52671 + Interview Questions in Chemistry in Class 11

1.	At low pressure Vander Waal's equation for 3 moles of a real gas will have its simplified from
Answer» `(PV)/(RT-(3a//V)) = 3` `(PV)/(RT + Rb) = 3` `(PV)/(RT - 3Pb) = 1` `(PV)/(RT - (9//V)) = 3` Solution :LOW `P IMPLIES` negligible b `(P + (an^2)/(V^2)) (V) = n^(RT) ,` for 3 moles `(P + (a XX 9)/(V^2)) (V) = 3RT` `PV + (9a)/V = 3RT , PV = 3 (RT - (3a)/(V))`

Discussion

2.	At low pressure, the fraction of the surface covered follows
Answer» zero-ORDER REACTION second-order reaction first-order reaction FRACTIONAL order ANSWER :C

Discussion

3.	At low pressure of 0.25 atm, 2 mole of a real gas has Boyl's temperature 100K. The approximate volume of gas at this temperature and pressure is:
Answer» 66 litre 33 litre 44.8litre none of these Answer :a

Discussion

4.	At K, K_(p) = 2*0 xx 10 ^(10)//"bar" for the given reaction at equilibrium 2 SO_(2) (g) + O_(2) (g) hArr 2 SO_(3) (g) What is K_(c) at this temperature ?
Answer» Solution :For the given reaction , ` Delta n_(g)= 2 -3 = -1 ` ` K_(p) = K_(c) (RT)^(Delta n) or K_(c) = K_(p) (RT)^(-Delta n)= K_(p) (RT) = ( 20 xx 10^(10)"bar"^(-1))(00831 L "bar" K^(-1) "MOL" ^(-1) ) ( 450 K)` ` = 7.48 xx 10^(10 )L"mol"^(-1) = 7*48 xx 10^(11) L "mol"^(-1)`

Discussion

5.	At its melting point ice is lighter than water because
Answer» `H_(2)O` molecules are more closely packed in solid state ICE crystals have hollow HEXAGONAL arrange-ment of `H_(2)O` molecules shrinks in size On MELTING of ice the `H_(2)O` molecules shrinks in size Ice forms MOSTLY heavy water on first melting ANSWER :B

Discussion

6.	At isoelectric point
Answer» colloidal PARTICLES migrate towards oppositely charged ELECTRODES coagulates becomes electrically neutral pH of the solution becomes 7 Solution :At isoelectric point, colloidal particles do not migrate towards oppositely charged electrodes as at isoelectric pH, colloidal particles do not carry any CHARGE. The pH at isoelectric point not 7.

Discussion

7.	At infinite dilution , the percentage ionisation of both strong and weak electrolytes is
Answer» `1%` `20%` ` 50%` ` 100%` Solution :` ALPHA = sqrt((K_a)/(C)) "as " c to 0, alpha to 1`

Discussion

8.	At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3sqrt(3) times that of a hydrocarbon having molecular formula C_(n)H_(2n-2). What is the value of n ?
Answer» Solution :`(gammaH_(2))/(gammaC_(n)H_(2n-2))=SQRT((m_(C_(n)H_(2n-2)))/(m_(H_(2))))` `3sqrt(3)=sqrt((m_(C_(n)H_(2n-2)))/(2))` Squaring on both SIDES and REARRANGING `27xx2-m_(C_(n)H_(2n-2))` `54=n_(12)+(2n-2)(1)` `54=12n+2n-2` `n=(54+2)//14` `=56/14=4` `:.n=4`

Discussion

9.	At identical conditions of temperature and pressure for complete combustion of 10 m^3 of sulphur dioxide volume of oxygen required is_______ .
Answer» `1m^(3)` `5m^(3)` `10M^(3)` `20m^(3)` SOLUTION :`{:(2SO_(2(G)),+O_(2(g)) to, 2SO_(3(g))),(2 "volume ",1 "volume",2 "volume"),(10 m^3,5 m^3,10m^3):}`

