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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.
| 801. |
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. `PCI_(3)(g)hArrPCI_(3)(g)+CI_(2)(g)`C. `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`D. `SO_(2)CI_(2)(g)hArrSO_(2)(g)+CI_(2)(g)` |
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Answer» Correct Answer - A In this reaction the ratio of number of moles of reactants to products in same `i.e., 2:2`, hence change in volume will not alter the number of moles. |
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| 802. |
Assertion: The dissociation of `CaCO_(3)` can be represented as, `CaCO_(3)(s)hArrCaO(s) + CO_(2)(g)` . Some solid `CaCO_(3)` is placed in an evacuted vessel enclosed by a piston and heated so that the volume of the vessel is doubled, while the temperature is held constant, the number of moles of `CO_(2)` in the vessel increase. Reason: The pressure ofv `CO_(2)` in the vessel will remain the same.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 - A Both assertionand reasion are true and the reason is the correct explanation of the assertion. |
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| 803. |
statement-1 : The dissociatiohn of `CaCO_(3)` can be represented as, `CaCO_(3)(s)hArrCaO(s)+CO_(2)(g)`. Some solid `CaCO_(3)` is placed in an evacurted vessel enclosed by a piston and heated so that a portion of it decomposes. If the piston is moved so that the volume of the vessel is doubled, while the tempratuere is held constant, the number moles of `CO_(2)` in the vessel increase. statement-2 : The pressure of `CO_(2)` in the vessel will remain the same.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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| 804. |
`C(s)+CO_(2)(g)hArr2CO(g) K_(p)=1atm` `CaCO_(3)(s)hArrCaO(s)+CO_(2)(s) K_(p)=4xx10^(-2)` Solid `C,CaO "and" CaCO_(3) "are mixed and allowed to attain equilibrium. Calculate final pressure of" CO`.A. `0.4`atmB. `0.2` atmC. `8` atmD. `0.01` atm |
| Answer» Correct Answer - B | |
| 805. |
The thermal dissociation of equilibrium of `CaCo_(3)(s)` is studied under different conditions `CaCO_(3)(s)hArrCaO(s)+CO_(2)(g)` For this equilibrium, the correct statement (s) is/areA. `DeltaH` is dependent on `T`B. `K` is independent of the initial amount of `CaCO_(3)`C. `K` is dependent on the pressure of `CO_(2)` at a given `T`D. `DeltaH ` is independent of the catalyst, if any |
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Answer» Correct Answer - A::B::D (A) `DeltaH_(2)-DeltaH_(1)=C_(P)_((rxb))(T_(2)-T_(1))` and `C_(P)` depends on temperature. Hence enthalpy also depends on temperature. (B) `CaCO_(3_(s))hArrCaO_(s)+CO_(2_((g))) K_(P)=(P_(CO_(2)))_("at equilibrium")` For a given reaction. `K_(eq)` depends only on temperature. (C) `K_(eq)` depends only on temperature. (D) Enthalpy of reaction is independent of the catalyst. Catalyst generaly changes activation energy. |
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| 806. |
In a gases reaction, `A_((g))+B_((g))hArrC_((g))`, predict the effect of addition of inert gas, if addition is made at : (`a`) constant volume, (`b`) constant pressure. |
| Answer» (`a`) No change, (`b`) In backward direction. | |
| 807. |
For the reaction `CaCO_(3)(s)hArrCaO(s)+CO_(2)(g)` the equilibrium amount of `CO_(2)` can be increased by `:`A. Adding a suitable catalystB. Adding more limestoneC. Increasing volumeD. Adding inert gas at constant volume |
| Answer» Correct Answer - 3 | |
| 808. |
For the gas phase exothermic reaction. `A_(2)+B_(2)hArrC_(2)`, carried out in a closed vessel, the equilibrium moles of `a_(2)` can be increased by:A. increasing the temperatureB. decreasing the pressureC. adding inert gas at constant pressureD. removing some `C_(2)` |
