94420 + Interview Questions in Chemistry in Class 12

1.	[Kr]4d^(10) 5s^(2) 5p^(4) is the electronic configuration of ___________.
Answer» SOLUTION :TELLURIUM

Discussion

2.	Kr and fluorine gases are irradiated with SbF_(5) it forms ……………. .
Answer» SOLUTION :`KrF_(2).2SbF_(3)`

Discussion

3.	K[ptCl_3(C_2H_4)] is Fischer salt.
Answer» SOLUTION :Zeise SALT

Discussion

4.	K_(p)andK_(c) are related by K_(p)=K_(c)(RT)^(Deltan). Under what practical condition/s, K_(p)=K_(c) ?
Answer» RT = 1 T = 12K `Deltan = 0` `Rprop1/T` Answer :C

Discussion

5.	K_p of the reaction, N_2O_4(g)hArr2NO(g) at 773K is 640 mm of Hg. If the equilibrium pressure is 160 mm of Hg, then whatpercent of N_2O_4 will be dissociated? At what pressure will its degree of dissociation be 50%?
Answer» Solution :Let initial NUMBER of MOLES of `N_O_2`=1 mol and the DEGREE of dissociation of `N_2O_4` at equilibrium`=prop` `N_2O_4(g) Number of moles at equilibrium: `1-prop""2prop` Total number of moles at equilibrium `=1-prop-2prop=1+prop` At equilibrium `P_(N_2O_4)=((1-prop)/(1+prop))xx160` and `P_(NO_2)=((2prop)/(1+prop))xx160` `K_p=(p_(NO_2)^2)/(p_(NO_2O_4))=(4prop^2xx160)/(1-prop^2)`or`640=(4prop^2xx160)/(1-prop^2)` `:.prop=0.707` `:.` PERCENT dissociation of `N_2O_4=70.7%` Let, at a pressure of p mm Hg, the degree of dissociation of `N_2O_4` is 0.5 `:p_(N_2O_4)=((1-0.5)/(1+0.5))p=p/3mmHg` and `p_(NO_2)=((2xx0.5)/(1+0.5))p=(2p)/3 mm Hg` `K_p=(p_(NO_2)^2)/(p_(N_2O_4))=((2p)/(3)^2)/(p/3)=(4p)/3`or`(4p)/3=640` `:.` p=480 atm

Discussion

6.	(K_(P))/(K_(C)) for the gaseous reaction – (a) 2 A + 3 B hArr2C (b) 2 AhArr 4B (c) A + B + 2ChArr4D would be respectively -
Answer» `(RT)^(-3) , (RT)^(2), (RT)^(@)` `(RT)^(-3) , (RT)^(-2), (RT)^(-1)` `(RT)^(-3) , (RT)^(2), (RT)` NONE of the above Answer :A

Discussion

7.	K_p/K_c for the reaction: CO(g)+1/2O_2(g) hArr CO_2(g) is :
Answer» 1 RT 1/(sqrtRT) RT^(1/2) ANSWER :C

Discussion

8.	The ratio of K_(p)// K_(c ) for the reaction :CO(g)+(1)/(2)O_(2)(g)hArr CO_(2)(g)is:
Answer» 1 RT `1//sqrt(RT)` `(RT)^((1//2)` ANSWER :C

Discussion

9.	K_(p) for the reaction PCl_(5)(g)ghArrPCl_(3)(g)+Cl_(2)(g) at 250^(@)C is 0.82. Calculate the degree of dissociation at given temperature under a total pressure of 5 atm. What will be the degree of dissociation if the equilibrium pressure is 10 atm, at same temperature.
Answer» `27.5^(@)` `23%` `35.5%` `40%` ANSWER :A

Discussion

10.	K_p for the reaction: N_2O_4(g) iff 2NO_2(g)is 0.66 atm at 320 K. Calculate the degreeof dissociation of N_2O_4 at 320 K and 380 mm. Also calculate the partial pressures of N_2O_4 and NO_2at equilibrium.
Answer» SOLUTION :0.497, 0.332 ATM, 0.168 atm

