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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. |
The reaction for which `K_(C) lt K_(P)` at a given temperature ( gt 50 k)is/are:A. `PCI_5(g)hArrPCI_3(g)+CI_2(g)`B. `CO_2(g)+C(s)hArr2CO(g)`C. `CaO(s)+CO_2(g)hArrCaCO_3(s)`D. `I_2(g)hArr2I(g)` |
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Answer» Correct Answer - A::B::D |
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| 102. |
When `K_(c)gt1` for a chemical reaction,A. the equilibrium would be achieved rapidlyB. the equilibrium would be achieved slowlyC. product concentrations would be much greater than reactant concentrations at equilibriumD. reactant concentraions would be much greater then product concentraions at equilibrium. |
| Answer» Correct Answer - C | |
| 103. |
The equilibrium constant in a reversible reaction at a given temperature whichA. Depends on the initial concentration of the reactantsB. Depends on the concentration of the products at equilibriumC. Does not depend on the initial concetrationsD. It is not characteristic of the reaction |
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Answer» Correct Answer - C Equilibrium constant is independent of original concentration of reactant. |
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| 104. |
In which of the following reaction `K_(p) gt K_(c)`A. `N_(2) + 3H_(2)hArr2NH_(3)`B. `H_(2) + I_(2)hArr2HI`C. `PC1_(3) + C1_(2)hArrPC1_(5)`D. `2SO_(3)hArrO_(2) + 2SO_(2)` |
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Answer» Correct Answer - D For reaction `2SO_(3)hArrO_(2) + 2SO_(2)` `Delta n` is + ve so `K_(p)` is more than `K_(c)` . By `K_(p) = K_(c)(RT)^(Delta n)` |
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| 105. |
Assertion : `K_(p) = K_(c)` for all reaction. Reason : At constant temperature, the pressure of the gas is proportional to its concentration.A. If both assertion and reason are true and reason is the correct explanation of the assertion.B. If both assertion and reason are true and 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. |
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Answer» Correct Answer - D Assertion is false but reason is true. `K_(p) != K_(c)` for all reaction `K_(p) = K_(c)(RT)^(Delta n)` `Delta n` = number of moles of products - number of moles of reactants in the balance chemicel equation. So, if for a reaction `Delta n = 0` . Then `K_(p) = K_(c)` |
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| 106. |
State and explain the Law of Mass Action. |
| Answer» Statement: At a given temperature, the product of concentrations of the reaction products reised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equlibrium Law or Law of Chemical Equilibrium. | |
| 107. |
At constant temperature, the equilibrium constant `(K_(p))` for the decomopsition reaction `N_(2)O_(4)hArr2NO_(2)` is expressed by `K_(p) = ((4x^(2)P))/((1-x^(2))`, where P = pressure, x = extent of decomposition. Which one of the following statement is true ?A. `K_(p)` increases with increase of PB. `K_(p)` increases with increase of xC. `K_(p)` increases with decreases of xD. `K_(p)` remains constant with change in P and x |
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Answer» Correct Answer - D `K_(p)` (equilibrium constant) is independent of pressure and concentration. |
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| 108. |
In the reaction, `A + B hArr2C`, at equilibrium, the concentration of A and B is `0.20` mol `L^(-1)` each and that of C was found to be `0.60` mol `L^(-1)`. The equilibrium constant of the reaction isA. `2.4`B. `18`C. `4.8`D. `9` |
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Answer» Correct Answer - D `A + B hArr 2C` `K_(c) = ([C]^(2))/([A][B]) = ([0.6]^(2))/([0.2][0.2]) = 9` |
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| 109. |
