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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Calculate the half-life period of a first order reaction, if the rate constant of the reaction is 6.93 times 10^(-3)S^(-1). |
| Answer» SOLUTION :`t_(1//2)=(0.693)/(6.93xx10^(-3))=100sec` | |
| 2. |
Calculate the half life of the first order reaction from their rate constants given as: |
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Answer» Solution :`(a)t_(1/2)=200S^-1` RATE CONSTANT,`K=0.693/(t_1/2)=0.693/(200S)=3.46xx10^-3s^-1` (b)`t_(1/2)=2MIN `K=0.693/(2min)=3.46xx10^-1min.` |
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| 3. |
Calculate the half-life of the first order reaction, C_(2)H_(4)O(g)rarrCH_(4)(g)+CO(g). If the initial pressure of C_(2)H_(4)O(g) is 80 mm and the total pressure at the end of 20 min is 120 mm. |
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Answer» 40 min |
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| 4. |
Calculate the half life of a first order reaction whose rate constant is 200 s^(-1) |
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Answer» SOLUTION :Here rate constant `k = 200 s^(-1)` `:.` Half - LIFE of a FIRST ORDER reaction is `t_(1//2)=(0.693)/k = (0.693)/200 = 3.46xx10^(-3)` sec |
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| 5. |
Calculate the half cell potential of a reaction Ag_(2)S + 2e rarr 2Ag + S^(2-) in a solution beffered at pH = 3 and also saturated with 0.1 M H_(2)S. K_(1) and K_(2) for H_(2)S are 10^(-8) and 1.1 xx 10^(-8) and 1.1 xx 10^(-13) respectively. (K_(SP_(Ag_(2)S)) = 2 xx 10^(-49), E_(Ag^(+)//Ag)^(@) = 0.8 V) |
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| 6. |
Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s^(-1)"" (ii) 2"min"^(-1)""(iii) 4"years"^(-1) |
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Answer» Solution :Half-life PERIOD of a first order REACTION is GIVEN by `t_(1//2)=(0.693)/(K)` (i) `t_(1//2)=(0.693)/(200s^(-1))=0.346 xx 10^(-2)s=3.46xx10^(-3)s` (ii) `t_(1//2)=(0.693)/(2"min"^(-1))=0.346` min (iii) `t_(1//2)=(0.693)/(4" year"^(-1))=0.173` year. |
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| 7. |
Calculate the half-life of a first order reaction from their rate constants given below: (i)200 s^(-1) (ii) 2 "min"^(-1) (iii) 4 "years"^(-1) |
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Answer» Solution :For any FIRST order Reaction `r_((1)/(2))=(0.693)/(k)` (i)`t_((1)/(2))(200 s)=(0.693)/(200 s)` Where ,k=`200^(-1) s` `=3.465xx10^(-3)s^(-1)` (ii)`t_((1)/(2)) (2min )=(0.693)/(2 min )=0.3465 min^(-1)` (III)`t_((t)/(2))(4 year)=(0.693)/(4year)=0.17325 year` |
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| 8. |
Calculate the H^(+) ion concentration in a 1.00(M) HCN litre solution (K_(a) = 4 xx 10^(-10)) |
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Answer» `4 xx 10^(-14)` mole/litre |
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| 9. |
Calculate the gram atoms in 2.3 g of sodium. |
| Answer» SOLUTION :No. of GRAM ATOMS `(2.3)/(23)=0.1` | |
| 10. |
Calculate the fuel efficiency in kJ/ gram of C_(2)H_(4) and C_(4)H_(10). The heats of formation of C_(2)H_(4), C_(4) H_(10), CO_(2) and H_(2)O are 52.3, -126.1, -393.5 and -285.8 kJ "mole"^(-1) respectively. |
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| 11. |
Calculate the frequency of the spectral line when an electron from the fifth orbit jumps to the second orbit in a hydrogen atom (R= 109737 cm^(-1)) |
| Answer» SOLUTION :`6.91 XX 10^(14)` | |
| 12. |