Discussion

11.	K_pfor the reactionN_2O_4(g) = 2NO_2(g)is 0.66 at 46^@C . Calculate the per cent dissociation of N_2O_4 at 46^@Cand a total pressure of 380 mm. What are the partial pressures of N_2O_4 and NO_2at equilibrium?
Answer» SOLUTION :0.168 ATM, 0.332 atm

Discussion

12.	K_(p) for the reaction H_(2)(g)+(1)/(2)O_(2)(g)hArrH_(2)O(l) is 8.0 "bar"^(-3//2) at T kelvin temperature. If vapour pressure of H_(2)O is 2.0 bar at same temperature then K_(P)^(0) for the reaction 2H_(2)(g)+O_(2)(g)hArr2H_(2)O(g) is
Answer» `8.0` 64 16 256 Solution :`H_(2)(g)+(1)/(2)O_(2)(g)hArrH_(2)O(L)` `k_(p_(i))=8"BAR"^(-3//2)` `H_(2)O(l)hArrH_(2)O(g)` `K_(p_(2))=V.P.` of `H_(2)O=2` bar `H_(2)(g)+(1)/(2)O_(2)(g)hArrH_(2)O(g)` `K_(P)=K_(p_(i))xxk_(p_(2))` `2H_(2)(g)+O_(2)(g)hArr2H_(2)O(g)` `K_(p)=(k_(p_(1))xxK_(p_(2))^(2)` `(8xx2)^(2)"bar"^(-1)=256 "bar"^(1)` `k_(p)^(0)=256`

Discussion

13.	K_(p)for the reaction :CO_(2)(g)+H_(2)(g) hArr CO(g)+H_(2)O(g)is found to be 16 at a given temperature. Originally equal number of moles of H_(2) andCO_(2)(g) were placed in the flask. At equilibrium, the pressure of H_(2) is 1.20 atm. What is the partial presure of CO_(2) and H_(2)O ?
Answer» 1.20 ATM. each 2.40 atm. each 4.80 atm each 9.60 atm each Answer :C

Discussion

14.	K_(p) for the reactionMgCO_(3_((s))) to MgO_((s)) + CO _(2_((g)))" is" 9 xx 10^(-10). CalculateDelta G^(@) for thereactionat 25^(@)C.
Answer» Solution :Given : `K_(p) = 9 xx 10^(-10)` Temperature`= T = 273+ 25 =298 K ` `Delta G^(@) = ? ` `Delta G^(@) =- 2 .303RT log_(10) K_(p)` `=-2.303x 8.314 xx 298 xx log_(10) 9 xx 10^(-10)` `=- 2.303 xx 8. 314 xx 298 xx [bar(10).9542]` `=- 2.303 xx 8.314 xx 298 xx [ - 9.0458]` `= 51683 J mol^(-1)` `=51. 683 kJ mol^(-1)`

Discussion

15.	K_(p)for thereaction2SO_(2(g))+ O_(2(g))to2SO_(3(g))" is" 7.1 xx 10^(24)at25^(@)C . CalculateDelta G^(@)for thereactions. (R= 8. 314KJ^(-1) mol^(-1))
Answer» Solution :`DELTA G^(@)=- 141.8 KJ mol^(-1)`

Discussion

16.	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 be
Answer» `1.1xx10^(-2)` `3.1xx10^(-2)` `5.2xx10^(-2)` `7.4xx10^(-2)` SOLUTION :`K_(p)=K_(C)(RT)^(Deltan)hArrR="Gas CONSTANT"` `K_(c)=(K_(p))/((RT)^(Deltan))=(1.3xx10^(-3))/((0.0821xx700)^(-4))=7.4xx10^(-2)`

Discussion

17.	K_(p) and K'_(p) are the equilibrium constants of the two reactions, given below 1/2N_(2(g))+3/2H_(2(g))hArrNH_(2(g))N_(2(g))+3H_(2(g))hArr2NH_(3(g)) Therefore, K_(p) and K'_(p) are related by
Answer» `K_(p)=K_(p)^(2)` `K_(2)sqrt(K'_(p))` `K_(p)=2K'_(p)` `K_(p)=K'_(p)` Solution :When a balanced equation having equlibrium constant `K_(p)` is MULTIPLIED by a certain VALUE n, the equlibrium constant for the new equation will be equal to `(K_(p))^(n).` In the GIVEN equation, `n=2` So, `K'_(p)=(K_(p))^(2)orK_(p)=sqrt(K'_(p))`