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Answer» Correct Answer - A::B::C |
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| 809. |
The gaseous reaction `A + B hArr2C + D, + Q `is most favoured atA. Low temperature and high pressureB. High temperature and high pressureC. High temperature and low pressureD. Low temperature and low pressure |
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Answer» Correct Answer - D Low temperature and low pressure. |
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| 810. |
In a two step exothermic reaction `A_(2)(g) + B_(2)(g) hArr 3C(g) hArr D(g)," "DeltaH=-ve` steps `1 & 2` are favoured respectively by .A. High pressure, high temperature & low pressure, low temperatureB. Low pressure, low temperature & high pressure, low temperatureC. High pressure, low temperature & low pressure, low temperatureD. Low pressure, high temperature & high pressure, high temperature |
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Answer» Correct Answer - 2 Follow the li-chatelier principle for both steps individually. |
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| 811. |
Assertion (A): The endothermic reactions are favoured at lower temperature and the exothermic reactions are favoured at higher temperature. Reason (R ) : when a system in equilibrium is disturbed by changing the temperature, it will tend to adjust itself so as to overcome the effect of the change.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - D |
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| 812. |
Consider the following equilibrium `N_(2)O_(4)(g)hArr2NO_(2)(g)` Then the select the correct graph , which shows the variation in concentratins of `N_(2)O_(4)` Against concentrations of `N_(2)`O_(4)`:A. B. C. D. |
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Answer» Correct Answer - b |
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| 813. |
The vapour pressure of mercury is 0.002 mm Hg at `27^(@)C` .`K_(c)` for the process `Hg(l)hArrHg(g) `is :A. `0.002`B. `8.12xx10^(-5)`C. `6.48xx10^(-5)`D. `1.068xx10^(-7)` |
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Answer» Correct Answer - d |
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| 814. |
For `A+BhArrC+D` , the equilibrium constant is `K_(1)` and for `C+DhArA+B`, the equilibrium constant is `K_(2)`. The correct relation between `K_(1)` and `K_(2)` isA. `K_(1)xxK_(2)=1`B. `K_(1)xx(K_(2)-1)=0`C. `K_(1)//K_(2)=1`D. All of these |
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Answer» Correct Answer - A `A+Boverset(K_(1))(hArr)C+D` `C+Doverset(K_(2))(hArr)A+B` `K_(1)=1/K_(2)rArr K_(1)xxK_(2)=1` |
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| 815. |
If we know the equilibrium constant for a particular reaction, we can calculate the concentration in the equilibrium mixture from the initial concentrations. Generally only the initial concentration of reactions are given. In above problem, what is `K_(p)` of the reaction?A. 1B. 2C. 4D. None of these |
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Answer» Correct Answer - C |
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| 816. |
The equilibrium constant for a reacton `N_(2)(g)+O_(2)(g)=2NO(g)` is `4xx10^(-4)` at `2000 K`. In the presence of catalyst, the equilibrium constant is attained `10` times faster. The equilibrium constant in the presence of catalyst, at `2000 K` isA. `40xx10^(-4)`B. `4xx10^(-4)`C. `4xx10^(-3)`D. difficult to compute without more data |
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Answer» Correct Answer - a |
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| 817. |
The equilibrium constant for a reacton `N_(2)(g)+O_(2)(g)=2NO(g)` is `4xx10^(-4)` at `2000 K`. In the presence of catalyst, the equilibrium constant is attained `10` times faster. The equilibrium constant in the presence of catalyst, at `2000 K` isA. `40xx10^(-4)`B. `4xx10^(-4)`C. `4xx10^(-2)`D. incomplete data |