Calculate the frequency, energy and wavelength of the radiation corresponding to the spectral line of lowest frequency in Lyman series in the spectra of hydrogen atom. Also, calculate the energy for the corresponding line in the spectra of Li^(2+) (R_(H)=1.09678 xx 10^(7) m^(-1), c= 3 xx 10^(8) m//s, h= 6.625 xx 10^(-24)J.s) |
| Answer» SOLUTION :`2.176 xx 10^(-18)J, 1.958 xx 10^(-17)J` | |
| 13. |
Calculate the freezing point of solution when 1.9 g of MgCl_(2) (M = "95 g mol"^(-1)) was dissolved in 50 g of water, assuming MgCl_(2) undergoes complete ionization (K_(f)" for water = 1.86 K kg mol"^(-1)) |
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Answer» `MgCl_(2) RARR MG^(2+)+2Cl^(-),i=3` `"Molality "=(1.9)/(95)xx(1)/(50)xx1000=0.4` `DeltaT_(f)=3xx1.86xx0.4=2.232` `therefore""T_(f)=273-2.232=270.77K` |
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| 14. |
Calculatethe freezingpoint of theone molaraqueoussolution(density 1.04 gL^(-1)) of KCl (k_(f) for wateer= 1.86 kg mol^(-1) , atomic mass of K = 39 ,Cl = 35.5) |
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Answer» Mass of solution` = 1000xx 1.04 = 1040 g` Mass of KCl= 74.5 g Mass of water= 1040-74.5 = 965.5 g . MOLALITYOF solution`= (1)/(965.5) XX 1000 = 1.0357` m SINCE KCl is strongelectrolyte, i = 2 `Delta T_(f) =i K_(f) xx m` `= 2 xx 1.86 xx 1.0357 = 3.852^(@)` Freezingpointof solution ` = 0 - 3.852 = - 3.852^(@)C` |
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| 15. |
Calculate the freezing point of solution when 2 g of Na-2SO_4 (M = 142 g "mol"^(-1)) was dissolved in 50 g of water, assuming Na_2 SO_4undergoes complete ionisation. |
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Answer» Solution :Molality of the solution = ` = (2//142)/(50) XX 1000m` For complete ionisation of`Na_2SO_4 , i= 3` Apply the RELATION ` Delta T_f = iK_f m` Substituting the values, we have ` Delta T_f = 2 xx 1.86 kg "mol"^(-1) xx (2g)/(142 "mol"^(-1)) xx (1000g kg^(-1))/(50G)` ` Delta T_f = 1.57` ` T_f = -1.57^@C " or " 271.43 K` |
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| 16. |
Calculate the freezing point of an aqueous solution having mole fraction of water 0.8. Latent heat of fusion of ice is 1436.3 cal mol^(-1). |
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| 17. |
Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr_(2)in 200 g of water (molar mass of MgBr_(2) = 184g. K_(f) for water = 1.86 K kg mol^(-1)) |
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Answer» Solution :`Delta T_(f) = 7.5^(@)` C `Delta T_(f) = iK_(f)`m `T_(f)^(@) = 3 XX 1.86^(@) C kg "mol"^(-1)` `xx (10.50 G )/(184 "g mol"^(-1)) xx (1000)/(200 kg)` `0^(@) C - T_(f) = 1.59^(@)` C `T_(f) = - 1.59^(@) `C or 271.41K |
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| 18. |
Calculate the freezing point of an aqueous solution containing 10.50 g ofMgBr_(2) in 200 g of water . (Molar mass of MgBr_(2) = 184 , K_(f) = 1.86 Kkg mol^(-1)) |
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Answer» Solution :`m = (n_(g) xx 1000)/(W_(A) (g))` <BR> `= (W_(B) xx 1000)/(M_(B) xx W_(A)) = (10.50 xx 1000)/(184 xx 200) = 0.2853` M `MgBr_(2)` IONIZES as `MgBr_(2) to Mg^(2+) + 2 Br^(-)` `i= 3` `DeltaT_(F) = ixx K_(f)xx M` `= 3 xx 1.86 xx 0.2855` `=1.59` Freezing point = `0 - 1.59^(@) C = -1.59^(@) C` |
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| 19. |
Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr_(2) in 200 g of water (Molar mass MgBr_(2)="184 g,", K_(f) for water = "1.80 K kg mol"^(-1)) |
| Answer» SOLUTION :(B) `-1.59^(@)C(i=3)` | |
| 20. |
Calculate the freezing point of an aqueous solution containing 10.5 g of Magnesium bromide in 200 g of water, assuming complete dissociation of Magnesium bromide. [Molar mass of Magnesium bromide = 184 g "mol"^(-1), K_f, for water = 1.86 K kg "mol"^(-1) ?] |