According to law of mass action rate of a chemical reaction is proportional toA. Concentration of reactantsB. Molar concentration of reactantsC. Concentration of productsD. Molar concentration of products |
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Answer» Correct Answer - B According to law of mass-acion, 'at a given temperature, the rate of a reaction at a particular instant is proportional to the product of the active masses of the reactants at that instant taised to powers which are numerically equal to the numbers of their respective molecules in the stoichiomtric equation decribing the reaction'. |
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| 110. |
High pressure and low temperature are favourable conditions for the synthesis of ammonia.A. High temperature and high pressureB. Low temperature and low pressureC. High temperature and low pressureD. Low temperature and high pressure |
| Answer» Correct Answer - D | |
| 111. |
In the reaction `A + 2B hArr2C`, if 2 moles of A, `3.0` moles of B and `2.0` moles of C are placed in a `2.0` L flask and equilibrium constant `(K_(c))` for the reaction isA. `0.073`B. `0.147`C. `0.05`D. `0.026` |
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Answer» Correct Answer - C `{:(,A,"+ ",2B,hArr,2C),("Initial conc.",2,,3,,2),("at etq.",2.5,,4,,1),("Molar",2.5/2=1.25,,4/2=2,,1/2=0.5):}` `K = ([0.5]^(2))/([1.25]xx[2]^(2)) = 0.05` |
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| 112. |
In a chemical equilibrium `A + B hArr` C + D, when one mole each of the two reactants are mixed, `0.6` mole each of the products are formed. The equilibrium constant calculated isA. 1B. `0.36`C. `2.25`D. `4//9` |
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Answer» Correct Answer - C `{:(,A,+,B,hArr,C,+,D),("Initial",1,,1,,0,,0),("Remaining at equiibrium",0.4,0.4,0.4,,0.6,,0.6):}` `K = ([C][D])/([A][B]) = (0.6xx0.6)/(0.4xx0.4) = (36)/(16) = 2.25` |
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| 113. |
The persentage of ammonia produced from nitrogen and hydrogen under conditions of temperature and pressure is given in the graph Use the graph answering the following questions: What happens to the percentage of ammonia produced when the temperature is increasedA. The `%` is decreasedB. The `%` is increasedC. No effectD. Cannot be predicted |
| Answer» Correct Answer - A | |
| 114. |
Physical and chemical equilibrium can respond to a change in their pressure, temperature, and concentration of reactants and products. To describe the change in the equilibrium we have a principle named Le Chatelier principle. According to this principle, even if we make some changes in equilibrium, then also the system even re-establishes the equilibrium by undoing the effect. If we add `SO_(4)^(2-)` ion to a saturated solution of `Ag_(2)SO_(4)`, it will result in a//anA. Increase in `Ag^(o+)` concentration.B. Decrease in `Ag^(o+)` concentrationC. It will shift `Ag^(o+)` ions from solid `ag_(2)SO_(4)` into solution.D. It will decrease the `SO_(4)^(2-)` ion concentration in the solution. |
| Answer» Correct Answer - B | |
| 115. |
In the reversible reaction A + B `hArr` C + D, the concentration of each C and D at equilobrium was `0.8` mole/litre, then the equilibrium constant `K_(c)` will beA. `6.4`B. `0.64`C. `1.6`D. `16.0` |
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Answer» Correct Answer - D Suppose 1 mole of A and B is each taken, then `0.8` mole/litre of C and D each formed remaining concentration of A and B will be `(1 - 0.8) = 0.2` mole/litre. `K_(c) = ([C][D])/([A][B]) = (0.8xx0.8)/(0.2xx0.2) = 16` |
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| 116. |
Under a given set of experiemental condition, with increase in the concentration of the reactants, the reate of a chemical reactionA. DecreasesB. IncreasesC. Remains unalteredD. First decreases and then increases |
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Answer» Correct Answer - C When rate of forward reaction is equal to the rate of backward reaction then reaction is said to be in equilibrium. |