Discussion

18.	K_(p)and K'_(p) are equilibrium constants of the two reactions given below :(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) hArr NH_(3)(g)N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)Therefore K_(p) and K'_(p) are related by :
Answer» `K_(p)=K'_(p)^(2)` `K_(p)=sqrt(K'_(p))` `K_(p)=2K'_(p)` `K_(p)=K'_(p)` Answer :B

Discussion

19.	Kolbe's synthesis of sodium salt of butanoic acid gives:
Answer» n-hexane Isobutane Butane Ethene Answer :A

Discussion

20.	Kolbes - Schmidt reaction is used to prepare
Answer» SALICYLIC acid salicylaldehyde phenyl acetate o-xylene Answer :A

Discussion

21.	Kolbe's electrolysis of potassium succinate gives CO_(2) and_____.
Answer» SOLUTION :ETHYLENE or ETHENE.

Discussion

22.	Kolbe-schmidt reaction is used for
Answer» SALICYLIC acid Salicylaldehyde Phenol Hydrocarbon Answer :A

Discussion

23.	Kolbe'ssynthesisof sodiumsaltof butaneacidgives :
Answer» N- HEXANE iso-BUTANE n- butane PROPANE ANSWER :A

Discussion

24.	Kolbe-Schmidt reaction is used for:
Answer» SALICYLIC acid Salicylaldehyde Phenol Hydrocarbon Answer :A

Discussion

25.	Kolbe 'seletrolyticmethodcan beapplied on :
Answer» `CH_(3)COONa` `{:(CH_(2)COOK),(\|),(CH_(2)COOK)` `{:(HC-COOK),(" "\|\|),(HC-COOK):}` `C_(6)H_(5)COONa` ANSWER :A,B,C

Discussion

26.	Kohlraush's law is applied to calculate
Answer» MOLAR conductance at INFINITE DILUTION of a weak electrolyte degree of dissociation of weak electrolyte solubility of a SPARINGLY soluble salt all of the above Answer :D

Discussion

27.	Kohlrausch's law is used to calculate
Answer» `^^_(m)^(OO)` `LAMBDA^(0)` c area both `lambda^(0) m C_(6)H_(12)O_(6)` Solution :Statement of LAW.

Discussion

28.	Kohlrausch's law is applied to calculate …………….. .
Answer» molar conductance at infinite dilution of a weak ELECTROLYTE degree of DISSOCIATION of weak electrolyte solubility of a SPARINGLY SOLUBLE salt all the above Answer :D

Discussion

29.	Kohlrausch's law helps to determine the degree of dissociation of a weak electrolyte at a given concentration.The molar conductivityLambda_mof .001M acetic acid is 4.95xx10^(_5)Scm^2mol^(-1).Calculate the degree of dissociation(alpha)at this concentration if limiting molar conductivitylamda_m^0for H+is 340xx10^(-5)Scm^2mol^(-1)and CH_3COO^-50.5xx10^(-5)Scm^2mol^(-1)
Answer» SOLUTION :`lamda_m^0CHCOOH=4.95XX10^(-5)Scm^2mol^(-1)` `lamda_m^0CH_3COOH=lamda_m^0CH_3COO^(-)+H^+`` alpha=lamda_m^0/lamda_x10^(-5)m^0=(4.95xx10^(-5))/(390.5xx^(-5))=0.0126`

Discussion

30.	Kohlrausch law states that at
Answer» infinite dilution, each ion MAKES definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte. Solution :At infinite dilution, when dissociation is complele eachion makes a definite contribution to WARDS molar conductance of the electrolye irrespective of the nature of the other ion with which its is associated and that the molar conductance of any electroylte at infinite dilution is given by the SUM of the contributions of two IONS. This is CALLED Kohlrausch law. `Lambda_(m)^(oo) = Lambda_(+)^(oo) + Lambda_(-)^(oo)` `Lambda_(1)^(oo) and Lambda_(-)^(oo)` are molar ionic conductance at infinite dilution,for cations and anions, respectively.