Discussion

10.	At idential tempreture and pressure , yhe rate of diffusion of hydrogen gas is 3sqrt3 times that of a hydrocarbon having moleculer formula C_(n)H_(2n-2) What is the value of n?
Answer» 8 4 3 1 Solution :`gamma_(H_(2))/(gammac_(n_(H_(2n+2))))=sqrt(mc_(n)H_(2n+2)/m_(H_(2)))Rightarrow3sqrt3=sqrt(m_(c_(n_(H_(2n+n))))/2)` SQUARING on both sides and rearring `27xx2=m_(c_(n_(H_(2n-n))))` `54=n(12)+(2n-2)(1)` `54=12n+2n-2=14n-2` `n=(54+2)/14=56/14=4` .

Discussion

11.	At high temperature , boron burns in air to give a mixture of B_2 O_3 and BN.
Answer» ANSWER :1

Discussion

12.	At half-neutralisation of a weak acid with a strong base, what is the relationship between pH and dissociation constant (K_(a)) of the weak acid ?
Answer» SOLUTION :at HALF- NEUTRALISATION, `PH = pK_(a)`.

Discussion

13.	At given temperature , these reaction tell about control of reaction which is :
Answer» ANSWER :A

Discussion

14.	Vapour pressure is an important property of liquids.pressure cooker is used for cooking food at higher altitudes. Give reason.
Answer» Solution :SINCE atmospheric PRESSURE is low at high attitudes, the NORMAL boiling point of water will be low. So food cannot be cooked properly. So pressure COOKER is used for cooking

Discussion

15.	At equilibrium, the value of equilibrium constant K is
Answer» `1` `2` `3` `0` Solution :At EQUILIBRIUM `DeltaG=0` `DeltaG=-nRT LN K` `0=- nRT ln K` or `K=1`

Discussion

16.	At equilibrium, the concentrations of N_(2) = 30 xx 10^(-3) M, O_(2)= 42 xx 10^(-3) M and No = 2*8 xx 10^(-3) M in a sealed vessel at 800 K. What will be K_(c) for the raction N_(2) (g) + O_(2) (g) hArr 2 NO (g) ?
Answer» ANSWER :`0*622`

Discussion

17.	At equilibrium , the mass of each of the reactants and products remains constant. Does it mean that the reaction has stopped ? Explain.
Answer» Solution :No, the reaction does not STOP. In CONTINUES totake place in the FORWARD as well as BACKWARD directions but at equal speeds.

Discussion

18.	At equilibrium, the concentration of : N_2=3.0xx10^(-3) M, O_2=4.2xx10^(-3) M and NO=2.8xx10^(-3) M in a sealed vessel at 800 K. What will be K_c for the reaction N_(2(g)) + O_(2(g)) hArr 2NO_((g))
Answer» Solution :`{:("REACTION:" ,N_(2(g)) + , O_(2(g))hArr , 2NO_((g))),("Concentration at equilibrium:" , 3.0xx10^(-3)M, 4.2xx10^(-3)M,2.8xx10^(3)M):}` Equilibrium CONSTANT of reaction is `K_c`, `K_c=[NO]^2/([N_2][O_2])=(2.8xx10^(-3)M)^2/((3.0xx10^(-3)M)(4.2xx10^(-3)M))` = 0.6222

Discussion

19.	At equilibrium state
Answer» `DeltaG^@=0` `DeltaG = "negative"` `DeltaG = "zero"` `DeltaG = "POSITIVE"` SOLUTION :at equilibrium, `DELTA G=0`

Discussion

20.	At equilibrium of the reaction,N_(2)O_(4) (g) hArr2 NO_(2) (g), the obsserved molecular weight ofN_(2)O_(4) is 80 g " mol "^(-1)at 350 K. The percentage of dissociation of N_(2)O_(4) (g) at 350 K is
Answer» 0.1 0.15 0.2 0.18 Solution :For the GIVEN dissociation REACTION , `alpha =(D-d)/d = (M_(t)-M_(o))/(M_(o))=(92 xx80)/80 xx 100 = 15%`

