant. |
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| 62. |
`k_(P)` for the following reaction at 700 K is `1.3xx10^(-3)atm^(-1) The K_(c)` at same temperature for the reaction `2SO_(2)+O_(2)hArr2SO_(3)` will beA. `1.1xx10^(-2)`B. `3.1xx10^(-2)`C. `5.2xx10^(-2)`D. `7.4xx10^(-2)` |
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Answer» Correct Answer - D `K_(p)=K_(c)(RT)^(Deltan)hArrR="Gas constant"` `K_(c)=(K_(p))/((RT)^(Deltan))=(1.3xx10^(-3))/((0.0821xx700)^(-4))=7.4xx10^(-2)` |
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| 63. |
Which of the following is not favourable for `SO_(3)` formation `2SO_(2(g))+O_(2(g))hArr2SO_(3(g)),DeltaH=-45.0` kcalA. High pressureB. High temperatureC. Decreasing `SO_(3)` concentrationD. Increasing reactant concentration |
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Answer» Correct Answer - B The reaction is exothermic so high temperature will favour backward reaction. |
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| 64. |
Given the reaction between 2 gases represented by `A_(2) and B_(2)` to give the compound AD(g). `A_(2)(g)+B_(2)(g)hArr2AB(g).` At equilibrium, the concentration `of A_(2)=3.0xx10^(-3)M` `of B_(2)=4.2xx10^(-3)M` `of AB=2.8xx10^(-3)M` If the reaction takes place in a scaled vassel at `527^(@)C,` then the value of `K_(c)` will beA. `2.0`B. `1.9`C. `0.62`D. `4.5` |
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Answer» Correct Answer - C `A_(2)+B_(2)hArr2AB` `K_(C)=((2.8xx10^(-3))^(2))/(3xx10^(-3)xx4.2xx10^(-3))=((2.8))/(3xx4.2)=0.62` |
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| 65. |
The dissociation constants for acetifc acid and HCN at `25^(@)C` are `1.5xx10^(-5) and 4.5xx10^(-10),` respectively. The equilibrium constant for the equilibrium `CN^(-)+CH_(3)COOHhArrHCN+CH_(3)COO^(-)` would beA. `3.0xx10^(5)`B. `3.0xx10^(-5)`C. `3.0xx10^(-4)`D. `3.0xx10^(4)` |
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Answer» Correct Answer - D `CH_(3)COOHhArrCH_(3)COO^(-)+H,K_(a)=1.5xx10^(-5)` `H^(+)+CN^(-)hArrHCN,(1)/(K_(a))=(1)/(4.5xx10^(-10))` `thereforeK_(a)"for"CN^(-)+CH_(3)COOHhArrCH_(3)COO^(-)+HCN` is `(1.5xx10^(-5))/(4.5xx10^(-10))=1/3xx10^(5)=3.33xx10^(4).` |
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| 66. |
Given that the equilibrium constant for the reaction `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` has a value of 270 at a particular temperaturee. What is the vlaue of the equilibrium constant for the following reaction at same temperaturee `SO_(3)(g)hArrSO_(2)(g)+1/2O_(2)(g)`A. `1.8xx10^(-3)`B. `3.6xx10^(-3)`C. `6.0xx10^(-2)`D. `1.3xx10^(-5)` |
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Answer» Correct Answer - C `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` `SO_(3)(g)hArrSO_(2)(g)+1/2O_(2)(g)` `K=278` `K=1/K=sqrt((1)/(278))=sqrt(35.97xx10^(-4))=6xx10^(-2)` |
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| 67. |
The reaction `2A_((g))+B_((g))hArr3C_((g))+D_((g))` is begun with the concentrations of A and B both at an initial value of `1.00` M. when equilibrium is reached, the concentration of D is measured and found to be `0.25` M. The value for the equilibrium constant for this reaction is given by the expressionA. `"["(0.75)^(3)(0.25)"]"div"["(1.00)^(2)(1.00)"]"`B. `"["(0.75)^(3)(0.25)"]"div"["(0.50)^(2)(0.75)"]"`C. `"["(0.75)^(3)(0.25)"]"div"["(0.50)^(2)(0.25)"]"`D. `"["(0.75)^(3)(0.25)"]"div"["(0.75)^(2)(0.25)"]"` |
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Answer» Correct Answer - B `2A+BhArr3C+D` `{:("Initial",1,1,0,0),(Eq.,1-0.50,1-0.25,0.75,0.25):}` `K=((0.75)^(3)(0.25))/((0.50)^(2)(0.75))` |
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| 68. |