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| 117. |
The equilibrium constant of a reaction is `20.0`. At equilibrium, the reate constant of forward reaction is `10.0`. The rate constant for backward reaction is :A. `0.5`B. `2.0`C. `10.0`D. `200.0` |
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Answer» `K_(c)=(K_(f))/(K_(b))` `:. 20=(10)/(K_(b))`, `:. K_(b)=(1)/(2)=0.5` |
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| 118. |
The value of `K_(c)` for the reaction `2A hArr B+C` is `2.0xx10^(-3)`. At a given time, the composition of reaction mixture is `[A]=[B]=[C]=3xx10^(-4)M`. In which direction the reaction will proceed? |
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Answer» For the reaction `2A hArr B+C` `K_(c )=2.0xx10^(-3)(given)` The reaction quotient `Q_(c )=([B][C])/([A]^(2))` Substituting the value of A, B, and C, we get `Q_( c)=((3xx10^(-4))(3xx10^(-4)))/((3xx10^(-4)))=1` Since `Q_(c ) gt K_(c )`, therefore, the reaction will proceed in the backward direction. |
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| 119. |
A reaction has a forward rate constant of `2.3xx10^6 s^(-1)` and an equilibrium constant of `4.0xx10^8` .What is the rate constant for the reverse reaction ?A. `1.1xx10^(-15) s^(-1)`B. `5.8xx10^(-3) s^(-1)`C. `1.7xx10^(2) s^(-1)`D. `9.2xx10^(14) s^(-1)` |
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Answer» Correct Answer - D |
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| 120. |
5 litre vessel contains 2 moles of each of gases A and B at equilibrium. If 1 mole each of A and B are removed. Calculate `K_(C)` for the reaction `A(g) iffB(g)` |
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Answer» Correct Answer - 1 |
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| 121. |
Calculate the degree of dissociation of HI at `450^(@)C` if the equilibrium constant for the dissociation reaction is `0.263`. |
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Answer» Correct Answer - B `{:(,HI,hArr,1/2H_(2),+,1/2I_(2)),("Initial",1,,0,,0),("At equilibrium",1-x,,x/2,,x/2),(,(1-x)/v,,x/(2v),,x/(2v)):}` `K_(c)=((x/(2v))^(1//2)(x/(2v))^(1//2))/((1-x)/v)` `0.263=(x xx v)/(2vxx(1-x))` `0.263=x/2(1-x~~1)` (x is very small) `x=0.526` |
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| 122. |
In the reaction `A_(2)(g) + 4B_(2)(g)hArr2AB_(4)(g)`, `Delta H lt 0` . The decomposition of `AB_(4)` (g) will be favoured atA. Low temperature and high pressureB. High temperature and low pressureC. Low temperature and low pressureD. High temperature and high pressure |
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Answer» Correct Answer - B `2AB_(4)(g)hArrA_(2)(g) + 4B_(2)(g)` , `Delta H = +ve` As decomposition of `AB_(4)` is endothermic, so reaction proceeds with increase in temperature. Secondly forward reaction proceeds with increase in number of moles of products, so low pressure favours it. |
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| 123. |
For the following reaction, formation of the prodcuts is favoured by : `Ag(g) +4B_(2)(g) hArr2AB_(4)(g)DeltaH lt 0`A. low temperature and high pressureB. high temperature and low pressreC. low temperature and low pressureD. high temperature and high pressure |
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Answer» Correct Answer - A |
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| 124. |
What is correct about the signs and magnitudes of the free energy, `DeltaG^circ` and the equilibrium constant, K, for a thermodynamically spontaneous reaction under standard conditons?A. `DeltaG^circ lt 0, K lt 0`B. `DeltaG^circ = 0, K gt 0`C. `DeltaG^circ lt 0, K = 0`D. `DeltaG^circ lt0, K gt 0` |
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Answer» Correct Answer - D |
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| 125. |