Discussion

21.	At equilibrium, Kp = 1 then the value of AG^(@) will be equal to.....
Answer» Solution :`DELTA G^(@)= -RT L N K=0`

Discussion

22.	At ......... distance the four oxygen atoms surrounded to each oxygen atom in ice.
Answer» `2.76xx10^(-10)` m 2.76 Å 276 PM All of SOLUTION :1M=`10^12` pm = `10^10` Å

Discussion

23.	At definite temperature total pressure is P bar derive equilibrium constant. CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))
Answer» SOLUTION :`CaCO_(3(s)) hArr CaO_((s)) + CO_(2(G))` `{:("Initial",-,-,),("At equilibrium",-,-,p_(CO_2)):}` `K_c=([CaO_((s))][CO_(2(g))])/([CaCO_(3(s))])` `CaCO_(3(s))` and `CaO_((s))`are solid so take its concentration CONSTANT. `therefore K_c=[CO_2]` and `K_p =p_(CO_2)`=P BAR

Discussion

24.	At definite temperature K_c is given by following equation, K_c=([I_2][H_5IO_6]^5)/([IO_3^(-)]^7[H_2O]^9[H^+]^7) Write the equilibrium equation.
Answer» Solution :`7IO_(3(aq))^(-) + 9H_2O_((L)) + 7H_((aq))^(+) HARR I_(2(aq)) + 5H_5IO_(6(aq))`)

Discussion

25.	At definite temperature in open vessel decomposition of Ammonium carbonate take place and total pressure is P bar derive K_p. (NH_4)_2CO_(3(s)) hArr 2NH_(3(g)) + CO_(2(g)) +H_2O_((g))
Answer» Solution :`{:("Reaction:",(NH_4)_2CO_(3(s)) hArr , 2NH_(3(g))+ , CO_(2(g)) + , H_2O_((g))),("Pressure at equilibrium:",,2x,X,x),("If total pressure P, So partial pressure",,(2P)/4,P/4,P/4):}` Total Pressure P=2x +x+ x = 4x `THEREFORE x=P/4` `therefore 2x=p_(NH_3)=(2P)/4` `x=p_(CO_2)=p_(H_2O)=P/4` `K_p=(p_(NH_3))^2(p_(CO)_2) (p_(H_2O))` `=((2P)/4)^2(P/4)(P/4)` `=P^4/64 "bar"^2`

Discussion

26.	At definite temperature if the volume of system decrease then what will be change in concentration of CO ? 2CO_((g)) + O_(2(g)) hArr 2CO_((g))
Answer» Solution :At the VOLUME decrease so the total pressure INCREASES so the reaction MOVES in such direction where moles are less and reaction moves in forward direction. `ubrace(2CO_((g))+O_(2(g)))_("3 mole") hArr UNDERSET"2 mole"(2CO_(2(g))) `

Discussion

27.	At definate temperature. The reaction,SO_(2(g)) + 1/2O_(2(g)) hArr SO_(3(g)) DeltaH=-94.7 kJ is in closed vessel. The equilibrium will be go in which direction by following changes ?(i) Increase temperature (ii) Addition of catalyst (iii) addition of SO_2 gas (iv) To decrease total pressure (v) If volume of vessel will increase (vi) Addition He gas at constant volume (vii) addition He gas at constant pressure.
Answer» Solution :(i) If the temperature will increase than reaction is endothermic therefore equilibrium shift in the reverse direction. (ii) Adding of catalyst, the reaction is not shift in any direction. The catalyst affect the same on speed of forward and backward reaction. (iii) `SO_2` gas is reactant. If the reactant will be added, it decrease and the equilibrium shift in forward direction. (iv) If TOTAL pressure decrease than such reaction take place in which pressure increase and moles of gas increase. In this reaction 1.5 mole gaseous reactant and 1 mole gases product. So on decreasing pressure mole charge from 1 to 1.5. (v) If the volume of vessel increase therefore pressure will be decrease and the reaction is reverse and decrease the product `SO_3` while increase the reactant `SO_2` (vi) If the reaction take place at CONSTANT volume, addition of inert gas like He will not change the molar concentrations of the REACTANTS and products hence, the state of equilibrium will remain unaffected. (VII) Reverse reaction is occurs. At constant temperature if inert gas is added the PARTIAL pressure of gases change. As gaseous reactants are more reaction proceed in reverse direction.