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Answer» Correct Answer - B Catalyst does not change the extent of reaction, and hence does not alter the value of K. |
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| 818. |
The equilibrium constant for a reacton `N_(2)(g)+O_(2)(g)=2NO(g)` is `4xx10^(-4)` at `2000 K`. In the presence of catalyst, the equilibrium constant is attained `10` times faster. The equilibrium constant in the presence of catalyst, at `2000 K` isA. `4xx10^(-3)`B. `4xx10^(-4)`C. `4xx10^(-5)`D. none of these |
| Answer» Catalyst simply helps in attaining equilibrium earlier. | |
| 819. |
`N_(2)O_(4)` is `25%` dissociated at `37^(@)C` and `1` atm pressure. Calculate a. `K_(p)` b. The percentage dissociation at `0.1` atm and `37^(@)C`. |
| Answer» (`a`) `0.267 atm` , (`b`) `63.26%` | |
| 820. |
`N_(2)O_(4)(g)` is dissociated to an extent of `20%` at equilibrium pressure of `1.0` atm and `57^(@)C`. Find the percentage of `N_(2)O_(4)` at `0.2` atm and `57^(@)C`. |
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Answer» Correct Answer - A::D `{:(,N_(2)O_(4)(g), hArr,2NO_(2)(g)),(underset(("moles"))(t=0),a,,0),(t=t_("eq"),3-aalpha,,2aalpha):}` `K_(p)=(P_(NO_(2))^(2))/(P_(N_(2)O_(4)))=(((2aalpha)/(a+aalpha).P)^(2))/((a-aalpha)/(a+aalpha)P)=(4alpha^(2)P)/(1-alpha^(2))` `K_(p)` will remain same as is constant `rArr (4alpha_(1)^(2)P_(1))/(1-alpha_(1)^(2))=(4alpha_(2)^(2)P_(2))/(1-alpha_(1)^(2))` `[alpha_(1)=0.2, P_(1)=1.0 "atm" P_(2)=0.2 "atm"]` `rArr alpha_(2)=0.41` |
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| 821. |
`Fe_2O_3(s)` may be converted to Fe by reaction : `Fe_2O_3(s) +3H_2(g)hArr 2Fe(s)+3H_2O(g)` For which `K_c=8` at temperature 800 K. What percentage of `H_2` remains unreacted at equilibrium ?A. `50%`B. `66.6%`C. `33.3%`D. `78%` |
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Answer» Correct Answer - C |
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| 822. |
`N_(2)O_(4)` is dissociated to `33%` and `50%` at total pressure `P_(1)` and `P_(2)atm` respectively. The ratio of `P_(1)//P_(2)` is:A. `7//4`B. `7//3`C. `8//3`D. `8//5` |
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Answer» `underset(1-alpha)underset(2)(N_(2)O_(4))hArrunderset(2alpha)underset(0)(2NO_(2))` `K_(p)=((n_(NO_(2)))^(2))/((n_(N_(2)O_(4))))xx[(P)/(sumn)]^(1)` For `33%` dissociation: `K_(p)=((2xx33)^(2))/(67)xx[(P_(1))/(133)]^(1)` For `50%` dissociation: `K_(p)=((2xx50)^(2))/(50)xx[(P_(2))/(150)]^(1)` `:. (P_(1))/(P_(2))=(133xx67xx(2xx50)^(2))/((2xx33)^(2)xx150xx50)=2.72` |
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| 823. |
For the reaction: `2Fe^(3+)(aq)+(Hg_(2))^(2+)(aq) hArr 2Fe^(2+)(aq)` `K_(c)=9.14xx10^(-6)` at `25^(@)C`. If the initial concentration of the ions are `Fe^(3+)=0.5 M, (Hg_(2))^(2+)=0.5 M, Fe^(2+)=0.03 M` and `Hg^(2+)=0.03 M,` what will be the concentration of ions at equilibrium. |
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Answer» Correct Answer - B::C::D `{:(2Fe^(3+)(aq),+,(Hg_(2))^(2+)(aq),hArr,2Fe^(2+)(aq),+,2Hg^(2+)(aq)),(0.5,,-.5,,0.03,,0.03),(,,,,,,("conc before reaction")),((0.5-a),,(0.5-a//2),,(0.03+a),,(0.03+a)),(,,,,,,"Concentration after reaction at equilibrium"):}` `K_(c)=9.14xx10^(-6)=([Fe^(2+)]^(2)[Hg^(+2)]^(2))/([Fe^(3+)]^(2)[Hg_(2)^(+2)])` `:. 9.14xx10^(-6)=((0.03+a)^(2)(0.03+a)^(2))/((0.5-a)^(2)(0.5-a//2))` `:. a=0.0027` `:. [Fe^(3+)]=0.5-0.0027=0.4973 M` `[Hg_(2)^(2+)]=0.5-0.0027/2=0.4987 M` `[Fe^(2+)]=0.03+0.0027=0.0327 M` `[Hg^(2+)]=0.03+0.0027=0.0327 M` |
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| 824. |