Discussion

31.	Kohlrausch's law helps to determine the degree of dissociation of a weak electrolyte at a given concentration.State Kohlrausch's law.
Answer» Solution :It states that the limiting molar CONDUCTIVITY of an electroylte is the sum of limiting molar ionic CONDUCTIVITIES of ANIONS and CATIONS.

Discussion

32.	Kohlrausch law can be used to find the molar conductivity of a weak electrolyte at infinite dilution.
Answer» ANSWER :1

Discussion

33.	Kodel polyester is formed by trans-esterification of dimethyl terephthalate with 1,4-di (hydroxymehyl) cyclohexane. Write its reaction ?
Answer» SOLUTION :

Discussion

34.	Kohlrausch’s law states that at
Answer» finite DILUTION, each ion MAKES definite contribution to the equivalent CONDUCTANCE of an electrolyte, whatever be the NATURE of the other ion of the electrolyte. infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte DEPENDING on the nature of the other ion of the electrolyte. infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte. infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte. Answer :D

Discussion

35.	KOH can be used as drying agent for
Answer» AMINES Phenols Acids Esters Solution :Amines is CORRECT because they do not reactwith KOH because both amines and KOH are BASES. Restcan REACT.

Discussion

36.	KO_2 (potassium superoxide) is used in oxygen cylinders in space and submarines because it
Answer» Absorbs `CO_(2)` and increases `O_(2)` CONTENT Eliminates moisture Absorbs `CO_(2)` Produce ozone Solution :`KO_(2)` is used in oxygen cylinder because it absorbs `CO_(2)` and increases `O_(2)` content. SUPER OXIDES reacts with water to give `H_(2)O_(2)& O_(2)`

Discussion

37.	KO_(2) (potassium superoxide) is used in oxygen cylinders in space and submarines because it :
Answer» 1. ABSORBS `CO_(2)` and increases `O_(2)` content 2. ELIMINATES moisture 3. absorbs `CO_(2)` 4. produces ozone Solution :`4KO_(2)+2CO_(2)to2K_(2)CO_(3)+3O_(2)`

Discussion

38.	KO_2 + CO_2 to ?(gas)
Answer» `H_(2)` `N_(2)` `O_(2)` `CO` Solution :`2KO_(2) + CO_(2) to K_(2)CO_(3) +3/2O_(2)`

Discussion

39.	KO_2 is used in oxygen cylinders in space and submarines because it
Answer» ABSORBS `CO_2` and INCREASES `O_2` CONTANT ELIMINATES moisture absorbs `CO_2` PRODUCE Ozone Answer :A

Discussion

40.	Known trihalide of N is:
Answer» `NF_3` `NCl_3` `NBr_3` `NI_3` ANSWER :A

Discussion

41.	Knowing the electron gain enthalpy values of O rarr O^(-) and O rarr O^(2-) as -141 and 702 kJ mol^(-1) respectively, how can you account for the formation of a large number of oxides having O^(2-) species and not O^(-) ?
Answer» Solution :Consider the reaction of a divalent metal (M) with oxygen. The formation of `MO_(2)` and MO involves the following steps : `{:(M(g) overset(Delta_(i)H_(1))RARR M^(+)(g) overset(Delta_(i)H_(2))rarr M^(2+)(g) ","O(g) overset(Delta_(eg)H_(1))rarr O^(-)(g)overset(Delta_(eg)H_(2))rarr O^(2-)(g)),(M^(2+)(g)+2O^(-)(g) overset("Lattice energy")rarr M^(2+)(O^(-))_(2)(s)","M^(2+)(g) overset("Lattice energy")rarr M^(2+) O^(2-)(s)):}` If electron gain enthalpy were the only factor involved in the formation of `O^(2-)` ions, we would expect that O will prefer to form `O^(-)` rather than `O^(2-)` ions. ACTUALLY a large number of oxides have `O^(2-)` species and not `O^(-)`. This is due to the following two reasons : (i) `O^(2-)` has the stable noble GAS configuration (ii) Due to higher charge on `O^(2-)` than on `O^(-)` ions, the lattice energy released during the formation of oxides containing `O^(2-)` species in the solid state is MUCH higher than lattice energy released during the formation of oxides having `O^(-)` species. In other words, it is the higher lattice energy of formation of oxides containing `O^(2-)` species which more than compensates the higher electron gain enthalpy `(Delta_(eg)H_(2))` needed during formation of `O^(2-)` from `O^(-)` ions. Thus, formation of oxides containing `O^(2-)` species is energyetically more favourable than oxides containing `O^(-)` species. It is because of this reason that oxygen forms a larger number of oxides having `O^(2-)` species and not `O^(-)`.