For which one of the following reactions `K_(p)=K_(c)`A. `N_(2)+3H_(2)hArr2NH_(3)`B. `N_(2)+O_(2)hArr2NO`C. `PCl_(5)hArrPCl_(3)+Cl_(2)`D. `2SO_(3)2SO_(2)+O_(2)` |
| Answer» Correct Answer - B | |
| 69. |
For the reaction `:H_(2(g))+CO_(2(g))hArr CO_((g))+H_(2)O_((g)),` if the initial concentraion of `[H_(2)]=[CO_(2)]and x` moles/litre of hydrogen is consumed at equlibrium, the correct expression of `K_(p)` isA. `(x^(2))/((1-x)^(2))`B. `(1+x)^(2)/((1-x)^(2))`C. `(x^(2))/((2+x)^(2))`D. `(x^(2))/(1-x^(2))` |
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Answer» Correct Answer - A `H_(2(g))+CO_(2(g))hArrCO_((g))+H_(2)O_((g))` `{:("Initial conc",1,1,0,0):}` `{:("At eqm",(1-x),(1-x),x,x):}` `K_(p)=(P_(CO).P_(H_(2)O))/(P_(H_(2)).P_(CO_(2)))=(x.x)/((1-x)(1-x))=(x^(2))/((1-x)^(2))` |
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| 70. |
For the gas phase reaction, `C_(2)H_(4)+H_(2)hArrC_(2)H_(6)(Deltah=-32.7kcal)` carried out in a vessel, the equlibrium concentration of `C_(2)H_(4)` can be increased byA. Increasing the temperatureB. Decreasing the pressureC. Removing some `H_(2)`D. Adding some `C_(2)H_(6)` |
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Answer» Correct Answer - A::B::C::D (abcd) The reaction is exothermic, hence increasing temperature will favour backward reaction (i.e., conc. Of `C_(2)H_(4)` increases). Again decreasing pressure, removing `H_(2)` and adding `C_(2)H_(6)` favours backward reaction. |
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| 71. |
If equlibrium constant for reaction `2ABhArrA_(2)+B_(2).is 49` then the equlibrium constant for reacton `ABhArr1/2A_(2)+1/2B_(2).` wll beA. 7B. 20C. 49D. 21 |
| Answer» Correct Answer - A | |
| 72. |
4 moles of A are mixed with 4 moles of B. At equilibreium for the reaction `A+B hArr C+D.2` moles of C and D are formed. The equilibrium constant for the reaction will beA. `1/4`B. `1/2`C. 1D. 4 |
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Answer» Correct Answer - C `A+BhArrC+D` `{:("Initial conc",4,4, 0,0),("After T time conc",(4-2),(4-2),0,0):}` Equlibrium constant `=([C][D])/([A][B])=(2xx2)/(2xx2)=1` |
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| 73. |
For the system `A_((g))+2B_((g))hArrC_((g)),` the equlibrium concentrations are (A) `0.06` mole/litre (B) 0.12 mole/litre (C ) `0.216` mole/litre. The `K_(eq)` for the reaction isA. 250B. 416C. `4xx10^(-3)`D. 125 |
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Answer» Correct Answer - A For reaction `A+2BhArrC` `K=([C])/([A][B]^(2))=(0.216)/(0.06xx0.12xx0.12)=250.` |
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| 74. |
Partial pressures of A, B, C and D on the basis of gaseous system `A+2B hArr C+3D are =0.20,B=0.10, C=0.30 and D=0.50 atm.` The numerical value of equilibrium constant isA. `11.25`B. `18.75`C. 5D. `3.75` |
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Answer» Correct Answer - D `A+2BhArrC+3D` `K=([pC][pD]^(3))/([pA][pB]^(2))=(0.30xx0.50xx0.50xx0.50)/(0.20xx0.10xx0.10)=18.75` |
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| 75. |
For the following homogeneous gas reaction `4NH_(3)+5O_(2)hArr 4NO+6H_(2)O,` the equilibrium constant `k_(c)` has the dimension ofA. Conc `""^*+(10)`B. Conc `""^(+1)`C. Conc`""^(-1)`D. It is dimensionless |
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Answer» Correct Answer - B K has the units of `(conc.)^(Deltan), where Deltan=10-+1` |
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| 76. |
For the system `3A+2BhArrC,` the expression for equilibrium constant isA. `([3A][2B])/(C)`B. `([C])/([3A][2B])`C. `([A]^(2)[B]^(2))/([C])`D. `([C])/([3A]^(3)[2B]^(2))` |
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Answer» Correct Answer - D Equlibrium constant for the reaction, `3A+2BhArrC isK=([C])/([A]^(3)[B]^(2)).` |