Discussion

28.	At definite temperature 3 atm pressure 75% PCl_5 decompose in PCl_3 and Cl_2. Find K_p.
Answer» SOLUTION :3.851 ATM

Discussion

29.	At definite temp the K_c of the following reaction is 0.18. PCl_(3(g))+Cl_(2(g)) hArr PCl_(5(g)) At a definite temp. in reactionmixture [PCl_3]= 0.042 M , [Cl_2]= 0.024 M and [PCl_5]= 0.005 M . Is this reaction in equilibrium ? In which direction reaction moves ?
Answer» Solution :`Q_c=4.96 gt K_c` THEREFORE not equilibrium. Equilibrium constant decreasing to 0.18 and the REACTION will be in reverse direction.

Discussion

30.	At definite temp if the volume of system decrease than what will be change in concentration of CO ? 2CO_((g)) + O_(2(g)) undersetlarrto 2CO_(2(g))
Answer» Solution :`{:(," "2CO_((g))+O_(2(g)) undersetlarrto,2CO_(2(g))),("total","3 MOLES reactant","2 moles PRODUCT"):}` As the volume of system decreases , pressure INCREASES and by decreasing the moles of gaseous substances forward REACTION will occurs and concentration of CO will decrease.

Discussion

31.	At critical temperature:
Answer» liquid passes into GASEOUS state imperceptibly and continously gas passes into liquid state imperceptibly and continously surface SEPARATING two PHASES DISAPPEARS all of the above Answer :d

Discussion

32.	At critical point, the surface tension is "___________________".
Answer» infinite zero same as that at ROOM temperature twice as that at room temperature Answer :B

Discussion

33.	At constant volume Z_(11) is directly proportional to:
Answer» <P>`SQRT(P)` `P` `T^(2)` T Answer :a

Discussion

34.	At constant volume, for a fixed number of moles of a gas, the pressure of the gas increases with a rise in temperature, due to
Answer» increases in average molecular SPEED decreased NUMBER of collisions amongst molecules increase in molecular ATTRACTIONS DECREASE in mean free path Solution :`P uarr` due to more collisions with walls of container which is due to increased average molecular speed.

Discussion

35.	At constant temperature, the equilibrium constant K_(p) N_(2)O_(4) hArr 2NO_(2) is given by k_(p) = (4x^(2)P)/(1 - x)were ,P = Pressure and X= Extent of reaction How does the value of K_(p)change on following changes (a) 'P' increases (b) X changes (c ) 'P' decreases
Answer» SOLUTION :`K_(p)`is equilibrium constant which does not change on CHANGING the P, X. `K_(p)`DEPENDS on temperature.

Discussion

36.	At constant temperature the vapour pressure of water, acetone and ether are respectively 23.3, 24.6 and 56 atm. state the order of boiling point.
Answer» Solution :Boiling point `ALPHA` 1/vapour pressure `therefore`ETHER, ACETONE, water are in increasing ORDER ( `because` vapour is more so, b.p. is less)

Discussion

37.	At constant temperature, the equilibrium constant (K_(p)) for the decomposition reaction N_(2)O_(4) hArr 2NO_(2) is expressed by K_(p) = 4x^(2) P//(1 - x^(2)) where Pis pressure, x is extent of decomposition. Which of the followingstatement is true ?
Answer» `K_(P)` increases with INCREASE of P `K_(P)` increases with increase of X `K_(P)` increases with decrease of x `K_(P)` REMAINS CONSTANT with change in P or x Solution :`K_(P)` does not change with P, x CHANGES according to Le-chatelier.s principle

Discussion

38.	At constant temperature, in a given mass of an ideal gas
Answer» the ratio of PRESSURE and VOLUME ALWAYS remains constant volume always remains constant pressure always remains constant the product of pressure and volume always remains constant. Solution :the product of pressure and volume always remains constant.