Discussion

42.	Knowing the electron gain enthalpy values for O to O^(-) and O to O^(2-)as -141 and 702 kj mol-1 respectively, how can you account for the formation of a large number of oxides having O^(2-)species and not O^(-)?
Answer» Solution :The VALUES of ELECTRON gain ENTHALPY suggest that formation of `O^(2-)`is endothermic. However, `O^(2-)`has most stable CONFIGURATION in `s_2 np_6`. In SOLID state, the lattice energies of the oxides having `O^(2-)`anions are very high which compensates the positive electron gain enthalpy of oxide (`O^(2-)`) formation. Thus, more oxides with `O^(2-)`ion are formed.

Discussion

43.	Knowing the electron gain enthalpy values forO to O^(-)and O to O^(2-) as- 141 and 702 kJ "mol"^(-1)respectively, how can you account for the formation of a large number of oxides having O^(2-)species and not O^(-) ?
Answer» SOLUTION :The second electron enthalpy of oxygen for the formation of `O^(2-)`is positive while first electron gain enthalpy of oxygen for the formation of `O^-`is negative. This means that if electron gain enthalpy is the only factor involved for the formation of divalent ions, we would expect that oxygen would prefer to form `O^-`ions rather than `O^(2-)`ions. ACTUALLY, a large number of oxides have `O^(2-)`species and not `O^-`. This is because (i) Divalent `O^(2-`) has the stable noble gas CONFIGURATION (II) In the solid state, large amount of energy KNOWN as lattice enthalpy is released to form divalent `O^(2-)`ions than monovalent `O^-` ions. It is the greater lattice enthalpy of `O^(2-)`ions which compensates for the high energy required to remove the second electron. This is responsible for greater stability of `O^(2-)`ions as compared to `O^-`ions.

Discussion

44.	Knowing the electron gain enthalpy values for O to O^(-) and O to O^(2-) as -141 and 702 kJ " mol"^(-1) respectively , how can you account for the formation of a large number of oxides having O^(2-) speciesand not O^(-) ?
Answer» Solution :Consider the reaction of a divalent metal ( M) with oxygen . The formation of `M_2O and MO` involves the following steps : `M(g) overset(Delta_i, H_1)to M^(+) (g) overset(Delta_i/ H_2) to M^(2+) (g)` `Deltai H_1 and Delta_i H_2` are first and second ionisation enthalpies of the metal M . `O(g) overset(Delta_(eg) H_1) to O^(-) ( g) overset(Delta_(eg)H_2)to O^(2-)(g)` `Delta_(eg) H_1 and Delta_(eg)H_2` are first and second ELECTRON gain enthalpies `2M^(+)(g) + O^(-)(g) overset("Lattice energy")to M_2O (s)` `M^(2+)(g) + O_(2)^(-)(g) overset("Lattice energy ")to MO(s)` Although `Delta_i H_2` is much more than `Delta_i H_1 and Delta_(eg) H_2` is much higher than `Delta_(eg)H_1`, yet the lattice nenergy of formation of MO(s) due to higher charges is muchmore than that of `M_2O(s)`. In other words, formation MO is energetically more favourable than `M_2O`. It is due to this reason that oxygen forms PREFERABLY oxides having the `O^(2-)` species and not `O^(-)`.

Discussion

45.	Knowing that the chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statement is incorrect?
Answer» The ionic sizes of LN(III) decreases in general with increasing atomic number Ln(III) COMPOUNDS are generally colorless Ln(III) hydroxides are mainly basic in character Because of the large size of the Ln(III) ions, the BONDING in its compounds is perdominantly ionic in character Solution :THEORY based

Discussion

46.	Knowing that the chemicstry of lanthanoids(Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect ?
Answer» The ionic sizes of Ln(III) DECREASE in GENERAL with increasing atomic number. Ln(III) compounds are generally colourless Ln(III)hydroxides are mainly basic in character Because of the large size of the Ln(III) ions, the bonding in its compounds is predominantly ionic in character. Solution :Ln(III)compounds are generally coloured due to PARTLY filled f-orbitals which PERMIT f-f transition.