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| 77. |
For the reaction `A+2BhArrC, `the expression for equlibrium constant isA. `([A][B]^(2))/([C])`B. `([A][B])/([C])`C. `([C])/([C][B]^(2))`D. `([C])/(2[B][A])` |
| Answer» Correct Answer - C | |
| 78. |
Equilibrium concentration of `HI,I_(2) and H_(2)` is 0.7, 0.1 and 0.1 M respectively. The equlibrium constant for the reaction `I_(2)+H_(2)hArr2HIis`A. 36B. 49C. `0.49`D. `0.36` |
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Answer» Correct Answer - B `K_(c)=([HI]^(2))/([H_(2)][I_(2)])=([0.7]^(2))/([0.1][0.1])=49` |
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| 79. |
In haber process 30 litres of dihydrogen and 30 littres of dinitorgen were taken for reaction which yieided only `50%` of the expected product. What will be the composition of gaseous mixture under these condition in the endA. 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogenB. 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogenC. 10 litres ammonia, 25 litres notrogen, 15 litres hydrogenD. 20n litres ammonia, 10 litres notrogen, 30 litres hydrogen |
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Answer» Correct Answer - C `N_(2)+2H_(2)hArr2NH_(3)` `{:(30,30,0),(30-x,30-3x,2x):}` `{:(,N_(2)=30-5=25 litre),(,H_(2)=30-3xx5=15litre),(,NH_(3)=2xx5=10 litre):}` |
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| 80. |
For the reaction `H_(2)+I_(2)hArr2Hl,` the equilibrium concentration of `H_(2),I_(2)and HI` are `8.0. 3.0 and 28.0` mol per litre respectively, the equlibrium constant of the reaction isA. `30.66`B. `32.66`C. `34.66`D. `36.66` |
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Answer» Correct Answer - B `K_(c)=([HI]^(2))/([H_(2)][I_(2)])=((28)^(2))/(8xx3)=32.66` |
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| 81. |
The equlibrium constant `(K_(c))` for the reaction `HA+B hArrBH^(+)+A^(-)` is 100. If the rate constant for the forward reactio is `10^(5).` then rate constant for the backward reaction isA. `10^(7)`B. `10^(3)`C. `10^(-3)`D. `10^(-5)` |
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Answer» Correct Answer - B `K_(c)(K_(f))/(K_(b))thereforeK_(b)=(K_(f))/(K_(c))=(10^(5))/(100)=100^(3)` |
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| 82. |
The reaction `A+BhArrC+D+` heat has reached equlibrium. The reaction may be made to proceed forward byA. Adding more CB. Adding more DC. Decreasing the temperatureD. Increasing the temperature |
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Answer» Correct Answer - C Exothermic reaction is faboured by low temperature to proceed in forward direction. |
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| 83. |
For the reactin `C_((s))+CO_(2(g))hArr2CO_((g)),` The partial pressure of `CO_(2) and CO` are 2.0 and 4.0 atm respectively at equlibrium. The `K_(p)` or the reaction isA. `0.5`B. `4.0`C. `8.0`D. `32.0` |
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Answer» Correct Answer - C `K_(p)=([P_(co)]^(2))/([P_(co_(2))])=(4xx4)/(2)=8.` |
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| 84. |
Indentifiy the incorrect statement regarding chemical equlibriumA. It can be attained from either sideB. `Q_(c)=K_(c)` at equlibriumC. Equlibreium is achieved when the reactant and product concentration become equalD. Presence of catalyst influences the position of equlibrium |
| Answer» Correct Answer - C | |
| 85. |
Assertion: Equlibrium constant has meaning only when the corresponding balanced chemical equation is given. Reason: Its value changes for the new equation obtained by multiplying or dividing the original equation by a numberA. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