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Answer» SOLUTION :Number of moles of `MgBr_2`= 10.5/184 =0.0571 mole Molality of the solution` = (0.571)/(200) xx 1000 = 0.2855 m` vant HOFF FACTOR i = 3`[ because MgBr_2 to Mg^(2+)+ 2Br^(-) ]` Applying the relation : ` Delta T_f = iK_f m` Substituting the values, we have ` Delta T_f = 2 xx 1.86 xx 0.2855 = 1.59 K` Freezing POINT of the solution = 273 |
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| 21. |
Calculate the freezing point of a solution that contains 30 g urea in 200 g water. Urea is a non-volatile, nonelectrolytic solid. K_(f) for H_(2)O=1.86^(@)C//m |
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Answer» `4.65^(@)C` |
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| 22. |
Calculate the freezing point of a solution that contains 30 g urea in 200 g water. Urea is a non-volatile, non-electrolytic solid. K_(f) for water =1.86^(@)C//m. |
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Answer» `4.65^(@)C` `=1.86xx(30)/(60)xx(1000)/(200)=4.65^(@)C` `:.T_(f)=0-4.65= - 4.65^(@)C` |
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| 23. |
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g "mol"^(-1) ) |
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Answer» Solution :APPLY the FORMULA ` DELTA T_f = (K_f xx w_2 xx 1000)/(M_2 xx w_1)` substituting the VALUES in the above equation , we have ` Delta T_f = (1.86 K kg "mol"^(-1) xx 60g xx 1000 g kg^(-1))/(180 g "mol"^(-1) xx 250 g)` `Delta T_f= (1.86 xx 60 xx 1000)/(180 xx 250) = 2.48 K` Freezing point of the solution ` = 273.15 K - 2.48 K = 270.67 K` |
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| 24. |
Calculate the freezing point of a containing 60g of glucose (Molar mass=180 g mol^-1) in 250g of water (K_l of water =1.86 K kg mol^-1) |
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Answer» SOLUTION :`DELTA T_f=k_tm` `=k_f (w_2 times 1000)/(M_2 times w_1)` `=(1.86 times 60 times 1000)/(180 times 250)` `=2.48K` `Delta T_f= T_f^@=T_f` `2.48=273.15-T_f` `T_f=270.67K//270.52K//-2.48^@C` |
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| 25. |
Calculate the freezing point of a solution conataining 54 g glucose (C_(6)H_(12)O_(6) in 250 gof water will freeze K_(f) for waer is 1.86 Km^(-1)) |
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Answer» `DeltaTK_(f)=1.86" K m"^(-1)=1.86" K kg mol"^(-1)`. `DeltaT_(f)=(W_(B)xxK_(f))/(M_(B)xxW_(A))=((0.52g)xx(1.86" K kg mol"^(-1)))/((180" g mol"^(-1))xx(0.08" kg"))=0.067 K` `"Freezing point of solution" = 273 K-0.067 K=272.93 K` |
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| 26. |
Calculate the freezing point depression expected for 0.0711M aqueoussolution of Na_(2) SO_(4). If this solution actually freezes at -0.320^(@) C , what would be the value of van't Hoff factor ? (K_(f) = 1.86^(@) C mol^(-1)) |
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Answer» Solution :`DeltaH_(F) = K_(f) xx M` `DELTA T_(f) = 1.86 xx 0.0711 = 0.132` OBSERVED FREEZING point = `0 - (-0.320) = 0.320^(@) C` `i= ("Observed freezing point")/("Calculate freezing point")` `(0.320)/(0.132) = 2.42` |
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| 27. |
Calculate the freezing pint of a solution contaning 0.5g KCl ( Molar mass = 74.5g//mol) dissolved in 100 g water , assuming KCl to be 92% ionised . K_(f) of water = 1.86Kkg/mol. |