Discussion

39.	At constant temperature & volume , 50% of ozone is decomposed out of 2 atm of ozone taken initially and the equilibrium 2O_(3 (g)) hArr 3O_(2 (g)) is established , K_(p) for the decomposition of ozone is
Answer» `((3)/(2))^(3)` `((3)/(2))^(2)` `((2)/(3))^(3)` `((2)/(3))^(2)` ANSWER :A

Discussion

40.	At constant temperature a gas occupies a volume of 200 mL at a pressure of 0.720 bar. It is subjected to an external pressure of 0.900 bar. What is the resulting volume ?
Answer» <P> SOLUTION :GIVEN `V_(1)=200mL` `P_(1)=0.720mL` `P_(2)=0.900`bar `V_(2)=?` `V_(1)P_(1)=P_(2)V_(2)` `thereforeV_(2)=(P_(1)V_(1))/(P_(2))=(0.720xx200)/(0.900)=160mL`

Discussion

41.	At constant temperature 80% AB dissociates into A_(2) and B_(2) then the equilibrium constant for 2AB_((g)) harr A_(2(g)) +B_(2(g))
Answer» 1 0.25 16 4 Solution : `K=(0.4 XX 0.4)/((0.2)^(2))=4`

Discussion

42.	At constant pressure, the volume of a given mass of a gas ......... by 1/273 of its volume at ........... for each one degree rise in temperature.
Answer» SOLUTION :INCREASES, `0^@C`

Discussion

43.	At constant pressure, the heat of combustion of carbon monoxide at 17^0 C is –284.5 kJ. Calculate its heat of combustion at constant volume
Answer» SOLUTION :` 283.3 KJ`

Discussion

44.	At constant pressure for a given amount of an ideal gas, will the graph obtained by plotting V vs t^(@)C and V vs TK be different ?
Answer» Solution :ACCORDING to Charles' law V=KT. . .[1] [at CONSTANT pressure for a given mass of a gas] But T=273+t, hence, V=K(273+t) . . [2] Both the equation [1] & [2] express equations for straight lines. Equation no [1] represents a straight line passing through the origin. equation no. [2] does not represent a straight line passing through the origin. the plot of V vs t results in a straight line that cuts the V-axis at `0^(@)C`. if this straight line is extrapolated BACKWARD, it meets the t-axis at `-273^(@)C`. LTBRGT From the, above two graphs it is evident that there are no actual defferences between the two graphs, because `-273^(@)C` and 0K express the same temperature.

Discussion

45.	At CMC the surfactant molecule
Answer» decomposes becomes COMPLETELY soluble associates dissociates. Answer :C

Discussion

46.	At chemical equilibrium ,
Answer» RATE of FORWARD reaction = rate of BACKWARD reaction rate of forward reaction `GT` rate of backward reaction rate of forward reaction `lt` rate of backward reaction rate of forward reaction `prop` rate of backward reaction Solution :rate of forward reaction = rate of backward reaction

Discussion

47.	At certain temperature the volume - pressure curves for four gases A , B , C and D are as shown below . The gas that deviates least from ideal nature is
Answer» B D C A Solution :Ideal gas can be HIGHLY compressed at HIGH pressure , because there are no intermolecular FORCES of ATTRACTION orrepulsion between is molecules .