For the reaction, `2Fe_((aq.))^(3+)+Hg_((aq.))^(2+)hArr2Fe_((aq.))^(2+)+2Hg_((aq.))^(2+)` `K_(c)=9.14xx10^(-6)` at `25^(@)C`. If the initial concentration of the ions are `Fe^(3+)=0.5M`, `(Hg_(2))^(2+)=0.5M`, `Fe^(2+)=0.03M` and `Hg^(2+)=0.03M`. What will be the concentrations of ions at equilibrium? |
| Answer» `[Fe^(3+)]=0.4973 M`, `[Hg_(2)^(2+)]=0.4987 M` , `[Fe^(2+)]=0.0327 M`, `[Hg^(2+)]=0.0327 M` , | |
| 825. |
Using given information in question provided calculate equilibrium constant of requried reaction. Calculate equilibrium constant of required reaction. `A(g)+2B(g) hArr4C(g)" " K_(P_(1))=X` `C(g) hArrD(g)" " K_(P_(2))=Y` Value of `K_(p)` for reaction `(1)/2A(g)+B(g)hArr2D(g)`A. `sqrtYxxX^2`B. `sqrtX/Y^2`C. `sqrtXxxY^2`D. `sqrtY/X^2` |
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Answer» Correct Answer - C |
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| 826. |
What quantiative information can you obtain from the value of the equilibrium constant? |
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Answer» (i) Large value of equlibium constant `(gt10^(3))` shows that forward reaction is favoured, ie. Concertration of products is much large than that of the reactants at equibrium. (ii) Intermediate value of `K(10^(-3)" to "10^(3))` shows that the concentration of the reactant and products are comparable. (iii) Low value of `K(lt10^(-3))` shows at backward reaction is favoured, i.e. concenctration of reactants is much large than that of the products. |
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| 827. |
Which of the following changes the value of the equilibrium constant ?A. change in concentrationB. change in pressureC. change in volumeD. None of these |
| Answer» Correct Answer - D | |
| 828. |
Explain why pure liquids and solids can ignored while writing the equilibrium constant expression? |
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Answer» [Pure liquid] or[Pure solid] `=("No.of moles")/("Volume of L")=("Mass"//"mol.mass")/("Volume")` `=(Mass)/(Volume)=(1)/(Mol.mass)=(Density)/(Mol.mass)` (Density & Moiecular weight are remain constant). |
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| 829. |
When `C_(2)H_(5)OH "and" CH_(3)COOH` are mixed in equivaient proportion, equilibrium is reached when `2//3` of acid and alchol are use. How much ester will be present when `3 `mole of acid were to react with `3` mole of alcohol. Express your anwer multiplying with `10`. |
| Answer» Correct Answer - 20 | |
| 830. |
Consider the equilibrium `Ni(s)+4CO(g)hArrNi(CO)_(4)(g), K_(p)=0.125atm^(-3)` If equal number of moles of `CO "and" Ni(CO)_(4)` (ideal gases) are mixed in a small container fitted with a piston, find the maximum total pressure (in atm) to which this mixture must be brough in order to just precipitate out metallic `Ni`? |
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Answer» Correct Answer - 4 `Ni(s)+4CO(g)hArrNi(CO)_(4)(g)` `P P` For backward reaction `Q_(P)geK_(P)` `(P)/(P^(4))geK_(P)` `therefore (1)/(P^(3))ge0.125atm^(-3)` `P^(3)le8atm^(3)` `Ple2atm` `P_(t otal)=2P=4atm`. |
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| 831. |
In an experiment starting with `1` mol `C_(2)H_(5)OH, 1` mol `CH_(3)COOH`, and `1` mol of water, the equilibrium mixture mixture of analysis showa that `54.3%` of the acid is eaterified. Calculate `K_(c)`. |
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Answer» Correct Answer - 4 `CH_(3)COOH(I)hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O(l)` `{:("initial",1,1,0,1),("at equilibrium",1-x,1-x,x,1+x):}` `1-0543 1-0.543 0.543 1+0.543` `(54.3% "of" 1 mol e=(1xx54.3)/(100)=0.543 mol e)` Hence given `x=0.543 mol e` Applying law of mass action `K_(C)=(["ester"]["water"])/(["acid][alcohol])=(0.543xx1.543)/(0.457xx0.457)=4.0` |
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| 832. |