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Answer» Solution : `KClrightarrowK^(+)+Cl^(-)` n=2 `i=1-alpha+nalpha` `i=1+alpha` APPLYING the reaction `DeltaT_(F)=iK_(f)m` `=(1+0.92)xx1.86xx(0.5xx1000)/(7.4.5xx100)` `DeltaT_(f)= 0.24` `DeltaT_(f)= t_(f)^(o)=T_(f)`or `t_(f)= -0.24@C` |
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| 28. |
Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na_(2)SO_(4). If this solution actually freezes at -0.320^(@)C, what would be the value of van't Hoff factor ? (K_(f) for water is "1.86 C mol"^(-1)). |
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Answer» Actual (OBSERVED )`DeltaT_(f)=0.320^(@)C` `i=((DeltaT_(f))_("obs"))/((DeltaT_(f))_("cal"))=(0.320)/(0.132)=2.42.` |
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| 29. |
Calculate the fractional void volume in the c.c.p. and h.c.p. structures of hard spheres. |
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| 30. |
Calculate the following: (i) Velocity of electron in first Bohr orbit of H-atom (r=a_(0)) (ii) De Broglie wavelengt of electron in first Bohr orbit of H- atom. (iii) Orbit angular momentum 2p-orbitals in terms of h/(2pi) unit. |
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Answer» Solution :(i)`"mu"=(nh)/(2pi)` `:.u=(nh)/(2pimr)=(1xx6.626xx10^(-34))/(2xx3.14xx9.108xx10^(-31)xx0.529xx10^(-10))` `=2.19xx10^(6)m//s` (II) `lamda =h/("mu")=(6.626xx10^(-34))/(9.108xx10^(-31)xx2.19xx10^(-6))=3.32xx10^(10)m` (iii) ORBITAL angular MOMENTUM for 2p -orbital `(L=1)` `sqrt(l(l+1))xxh/(2pi)` `sqrt(2 pi)xxh/2=sqrt(2H)` |
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| 31. |
Calculate the first dissociation constant of H_(3)PO_(4) if the e.m.f. of the cell, Hg|Hg_(2)Cl_(2)(s)|KCl("salt")(conc.=4N)||H_(3)PO_(4)(0.1M)|H_(2)(1atm)|Pt,-0.3665V.E_(red)^(@) of SCE=0.2412V.(10^(-212)=131.82) |
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| 32. |
Calculate the final temperature of the gas, if one mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27^@C and the work done Given (C_V=20J//K) |
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Answer» 100 K |
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| 33. |
Calculate the final concentration of all ions in solution after 2 litres of 1.3 M Ba(OH)_2is treated with 3 litres of 2.0 M HCl. |
| Answer» SOLUTION :`0.16 MH^(+) , 0.52 M BA^(2+) , 1.2 M CL^-` | |
| 34. |
Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that K_a=K_b=1.8xx10^(-3). |
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Answer» Solution :`h=sqrt(K_b)=sqrt((K_w)/(K_(a)K_(B)))=sqrt((1XX10^(-14))/(1.8xx10^(-6)xx1.8xx10^-6))` `=sqrt((1)/(1.8)xx10^(-4))=sqrt(0.5555xx10^(-4))` `0.7453xx10^(-2)` `pH=(1)/(2)pK_w+(1)/(2)pK_(a)-(1)/(2)pK_(b)` Given that `K_a=K_b=1.8xx10^(-5)` If `K_a=K_b`, then, `pK_(a)=pK_(b)` `therefore pH=1/2pK_(w)=1/2(14)=7`. |
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| 35. |
Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that K_a=K_b=1.8 times 10^-4 |
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Answer» SOLUTION :`h=sqrtK_a=SQRT((K_w)/(K_aK_b))=sqrt((1 times10^-14)/(1.8 times10^-5 times1.8 times10^-5))=sqrt(1/1.8 times10^-4)=sqrt(0.5555 times10^-4)` `=0.7453 times 10^-2` `pH=1/2 pK_w+1/2 pK_a-1/2 pK_b` Given that `K_a=K_b=1.8 times 10^-5` if `K_a=K_b`, if then `pK_a=pK_b` `therefore pH=1/2 pK_w=1/2(14)=7` |
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| 36. |
Calculate the equivalent weight of the following: (i) KMnO_4 in acidic medium (ii) KMnO_4 in alkuline medium (iii) FeSO_4 (NH_4)_2 SO_4. 6H_(2) O (converting to Fe^(3+) ) (iv) K_2 Cr_2 O_7 in acidic medium (v) H_2 C_2 O_4 (converting to CO_2 ) (vi) Na_(2) S_(2) O_(3). 5H_(2) O ( reacting with I_2 ) |