Discussion

48.	At certain temperature the proton concentration of pure water is 2.5 xx10^(-7) M. What is the hydroxyl ion concentration ? What is the magnitude of K_w ?
Answer» Solution :For pure WATER, `[H^+] = [OH^(-)]` Concentration of `[OH^(-) ] = 2.5 XX 10^(-7) M` Ionic product of water `= K_w= [H^+ ][OH^(-)] = (2.5xx10^(-7) M)^2 = 6.25 xx 10^(-14)mol^2 L^(-2)`

Discussion

49.	At certain temperature the K_(w) of D_(2),O is 10^(=-16) M. Then the pD of pure D_(2),O at that temperature is
Answer» <P>7 16 8 6 Solution :` K_w " of "D_2O =10 ^(-16), [D^(+) ] [OD^(-)] = 10 ^(-16) ` ` [D^(+)]= 10^(-8)M RARR P^(D) = 8`

Discussion

50.	At certain temperature, the hydroxyl ion concentration of pure water is 8.4 xx 10^(-8) M. Calculate the ionic product at this temperature.
Answer» SOLUTION :`7.0 XX 10^(-15)`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

At low pressure Vander Waal's equation for 3 moles of a real gas will have its simplified from

At low pressure, the fraction of the surface covered follows

At low pressure of 0.25 atm, 2 mole of a real gas has Boyl's temperature 100K. The approximate volume of gas at this temperature and pressure is:

At K, K_(p) = 2*0 xx 10 ^(10)//"bar" for the given reaction at equilibrium 2 SO_(2) (g) + O_(2) (g) hArr 2 SO_(3) (g) What is K_(c) at this temperature ?

At its melting point ice is lighter than water because

At isoelectric point

At infinite dilution , the percentage ionisation of both strong and weak electrolytes is

At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3sqrt(3) times that of a hydrocarbon having molecular formula C_(n)H_(2n-2). What is the value of n ?

At identical conditions of temperature and pressure for complete combustion of 10 m^3 of sulphur dioxide volume of oxygen required is_______ .

At idential tempreture and pressure , yhe rate of diffusion of hydrogen gas is 3sqrt3 times that of a hydrocarbon having moleculer formula C_(n)H_(2n-2) What is the value of n?

At high temperature , boron burns in air to give a mixture of B_2 O_3 and BN.

At half-neutralisation of a weak acid with a strong base, what is the relationship between pH and dissociation constant (K_(a)) of the weak acid ?

At given temperature , these reaction tell about control of reaction which is :

Vapour pressure is an important property of liquids.pressure cooker is used for cooking food at higher altitudes. Give reason.

At equilibrium, the value of equilibrium constant K is

At equilibrium, the concentrations of N_(2) = 3*0 xx 10^(-3) M, O_(2)= 4*2 xx 10^(-3) M and No = 2*8 xx 10^(-3) M in a sealed vessel at 800 K. What will be K_(c) for the raction N_(2) (g) + O_(2) (g) hArr 2 NO (g) ?

At equilibrium , the mass of each of the reactants and products remains constant. Does it mean that the reaction has stopped ? Explain.

At equilibrium, the concentration of : N_2=3.0xx10^(-3) M, O_2=4.2xx10^(-3) M and NO=2.8xx10^(-3) M in a sealed vessel at 800 K. What will be K_c for the reaction N_(2(g)) + O_(2(g)) hArr 2NO_((g))

At equilibrium state

At equilibrium of the reaction,N_(2)O_(4) (g) hArr2 NO_(2) (g), the obsserved molecular weight ofN_(2)O_(4) is 80 g " mol "^(-1)at 350 K. The percentage of dissociation of N_(2)O_(4) (g) at 350 K is

At equilibrium, Kp = 1 then the value of AG^(@) will be equal to.....

At ......... distance the four oxygen atoms surrounded to each oxygen atom in ice.

At definite temperature total pressure is P bar derive equilibrium constant. CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))

At definite temperature K_c is given by following equation, K_c=([I_2][H_5IO_6]^5)/([IO_3^(-)]^7[H_2O]^9[H^+]^7) Write the equilibrium equation.