Discussion

47.	Knowing that the Chemistry of lanthanoids (Ln) is dominated by its+3 oxidation state, which of the following statement is incorrect?
Answer» Because of the LARGE size of the Ln (III) ions the bonding in its compounds is predominantly ionic in character The ionic sizes of LU (III) DECREASE in general with increasing atomic number Lu (III) compounds are generally colourless Lu (III) hydroxide are mainly basic in character Solution :The most common oxidation state of lanthanoid is +3. Lanthanoids in +3 oxidation state USUALLY have unpaired electron in f-subshell and impart characterristic colour in SOLID as well as in solution state due to f-f transition. (Except lanthanum and lutetium).

Discussion

48.	Knocking sound occurs in engine when fuel
Answer» Ignites slowly Ignites rapidly CONTAINS water Is mixed with machine oil Solution :Knocking -Sudden and IRREGULAR burning of the fuel mixture causing JERKS against the piston and give rise to violent sound. This is KNOWN as knocking.

Discussion

49.	Knowing that average kinetic energy of an ideal gas (X) is directly proportinal to absolute temperature, if T_(1)=273K, which statements describe the other curves?
Answer» CURVE A is for heavier GAS but at same temperature Curve B is for the same gas but at `373K Curve A is for the same gas but at `373K Curve B is for lighter gas but a same temperature Answer :A::B::D

Discussion

50.	KNO_(2)+AgF to AgNO_(2)darr+KF
Answer» For PRECIPITATE FORMATION formation reaction For precipitate DISSOLUTION reaction For precipitate EXCHANGE reaction For no reaction Answer :a

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

[Kr]4d^(10) 5s^(2) 5p^(4) is the electronic configuration of ___________.

Kr and fluorine gases are irradiated with SbF_(5) it forms ……………. .

K[ptCl_3(C_2H_4)] is Fischer salt.

K_(p)andK_(c) are related by K_(p)=K_(c)(RT)^(Deltan). Under what practical condition/s, K_(p)=K_(c) ?

K_p of the reaction, N_2O_4(g)hArr2NO(g) at 773K is 640 mm of Hg. If the equilibrium pressure is 160 mm of Hg, then whatpercent of N_2O_4 will be dissociated? At what pressure will its degree of dissociation be 50%?

(K_(P))/(K_(C)) for the gaseous reaction – (a) 2 A + 3 B hArr2C (b) 2 AhArr 4B (c) A + B + 2ChArr4D would be respectively -

K_p/K_c for the reaction: CO(g)+1/2O_2(g) hArr CO_2(g) is :

The ratio of K_(p)// K_(c ) for the reaction :CO(g)+(1)/(2)O_(2)(g)hArr CO_(2)(g)is:

K_(p) for the reaction PCl_(5)(g)ghArrPCl_(3)(g)+Cl_(2)(g) at 250^(@)C is 0.82. Calculate the degree of dissociation at given temperature under a total pressure of 5 atm. What will be the degree of dissociation if the equilibrium pressure is 10 atm, at same temperature.

K_p for the reaction: N_2O_4(g) iff 2NO_2(g)is 0.66 atm at 320 K. Calculate the degreeof dissociation of N_2O_4 at 320 K and 380 mm. Also calculate the partial pressures of N_2O_4 and NO_2at equilibrium.

K_pfor the reactionN_2O_4(g) = 2NO_2(g)is 0.66 at 46^@C . Calculate the per cent dissociation of N_2O_4 at 46^@Cand a total pressure of 380 mm. What are the partial pressures of N_2O_4 and NO_2at equilibrium?