For the reaction, `ADP+"phosphate" hArr ATP, DeltaG^circ=30.50kJmol^-1`. What is the value of a equilibrium constant, K for this process under physiological conditions of `37.5^circC`?.A. `4.5xx10^-6`B. `7.4xx10^-6`C. `1.3xx10^5`D. `2.2xx10^5` |
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Answer» Correct Answer - B |
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| 126. |
Hydrolysis of phophodiester groups is the back bone of DNA, has `DeltaG^circ=-5.5kcal//mol` at `27^circC`. Approximate equilibrium costant for the hydorysis reaction is :A. `10^9`B. `10^4`C. `10`D. `10^10` |
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Answer» Correct Answer - B |
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| 127. |
Ammoina carbonate when heated to `200^(@)C` gives a mixture of `NH_(3)` and `CO_(2)` vapour with a density of `13.0` What is the degree of dissociation of ammonium carbonate ?A. `3//2`B. `1//2`C. 2D. 1 |
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Answer» Correct Answer - D `NH_(2)COONH_(4)hArr2NH_(3) + CO_(2)` `alpha =(D-d)/((n-1)d)` where D is the density (initial) `D=(mol.wt)/(2) =(78)/(2)=39` n=no. of product = 3d = final density `alpha =(39-13)/((3-13)13)=1` , so `alpha =1` |
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| 128. |
The degree of dissociation of `HI` at a particualr temperature is `0.8`. Calculate the volume of `2 M Na_(2)S_(2)O_(3)` solution required to neutralise the iodine present in an equilibrium mixture of a reaction when 2 mol each of `H_(2)` and `I_(2)` are heated in a closed vessel of `2 L` capacity and the equilibrium mixture is freezed. |
| Answer» Correct Answer - `1.6 litre` , | |
| 129. |
As `DeltaG^(circ)` for a reaction changes form a large negative value to a large positive value, K for the reaction will change form :A. a large positive value to a large negative value.B. a large positive value to a small positive value.C. a large negative value to a large positive value.D. a large negative value to a small negative value. |
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Answer» Correct Answer - B |
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| 130. |
The equilibrium constant `(K_(p))` for the reaction, `2SO_(2)+O_(2)hArr2SO_(3)` at `1000 K` is `3.5`. The partial pressure of oxygen gas to give equal mole of `SO_(2)` and `SO_(3)` is :A. `0.29 atm`B. `35 atm`C. `0.53 atm`D. `1.87 atm` |
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Answer» `2SO_(2)+O_(2)hArr2SO_(3)`, Let at equilibrium, mole of `SO_(2)` and `SO_(3)` be same, then `P_(SO_(3))=P_(SO_(2))` `K_(p)=((P_(SO_(3)))^(2))/((P_(SO_(2)))^(2)*P_(O_(2)))` or `P_(O_(2))=(1)/(K_(P))=(1)/(3.5)=0.29 atm` |
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| 131. |
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 towards right side if where `K_(c)` is the equilibrium constant.A. (A) `Q=K_(C)`B. (B) `QltK_(C)`C. (C) `QgtK_(C)`D. (D) `Q=0` |
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Answer» Correct Answer - C When`QgtK`, reaction will favour backward direction and when`QltK`, it will favour forward direction |
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| 132. |
Calculate the value of `log K_(p)` for the reaction, `N_(2(g))+3H_(2(g))hArr2NH_(3(g))` at `25^(@)C`. The standard enthalpy of formation of `NH_(3(g))` is `-46kJ` and standard entropies of `N_(2(g))`, `H_(2(g))` and `NH_(3(g))` are `191`, `130`, `192` `JK^(-1) mol^(-1)`. respectively. (`R=8.3JK^(-1)mol^(-1)`) |
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Answer» `N_(2(g))+3H_(2(g))hArr2NH_(3(g))` At equilibrium, `-DeltaG^(@)=2.303RT log_(10)K_(p)` ….(`1`) Also `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` (Given : `DeltaH^(@)` for `NH_(3)=-46 kJ`) and `DeltaS_(reaction)^(@)=2xxS_(NH_(3))^(@)-S_(N_(2))^(@)-3xxS_(H_(2))^(@)` `=2xx192-191-3xx130=-197J` Also `T=273+25=298 K` Thus, `DeltaG^(@)= -92xx10^(3)-298xx(-197)` (`DeltaH^(@)` for reaction`= -46xx2 kJ`) `=-92000+58706= -33294 J` Thus, from eq. (`1`), `+33294=2.303xx298xx8.3 log_(10) K_(p)` `:. log_(10) K_(p)=5.845` |