`CH_(3)COOH_((l)) + C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5(l)) + H_(2)O_((l))` In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of little cone. `H_(2)SO_(4)`. On equilibrium being attainedA. 1 mole of ethyl acetate is formedB. 2 mole of ethyl acetate are formedC. `1//2` mole of ethyl acetate is formedD. `2//3` moles of ethyl acetate is formed |
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Answer» Correct Answer - D In presence of little `H_(2)SO_(4)` (as catalyst) about `2//3` mole of each of `CH_(3)COOH` and `C_(2)H_(5)OH` react to from `2//3` mole of the product at equilibrium. |
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| 833. |
consider the following reversible gaseous reaction (at 298 K): `(A) N_(2)O_(4)hArr` (B) `2SO_(2) + O_(2)hArr2SO_(3)` (C) `2HI hArrH_(2) + I_(2)` (D) `X + Y hArr4Z ` Highest and lowest value of `(K_(p))/(K_(c))` will be shown by the equilibriumA. (A) d,bB. (B) a,cC. (C) a,bD. (D) b,c7 |
| Answer» Correct Answer - A | |
| 834. |
If the equilibrium constant of the reaction `2HI(g)hArrH_(2)(g)+I_(2)(g)` is `0.25`, find the equilibrium constant of the reaction. `(1)/(2)H_(2)+(1)/(2)I_(2)hArrHI(g)` |
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Answer» Correct Answer - 2 For reaction `H_(2)(g)I_(2)(g)hArr2HI(g) K"=(1)/(0.25)=4` For reaction `(1)/(2)H_(2)+(1)/(2)I_(2)hArrHI(g) K"=sqrt4` |
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| 835. |
If the equilibrium constant of the reaction `2HI hArrH_(2) + I_(2)` is `0.25` , then the equilibrium constant 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 `2HI hArrH_(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 second reaction is reverse of first. |
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| 836. |
`Delta G^(@)(HI, g) ~= + 1.7 kJ`. What is the equilibrium constant at `25^(@)C` for `2HI(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 `Delta G^(@) = - 2.303xx8.314xx10^(-3)xx298 log K_(p)` `1.7 = - 2.303xx8.314xx10^(-3)xx298xxlog K_(p)` `K_(p) = 0.5` |
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| 837. |
At a certain temp. `2HI hArrH_(2) + I_(2)` . Only `50%` HI is dissociated at equilibrium. The equilibrium constant isA. `0.25`B. `1.0`C. `3.0`D. `0.50` |
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Answer» Correct Answer - A `underset(50)underset(100)(2HI)hArrunderset(25)underset(0)(H_(2))+underset(25)underset(0)(I_(2))` `([H_(2)][I_(2)])/([HI]^(2)) = (25xx25)/(50xx50) = 0.25` |
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| 838. |
The equilibrium constant for the reaction `H_(2)(g)+S(s) hArr H_(2)S(g)` is `18.5` at `925 K` and `9.25` at `1000 K`, respectively. Calculate the enthalpy of the reaction. |
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Answer» Given `K_(1)=18.5` at `T_(1)=925` `K_(2)=9.25 at T_(2)=1`. Using the relation, `"log"K_(2)/K_(1)=(DeltaH)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `"log" 9.25/18.5=(DeltaH)/(2.303xx8.314)xx(75)/(925xx1000)` `-0.301=(DeltaHxx75)/(2.303xx8.314xx925xx1000)` or `DeltaH=-71080.57 J "mol"^(-1)` |
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| 839. |
An exothermic reaction is represented by the greph :A. B. C. D. |
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Answer» Correct Answer - c |
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| 840. |
The volume of the reaction vessel containing an equilibrium mixture is increased in the following reaction `SO_(2)Cl_(2)(g)hArrSO_(2)(g)+Cl_(2)(g)` When equilibrium is re-established :A. The amount of `Cl_(2)(g)` remains unchangedB. the amount of `Cl_(2)(g)` increasesC. The amount of` SO_(@)Cl_(2)(g)`decreasesD. The amount of `SO_(@)(g)` decrsases |
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Answer» Correct Answer - b |
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| 841. |