K_(p) for the reaction H_(2)(g)+(1)/(2)O_(2)(g)hArrH_(2)O(l) is 8.0 "bar"^(-3//2) at T kelvin temperature. If vapour pressure of H_(2)O is 2.0 bar at same temperature then K_(P)^(0) for the reaction 2H_(2)(g)+O_(2)(g)hArr2H_(2)O(g) is

K_(p)for the reaction :CO_(2)(g)+H_(2)(g) hArr CO(g)+H_(2)O(g)is found to be 16 at a given temperature. Originally equal number of moles of H_(2) andCO_(2)(g) were placed in the flask. At equilibrium, the pressure of H_(2) is 1.20 atm. What is the partial presure of CO_(2) and H_(2)O ?

K_(p) for the reactionMgCO_(3_((s))) to MgO_((s)) + CO _(2_((g)))" is" 9 xx 10^(-10). CalculateDelta G^(@) for thereactionat 25^(@)C.

K_(p)for thereaction2SO_(2(g))+ O_(2(g))to2SO_(3(g))" is" 7.1 xx 10^(24)at25^(@)C . CalculateDelta G^(@)for thereactions. (R= 8. 314KJ^(-1) mol^(-1))

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 be

K_(p) and K'_(p) are the equilibrium constants of the two reactions, given below 1/2N_(2(g))+3/2H_(2(g))hArrNH_(2(g))N_(2(g))+3H_(2(g))hArr2NH_(3(g)) Therefore, K_(p) and K'_(p) are related by

K_(p)and K'_(p) are equilibrium constants of the two reactions given below :(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) hArr NH_(3)(g)N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)Therefore K_(p) and K'_(p) are related by :

Kolbe's synthesis of sodium salt of butanoic acid gives:

Kolbes - Schmidt reaction is used to prepare

Kolbe's electrolysis of potassium succinate gives CO_(2) and_____.

Kolbe-schmidt reaction is used for

Kolbe'ssynthesisof sodiumsaltof butaneacidgives :

Kolbe-Schmidt reaction is used for:

Kolbe 'seletrolyticmethodcan beapplied on :

Kohlraush's law is applied to calculate

Kohlrausch's law is used to calculate

Kohlrausch's law is applied to calculate …………….. .

Kohlrausch law states that at

Kohlrausch's law helps to determine the degree of dissociation of a weak electrolyte at a given concentration.State Kohlrausch's law.

Kohlrausch law can be used to find the molar conductivity of a weak electrolyte at infinite dilution.

Kodel polyester is formed by trans-esterification of dimethyl terephthalate with 1,4-di (hydroxymehyl) cyclohexane. Write its reaction ?

Kohlrausch’s law states that at

KOH can be used as drying agent for

KO_2 (potassium superoxide) is used in oxygen cylinders in space and submarines because it

KO_(2) (potassium superoxide) is used in oxygen cylinders in space and submarines because it :

KO_2 + CO_2 to ?(gas)

KO_2 is used in oxygen cylinders in space and submarines because it

Known trihalide of N is:

Knowing the electron gain enthalpy values of O rarr O^(-) and O rarr O^(2-) as -141 and 702 kJ mol^(-1) respectively, how can you account for the formation of a large number of oxides having O^(2-) species and not O^(-) ?

Knowing the electron gain enthalpy values for O to O^(-) and O to O^(2-)as -141 and 702 kj mol-1 respectively, how can you account for the formation of a large number of oxides having O^(2-)species and not O^(-)?

Knowing the electron gain enthalpy values forO to O^(-)and O to O^(2-) as- 141 and 702 kJ "mol"^(-1)respectively, how can you account for the formation of a large number of oxides having O^(2-)species and not O^(-) ?

Knowing the electron gain enthalpy values for O to O^(-) and O to O^(2-) as -141 and 702 kJ " mol"^(-1) respectively , how can you account for the formation of a large number of oxides having O^(2-) speciesand not O^(-) ?

Knowing that the chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statement is incorrect?

Knowing that the chemicstry of lanthanoids(Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect ?

Knowing that the Chemistry of lanthanoids (Ln) is dominated by its+3 oxidation state, which of the following statement is incorrect?

Knocking sound occurs in engine when fuel

Knowing that average kinetic energy of an ideal gas (X) is directly proportinal to absolute temperature, if T_(1)=273K, which statements describe the other curves?

KNO_(2)+AgF to AgNO_(2)darr+KF