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| 133. |
For the reaction : `2SO_(2)(g)+O_(2)(g) hArr2SO_(3)(g),DeltaH=ve`. An increase in temperature shows:A. more dissociation of `SO_(3)` and a decreases in `K_(c)`B. less dissociation of `SO_(3)` and an increases in `K_(c)`C. more dissociation of `SO_(3)` and an increase in `K_(c)`D. less dissociation of `SO_(3)` and a decrease in `K_(c)` |
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Answer» Correct Answer - A |
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| 134. |
The equilibrium constant, `K_(p)` for the reaction `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` is `44.0atm^(-1) "at" 1000 K`. What would be the partial pressure of `O_(2)` if at equilibrium the amound of `SO_(2)` and `SO_(3)` is the same?A. (A) `16.0 atm`B. (B) `0.25atm`C. (C) `1atm`D. (D) `0.75atm` |
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Answer» Correct Answer - B |
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| 135. |
In the equilibrium `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` , the partial pressure of `SO_(2),O_(2)` and `SO_(3)` are 0.662,0.10 and 0.331 atm respectively . What should be the partial pressure of Oxygen so that the equilibrium concentrations of `SO_(3)` are equal ?A. `0.4atm`B. `1.0atm`C. `0.8atm`D. `0.25 atm` |
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Answer» Correct Answer - a |
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| 136. |
The reaction quotient (Q) for the reaction `N_(2)(g) + 3H_(2)(g)hArr2NH_(2)(g)` is given by Q `= ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`. The reaction will proceed towards right side if where `K_(c)` is the equilibrium constant.A. `Q gt K_(c)`B. `Q = 0`C. `Q = K_(c)`D. `Q lt K_(c)` |
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Answer» Correct Answer - A `N_(2)(g) + 3H_(2)(g)hArr2NH_(3)(g)` `.:` (Quotient) `=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))` , `Delta n = 2-4 = -2` At equilibrium Q is equal to `K_(c)` but for the progress of reaction towards right side `Q gt K_(c)` . |
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| 137. |
The reaction, `2SO_(2(g)) + O_(2(g))hArr2SO_(3(g))` is carried out in a 1 `dm^(3)` and ` 2 dm^3` vessel separately. The ratio of the reaction velocity will beA. `1 : 8`B. `1 : 4`C. `4 : 1`D. `8 : 1` |
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Answer» Correct Answer - D `2SO_(2)(g) + O_(2)(g)hArr2So_(3)(g)` For 1 `dm^(3) R = K[SO_(2)]^(2) [O_(2)]` `R = K[(1)/(1)]^(2) [(1)/(1)] = 1` For 2 `dm^(3) R = K[(1)/(2)]^(2) [(1)/(2)] = (1)/(8)` So, the ratio is `8 : 1` |
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| 138. |
A reaction system in equilibrium according to reaction `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` in one litre vessel at a given temperature was found to be `0.12` mole each of `SO_(2)` and `SO_(3)` and `5` mole of `O_(2)` In another vessell of one litre contains `32` g of `SO_(2)` at the same temperature. What mass of `O_(2)` must be added to this vessel in order that at equilibrium `20%` of `SO_(2)`is oxidized to `SO_(3)`?A. `0.4125`B. `11.6` gC. `1.6` gD. None of these |
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Answer» Correct Answer - B |
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| 139. |
For the equlibrium `H_(2)O(s)hArrH_(2)O(l)` which of the following statements is true?A. The pressure changes do not affect the equilibriumB. More of ice melts if pressure on the system is increasedC. More of liquid freezes if pressure on the system is increasedD. The degree of advancement of the reaction do not depend on pressure. |