At definite temperature in open vessel decomposition of Ammonium carbonate take place and total pressure is P bar derive K_p. (NH_4)_2CO_(3(s)) hArr 2NH_(3(g)) + CO_(2(g)) +H_2O_((g))

At definite temperature if the volume of system decrease then what will be change in concentration of CO ? 2CO_((g)) + O_(2(g)) hArr 2CO_((g))

At definite temperature 3 atm pressure 75% PCl_5 decompose in PCl_3 and Cl_2. Find K_p.

At definite temp the K_c of the following reaction is 0.18. PCl_(3(g))+Cl_(2(g)) hArr PCl_(5(g)) At a definite temp. in reactionmixture [PCl_3]= 0.042 M , [Cl_2]= 0.024 M and [PCl_5]= 0.005 M . Is this reaction in equilibrium ? In which direction reaction moves ?

At definite temp if the volume of system decrease than what will be change in concentration of CO ? 2CO_((g)) + O_(2(g)) undersetlarrto 2CO_(2(g))

At critical temperature:

At critical point, the surface tension is "___________________".

At constant volume Z_(11) is directly proportional to:

At constant volume, for a fixed number of moles of a gas, the pressure of the gas increases with a rise in temperature, due to

At constant temperature, the equilibrium constant K_(p) N_(2)O_(4) hArr 2NO_(2) is given by k_(p) = (4x^(2)P)/(1 - x)were ,P = Pressure and X= Extent of reaction How does the value of K_(p)change on following changes (a) 'P' increases (b) X changes (c ) 'P' decreases

At constant temperature the vapour pressure of water, acetone and ether are respectively 23.3, 24.6 and 56 atm. state the order of boiling point.

At constant temperature, the equilibrium constant (K_(p)) for the decomposition reaction N_(2)O_(4) hArr 2NO_(2) is expressed by K_(p) = 4x^(2) P//(1 - x^(2)) where Pis pressure, x is extent of decomposition. Which of the followingstatement is true ?

At constant temperature, in a given mass of an ideal gas

At constant temperature & volume , 50% of ozone is decomposed out of 2 atm of ozone taken initially and the equilibrium 2O_(3 (g)) hArr 3O_(2 (g)) is established , K_(p) for the decomposition of ozone is

At constant temperature a gas occupies a volume of 200 mL at a pressure of 0.720 bar. It is subjected to an external pressure of 0.900 bar. What is the resulting volume ?

At constant temperature 80% AB dissociates into A_(2) and B_(2) then the equilibrium constant for 2AB_((g)) harr A_(2(g)) +B_(2(g))

At constant pressure, the volume of a given mass of a gas ......... by 1/273 of its volume at ........... for each one degree rise in temperature.

At constant pressure, the heat of combustion of carbon monoxide at 17^0 C is –284.5 kJ. Calculate its heat of combustion at constant volume

At constant pressure for a given amount of an ideal gas, will the graph obtained by plotting V vs t^(@)C and V vs TK be different ?

At CMC the surfactant molecule

At chemical equilibrium ,

At certain temperature the volume - pressure curves for four gases A , B , C and D are as shown below . The gas that deviates least from ideal nature is

At certain temperature the proton concentration of pure water is 2.5 xx10^(-7) M. What is the hydroxyl ion concentration ? What is the magnitude of K_w ?

At certain temperature the K_(w) of D_(2),O is 10^(=-16) M. Then the pD of pure D_(2),O at that temperature is

At certain temperature, the hydroxyl ion concentration of pure water is 8.4 xx 10^(-8) M. Calculate the ionic product at this temperature.

At equilibrium, the concentrations of N_(2) = 30 xx 10^(-3) M, O_(2)= 42 xx 10^(-3) M and No = 2*8 xx 10^(-3) M in a sealed vessel at 800 K. What will be K_(c) for the raction N_(2) (g) + O_(2) (g) hArr 2 NO (g) ?