| Answer» Correct Answer - A | |
| 86. |
In any chemical reaction, equlibrium is supposed to be estabilished whenA. Mutual opposite reaction undergoB. Concentration of reactants and resulting products are equalC. Velocity of mutual reactions become equalD. The temperaturee of mutual opposite reactions become equal |
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Answer» Correct Answer - B When rate of forward recton is requal to the rate of backeard reaction then equlibrium is supposed to be estblished. |
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| 87. |
Given reaction is `2X_((gas))+Y_((gas))hArr2Z_((gas))+80` kcal Which combination of pressure and temperatue gives the highest yield of Z at equlibriumA. `1000atm and 500^(@)C`B. `500atm and 500^(@)C`C. `1000atm and 100^(@)C`D. `500atm and 100^(@)C` |
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Answer» Correct Answer - C The reaction takes place with a reduction in number of moles (volume) and is exothermic. So high pressrue and low temperatue, will favour the reaction in forward direction. |
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| 88. |
Chemical equlibrium is dynamic in nature becauseA. Equlibrium is maintained rapidlyB. The concentraion of reactants and products become some at equlibriumC. The concentration of reactants and products are constant but differentD. Both forward and backward reaction occur at all time which same speed |
| Answer» Correct Answer - D | |
| 89. |
An equlibrium constant of `10^(-4)` for a reaction means, the equlibrium isA. Largely towards backward directionB. Largely towards forward directionC. Equally poisedD. Never established |
| Answer» Correct Answer - A | |
| 90. |
If pressure increases then its effect on given equlibrium `C_((s))+H_(2)O_((g))hArrCO_((g))+H_(2(g))` it is satisfied inA. Forward direactionB. Backward directionC. No effectD. None of these |
| Answer» Correct Answer - B | |
| 91. |
`XY_(2)` diddociates as `XY_(2)hArrXY_((g))+Y_((g))` when the initial pressure of `XY_(2)` is 600mm Hg, the total equlibrium pressure is 500 mm Hg. Calculate K for the reaction assuming that the volume of the system remain unchangedA. 50B. `100.0`C. `166.6`D. `400.0` |
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Answer» Correct Answer - B `XY_(2)hArrXY+Y` Initial pressure of `XY_(2)=600mm Hg` According to equation `XY_(2)hArrXY+Y` `{:("Initial",600,0,0),("At equlibrium",600-P,P,P):}` Total `=600+P=800P=200` `K=(200xx200)/(400)=100.0` |
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| 92. |
The equlibrium constant `K_(P)` for the thermal dissociation of `PCl_(5)at 200^(@)C` is 1.6 atm. The pressure (in atm) at which it is `50%` dossociated at that temperatureeA. `4.8`B. `4.2`C. `3.2`D. 6.4` |
| Answer» Correct Answer - A | |
| 93. |
The equilibrium constant `K_(c)` for the reaction `SO_(2)(g)+NO_(2)(g)hArrSO_(3)(g)+NO(g) ` is 16. If one mole of all four gases is taken in a one litre container, the equilibrium concentration of `SO_(3)` would beA. `0.4m`B. `0.6m`C. `1.4m`D. `1.6m` |
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Answer» Correct Answer - D Given `SO_(2(g))+NO_(2(g))to SO_(3(g))+NO_((g))` `{:("At initial concentration", "1 mole", "1 mole", "1 mole", "1 mole"),("At equlibrium concentration","1-x mole","1-x mole", "1+ mole", "1+ mole"):}` Applying law of mass action, `K_(c)=([SO_(3)][NO])/([SO_(2)][NO_(2)])=((1+x)(1+x))/((1-x)(1-x))=16` `or ((1+x)^(2))/((1-x)^(2))=16or(1+x)/(1-x)=4` `1+x=4-4x` `5x=3, i.e.,x=3//5=0.6` `"conc. of" SO_(3) "at equlibrium"=1+x=1+0.6=1.6"mole"` |
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| 94. |