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Answer» Correct Answer - B `N_(2)(g) + O_(2)(g) hArr 2NO(g)` `1mol e 2mol e 3mol e` `K_(c)=((3))^(2)/(1xx2)=((g)/(2))`. Let a mole of `O_(2)` is added, Then, `t=0{:(N_(2)(g),+,O_(2)(g),hArr,2NO(g)),(1"mole",,2"mole",,3"mole"),(1,,(2+a),,3),((1-x),,(2+a)-x,,(3+2x)):}` `[NO]=[(3+2x)/(100)]=0.04 , (3+2x)=4`. `2x=1, x=0.5`. `((3+2x)^2) /((1-x)(2+a-x))=(9)/(2)` `K_(c)=((4)^2))^(2)/(0.5[(1.5)-a])=(9)/(2)` `=(16)/(0.5(1.5+a))=(9)/(2)` `=(35)/(4.5)=[1.5+a]` `7.11=1.5+a`. `a=(101)/(18)=5.61` |
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| 140. |
When 3 moles of A and 1 mole of B are mixed in 1 litre vessel, the following reaction takes place `A_((g)) + B_((g))hArr2C_((g)). 1.5` moles of C are formed. The equilibrium constant for the reaction isA. `0.12`B. `0.25`C. `0.50`D. `4.0` |
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Answer» Correct Answer - D A + B `rarr` 2C `(3-0.75)` `(1-0.75)` `1.5` `K = ([C]^(2))/([A][B]) = ((1.5)^(2))/(2.25xx0.25) = (2.25)/(2.25xx0.25) = 4.0` |
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| 141. |
For the reaction `A_(2)(g) + 2B_(2)hArr2C_(2)(g)` the partial pressure of `A_(2)` and `B_(2)` at equilibrium are `0.80` atm and `0.40` atm respectively.The pressure of the system is `2.80` atm. The equilibrium constant `K_(p)` will beA. (A) `20`B. (B) `5.0`C. (C) `0.02`D. (D) `0.02` |
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Answer» Correct Answer - A `A_(2)(g)+2B_(2)(g)hArr2C_(2)(g)` `P_(A_(2)=0.80atm. P_(B_(2)=0.4` atm. Total pressure of the system`=2.8` atm. `thereforeP_(C_(2)=2.8-0.8-0.4=1.6` `Kp=(P_(C_(2)^(2))/(P_(A_(2)+P_(B_(2)^(3))))=((1.6)^(2))/(0.8+(0.4))^(2)=20` |
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| 142. |
If `10^(-4) dm^(3)` of water is introduced into a `1.0 dm^(3)` flask to `300K` how many moles of water are in the vapour phase when equilibrium is established ? (Given vapour pressure of `H_(2)O` at `300K` is `3170Pa R =8.314 JK^(-1) mol^(-1))` .A. `5.56xx10^(-3) mol`B. `1.53xx10^(-2) mol`C. `4.46xx10^(-2) mol`D. `1.27xx10^(-3) mol` |
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Answer» Correct Answer - D |
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| 143. |
The equilibrium constant `K_(p_(1))` and `K_(p_(2))` for the reactions `XhArr2Y` and `ZhArrP+Q`, respectively are in the ratio of `1:9`. If the degree of dissociation of `X` and `Z` be equal, then the ratio of total pressure at these equilibrium is:A. `1:1`B. `1:3`C. `1:9`D. `1:36` |
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Answer» Correct Answer - D `underset(x - alpha)(X) hArr underset(2alpha)(2Y)` ` underset(1 - alpha)(Z) hArr underset(alpha)(P) + underset(alpha)(Q)` `K_(P_(1))=((2alpha)/(1+alpha)P_(T_1))^(2)/(((1-alpha)/(1+alpha)P_(T_2))) K_(P_(2))=(((alpha)/(1+alpha)Pr_(2))((alpha)/(1+alpha)Pr_(2)))/(((1-alpha)/(1+alpha)P_(T_(2))))` `(K_(P_(1)))/(K_(P_2))=((2alpha)/(1+alpha)P_(T_(1)))^(2)/(((1-alpha)/(1+alpha)P_(T_(2))))xx(((1-alpha)/(1+alpha)P_(T_(2))))/(((alpha)/(1+alpha)P_(T_(2)))((alpha)/(1+alpha)P_(T_(2))))` `(1)/(9)=(4P_(T_(1)))/(P_(T_(2))), (P_(T_(1)))/(P_(T_(2)))=(1)/(36)` |
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| 144. |
The standard Gibbs energy change at `300K` for the reaction `2AhArrB+C` is 2494. `2J`. At a given time, the composition of the reaction mixture is `[A]=1/2, [B]=2` and `[C]=1/2`. The reaction proceeds in the `(R=8.314JK//"mol"e=2.718)`A. forward direction because `QgtK_(C)`B. reverse direction because `QgtK_(C)`C. forward direction because `QltK_(C)`D. reverse direction because `QltK_(C)` |
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Answer» Correct Answer - B `DeltaG=DeltaG^(@)+RT "in" Q=2494.2+8.314xx300 "in" 4=` positive DeltaG=RTln(Q)/(K)` Since, `DeltaG` is positive so, `QgtK`, so reaction shifts in reverse direction. |
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| 145. |