A Syatem at equilibrium is described by the equation of fixed temperature T. `SO_(2)Cl_(2)(g)hArrSO_(2)(g)+Cl_(2)(g)` What effect will be the effect on equilibrium, if total pressure is resucing volume?A. Concentration of `SO_(2)Cl_(2)(g)` increasesB. Concentrations of `SO_(2)(g)` increasesC. Concentration of `Cl_(2)(g)` increasesD. Concentration of all gases increaseses |
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Answer» Correct Answer - d |
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| 842. |
Both matels Mg and Fe can reduce copper metal from a solution having copper ions `(Cu^(2+))`. According to the equilibria: `Mg(s)+Cu^(2+)hArrMg^(2+)+Cu(s), K_(1)=6xx10^(90)` `Fe(s)+Cu^(2+)hArrFe^(2+)+Cu(s), K_(2)=3xx10^(26)` Which metal will remove cupric ion from the solution to a greater extent? |
| Answer» Since `K_(1) gt K_(2)` the product in the first reaction is much more favoured than in the second one. `Mg` thus removes more `Cu^(2+)` from solution than `Fe` does | |
| 843. |
Addition of an inert gas at constant volume to the equilibrium mixture does not affect the position of equilibrium. |
| Answer» Correct Answer - True | |
| 844. |
Introduction of inert gas (at the same temperature) will affect the equilibrium if :A. volume is constant and `Deltan_(g)ne0`B. pressure is constant and `Deltan_(g)ne0`C. volume is constant and `Deltan_(g)=0`D. pressure is constant and `Deltan_(g)=0` |
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Answer» Correct Answer - B |
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| 845. |
For the following gaseous equilibrium, `N_2O_4(g)hArr 2NO_2(g)` `K_P` is found to be equal to `K_c` This is attained when temperature is :A. `0^@C`B. 273 KC. 1 KD. 12.19 K |
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Answer» Correct Answer - D |
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| 846. |
For the reaction `N_2O_4 hArr 2NO_2(g)`, if percentage dissociation of `N_2O_4` are 20%, 45%, 65%, 80% then the sequence of observed vapour densities wil be :A. `d_(20)gtd_(45)gtd_(65)gtd_(80)`B. `d_(80)gtd_(63)gtd_(45)gtd_(20)`C. `d_(20)=d_(45)=d_(65)=d_(80)`D. `d_(2)=d_(45)gt d_(65)=d_(80)` |
| Answer» Correct Answer - A | |
| 847. |
For the reaction, `SnO_(2)(s)+2H_(2)(g) hArr Sn(l)+2H_(2)O(g)` the equilibrium mixture of steam and hydrogen contained `45%` and `24% H_(2)` at `900 K` and `1100 K` respectively. Calculate `K_(p)` at both the temperature. Generally should it be higher or lower temperatures for better reduction of `SnO_(2)`? |
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Answer» Correct Answer - A::C::D `{:(,SnO_(2)(s),+,2H_(2)(g),hArr,Sn(l),+,2H_(2)O(g)),("Case I:",900 K,,,,45,,55):}` `:. K_(C_(1))=((55)^(2))/((45)^(2))=1.494` `{:(,SnO_(2)(s),+,2H_(2)(g),hArr,Sn(l),+,2H_(2)O(g)),("Case I:",1100 K,,24,,,,76):}` `:. K_(C_(2))=(76xx76)/(24xx24)=10.03` `K_(C)` increases with temperature and thus the reaction should be endothermic. Thus, high temperature is perferred for reduction of `SnO_(2)`. |
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| 848. |
The active mass of `64 g` of `HI` in a `2-L` flask would beA. 2B. 1C. 5D. 0.25 |
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Answer» Correct Answer - D |
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| 849. |
The active mass of `64 g` of `HI` in a `2-L` flask would beA. (A) `2`B. (B) `1`C. (C) `5`D. (D) `0.25` |
| Answer» Correct Answer - D | |
| 850. |
The active mass of `64 g` of `HI` in a `2-L` flask would beA. (A) `22:3:7`B. (B) `0.5:3:7`C. (C) `1:3:1`D. (D) `1:3:0.5` |
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Answer» Correct Answer - D Molar conc.`=(22)/(44)=(1)/(2)` Moles of `H_(2)=(3)/(2) rArr "Moles of" N_(2)=(7)/(28)=(1)/(4)` Ratio of active massess`=1//2:3//2:1//4 "or"1:3:0.5`. |
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