Which of the following lines coorrectly show the temperature dependence of equlibrium constant, K, for an exothermic reaction A. B and CB. C and DC. A and DD. A and B |
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Answer» Correct Answer - D `DeltaG^(@)=-RTlnK` `DeltaH^(@)-T DeltaS^(@)=-RTlnK` `-(DeltaH^(@))/(RT)+(DeltaS^(@))/(R)=lnK` Therefore ln k vs `1/T` graph will be a straight line with slope equal to `-(DeltaH^(@))/(R),` Since reactio is exothermic, therefore `DeltaH^(@)` itself will be negative resulting in positive slope. |
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| 95. |
The equlibrium constant at 298 K for a reaction `A+BhArr C+D` is 100. If th initial concentration of all the foru species whre 1 M each, then equlibrium concentration of D (in mol`L^(-1))`will beA. `0.818`B. `1.818`C. `1.182`D. `0.182` |
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Answer» Correct Answer - B `A+BhArrC+D` `{:(t=0,1,1,1,1),(t_(eq),1-x,1-x,1+x,1+x):}` `implies((1+x)^(2))/((1-x)^(2))=100" "implies(1+x)/(1-x)=10` `implies1+x=10-10ximplies11x=9` `impliesx=9/11implies[D]=1+9/11implies[D]=1.818` |
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| 96. |
Two moles of `NH_(3)` when put into a previously evacuated vessel ( one litre), partially dossociate into `N_(2) and H_(2).` If at equlib rium one mole of `NH_(3)` is present, the equlibrium constant isA. `3//4 mol^(2) litre^(-2)`B. `27//64 mol^(2) litre^(-2)`C. `27//32 mol^(2) litre^(-2)`D. `27//16 mol^(2) litre^(-2)` |
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Answer» Correct Answer - D `2NH_(3)hArrN_(2)+3H_(2)` `{:("Initial concentration",2,0,0),("At equlibrium concentration",1,1//2,3//2):}` `K_(c)=([N_(2)][H_(2)]^(3))/([NH_(3)]^(2))=(1//2xx[3//2]^(3))/(I^(2))=27/16` `[Given 2-2x=1,2x=1,x=1//2]` |
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| 97. |
In the reaction `AB_((g))hArrA_((g))+B_((g))at 30^(@)C, K_(p)` for the dissociation equlibrium is `2.56xx10^(-2)` atm. If the total pressure at equlibrium is I atm, then the precentage dissociation of ABA. `87%`B. `13%`C. `43.5%`D. `6%` |
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Answer» `AB_((g)) to A_((g))+B_((g))` Applying law of mass action `K_(p)=(alpha^(2)P)/(a-alpha^(2))` `(alpha^(2))/(1-alpha)=2.56xx10^(-2)" "[therefore(1-alpha)=1]` `impliesalpha^(2)=2.56xx10^(-2)` `impliesalpha=sqrt(2.56xx10^(-2))impliesalpha=0.16` `%` age dissociation `=16%` |
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| 98. |
An equlibrium mixture of the reaction `2H_(2)S_((g))hArr2H_(2(g))+S_(2(g))"had 0.5 mole"H_(2)S, 0.10 "mole" H_(2)and 0.4 "mole" S_(2)` in one litre vessel. The vlaue of equlibrium constant (K) in mole `"litre"^(-1)` isA. `0.004`B. `0.008`C. `0.016`D. `0.160` |
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Answer» Correct Answer - C `K=([H_(2)]^(2)[S_(2)])/([H_(2)S]^(2))=([0.10]^(2)[0.4])/([0.5]^(2))=0.016` |
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| 99. |
The compound A and B are mixed in equimolar proportionto from the products, `A+BhArr C+D.` At equlibrium, one third of A and B are consumed. The equlibrium constant for the reaction isA. `0.5`B. `4.0`C. `2.5`D. `0.25` |
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Answer» Correct Answer - D `K=([C][D])/([A][B])=(1/3xx1/3)/(2/3xx2/3)=1/4=0.25` So, `K=0.25` |
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| 100. |
For the reaction `Pcl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`A. `K_(p)=K_(c)`B. `K_(p)=K_(c)(RT)^(-1)`C. `K_(p)=K_(c)(RT)`D. `K_(p)=K_(c)(RT)^(2)` |
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Answer» Correct Answer - C `Deltan=2-1=1` `K_(p)=K_(c)(RT)` |
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