If `10^(-4) dm^(3)` of water is introduced into a `1.0 dm^(3)` flask to `300K` how many moles of water are in the vapour phase when equilibrium is established ? (Given vapour pressure of `H_(2)O` at `300K` is `3170Pa R =8.314 JK^(-1) mol^(-1))` .A. `5.56xx10^(-3)mol`B. `1.56xx10^(-2) mol`C. `4.46xx10^(-2) mol`D. `1.27xx10^(-3) mol` |
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Answer» Correct Answer - D `PV=nRT rArr V=1dm^(3)=10^(-3)m^(3) rArr P=3170Pa` `R=8.314JK^(-1)mol^(-1) rArr T=300K rArr 3170xx10^(-3)=nxx8.314xx300` `n=(3170xx10^(-3))/(8.314xx300)=1.27xx10^(-3)mol`. |
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| 146. |
statement-1 : The reaction quotient,Q has the same form as the equilibrium constant `K_(eq)`, and is evaluated using any given concentration of the species involved in the raction, and not necessarily equilibrium concentrations. statement-2 : If the numerical value of Q is not the same as the value of equillibrium constant, a reaction will occur.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 - B |
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| 147. |
In a container `H_(2)O(g),CO(g) "and" H_(2)(g)` are present in the molar ration of `1:2:3` respectively at temperature of `300K`, Find the pressure in the container at which solid carbon (graphile) will star forming in the container given that: `C(S)+H_(2)O(G)hArrCO(g)+H_(2)(g) K_(p)=3` atm |
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Answer» Correct Answer - `3` `C(S)+H_(2)O(g)hArrCO(g)+H_(2)(g)` `t=0 - 1 2 3` carbon solid will start forming when there will be equilibrium in the container `K_(p)=(P_(co).P_(H_(2)))/(P_(H_(2)O))=3 atm=((2P)(3P))/(P)=6P=3`atm So `P=(1)/(2)atm` So Total pressure in the container will be `P_(t)=P+2P+3P+=6P=3`atm |
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| 148. |
Assertion: The reaction quotient, Q has the same form as the equilibrium constant `K_(eq)`, and is evaluated using any given concentrations of the species involved in the reaction, and not necessarily equilibrium constrations. Reason: If the numerical value of Q is not the same as the value of equilibrium constant, a reaction will occur.A. 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. |
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Answer» Correct Answer - B Both assertion and reason are true but reason is not the correct explanation of the assertion. |
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| 149. |
The equilibrium constant for a reaction is K, and the reaction quotient is Q . For a particular reaction mixture , the ration `(K)/(Q)` is 0.33. this means that:A. the reaction mixture will equilirate to from more reactant speciesB. the rection mixture will equilirate to from more product speciesC. the equlibrium ratio of reactant to product concentration will be 3D. the equilibrium ratio of reactant to product concentrations will be 0.33 |
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Answer» Correct Answer - b |
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| 150. |
`K_(c)` for the reaction `N_(2)O_(4) hArr 2NO_(2)` in chloroform at `291 K` is `1.14`. Calculate the free energy change of the reaction when the concentration of the two gases are `0.5` mol `dm^(-3)` each at the same temperature. `(R=0.082 L atm K^(-1) mol^(-1))` |
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Answer» From the given data `T=291 K, R=0.082 L "atm" K^(-1) "mol"^(-1)` `K_(c )=1.14, C_(NO_(2))=C_(N_(2)O_(4))=0.5 "mol" dm^(-3)` The reaction quotient `Q_(c )` for the reaction `N_(2)O_(4) hArr 2NO_(2)`, `Q_(c )=([NO_(2)]^(2))/([N_(2)O_(4)])=(0.5xx0.5)/0.5=0.5` Since `Q_(p)=Q_(c ) (RT)^(Deltan)` and `Deltan=2-1=1` in this case `:. Q_(p)=0.5(0.082xx291)=11.93` `K_(p)=K_(c )(RT)^(Deltan)=1.14 (0.082xx291)=27.1` Substituting these values in the following equation, we get `DeltaG=DeltaG^(ɵ)+RT ln Q_(p)` `=-RT ln K_(p)+RT 1n Q_(p)` `=-2.303 RT (log K_(p)-log Q_(p))` `DeltaG=-(0.082xx291xx2.303)(log 27.2-log 11.93)` `-54.95 (1.4346-1.0766)=-19.67 L "atm"` |
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