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Answer» SOLUTION :(i) `underset(+7)overset("1 MOLE")(KMnO_4)to underset(+2)(MN^(2+))"(acidic medium)"` Equivalent weight of `KMnO_4= ("mol. weight of" KMnO_4)/("change in ON per mole")` `= (158)/( 5) = 31.6` (ii) `underset(+7) overset("1 mole") (KMnO_(4)) to underset(+6)( MnO_(4)^(2-)"(alkaline medium)"` Equivalent weight of `KMnO_4 = ( 158)/(1) = 158`. (iii) `Fe^(2+) to Fe^(3+)` Equivalent weight of `FeSO_(4) (NH_(4) )_(2) SO_(4) . 6H_(2) O = ( 392)/( 1) = 392`. (iv) `underset(+12) overset("1 mole")(K_(2) Cr_(2) O_(7)) to underset(+6) (2Cr^(3+)) ` Equivalent weight of `K_2 Cr_2 O_7 = (294.2)/( 6) = 49.03`. (v) `underset(+6) overset("1 mole") (H_(2) C_(2) O_(4)) to underset(+8) (2CO_(2))` Equivalent weight of `H_(2) C_(2) O_(4) = (90)/(2) = 45`. (vi) `Na_(2) S_(2) O_(3). 5H_(2) O + I_(2)to Na_(2) S_(4) O_(6) + 2I^(-)` or `underset(+4)(S_(2) O_(3)^(2-)) tounderset(+5)((1)/(2) S_(4) O_(6)^(2-))` Equivalent weight of `Na_(2) S_(2) O_(3) . 5H_(2) O = ( 248)/( 1) = 248.` |
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| 37. |
Calculate the equivalent weight of KMnO_(4) in (i) Acidic medium (in) Basic medium (iii) Neutral medium |
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Answer» Solution :(i) `"Equivalent weight of" KMnO_(4) "in acidic medium" = ("Molecular weight of" KMnO_(4) )/("No. of moles of ELECTRONS TRANSFERRED")= (158)/(5)= 31.6 ` (ii) `"Equivalent weight of" KMnO_(4) "in basic medium" = ("Molecular weight of" KMnO_(4) )/("No. of moles of electrons transferred")= (158)/(1)= 158` (iii) `"Equivalent weight of" KMnO_(4) "in neutral medium"= ("Molecular weight of" KMnO_(4))/("No. of moles of electrons transferred")= (158)/(3)= 52.67` |
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| 38. |
Calculate the equivalent weight of K_(4) [Fe(CN)_(6)] for the following reaction K_(4) [Fe(CN)_(6)] rarr K^(o+) + Fe^(3+) + CO_(2) + NO_(3)^(-) (M = M. Wt. of K_(4) [Fe( CN)_(6) ]) |
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Answer» `M//1` |
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| 39. |
Calculate the equivalent weight and electrochemicalequivalent of copper deposited from cupric salt (At. Wt of Cu = 63.5). |
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Answer» SOLUTION :DEPOSITION of Cu from cupric salt electrolytically is GIVEN as, `Cu^(2+)+2e^(-) rarr Cu` EQUIVALENT weight of copper `= ("Atomic weight")/("Valency") = (63.5)/(2) = 31.75` ELECTROCHEMICAL equivalent of copper `= ("Equivalent weight")/(96,500)=(31.75)/(96,500)=3.29xx10^(-4)"g coul"^(-1)`. |
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| 40. |
Calculate the equivalent conductivity of 1 M H_(2)SO_(4) solution, if its conductivity is 26xx10^(-2)ohm^(-1)cm^(-1) (Atomic weight of sulphur=32). |
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| 41. |
Calculate the equivalent conductance of NH_4OH at infinite dilution using Kohlrausch law. Given that Lambda_(0) values of NaOH, NaCI and NH_4Cl are respectively 217.4, 108.9 and 129"ohm"^(-1) cm^(2). |
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| 42. |
Calculate the equilibrium constant of the reaction: Cu_((s))+2Ag_((ag))^(+)toCu_((aq))^(2+)+2Ag_((s))""[E_(cell)^(Theta)=0.46V] |
Answer» SOLUTION : ELECTRON CHANGE in this reaction =2=n `E_((cell))^(Theta)=0.46V` `K_(C)=`EQUILIBRIUM constant `E_((cell))^(Theta)=(0.059V)/(n)"log "K_(C)=0.46V` `therefore logK_(C)=(0.46Vxx2)/(0.059)=15.5932` `therefore K_(C)=`Antilog 15.5932 `=3.9194xx10^(15)=3.92xx10^(15)`. |
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| 43. |
Calculate the equilibrium cosntant for the reaction, Zn+Cd^(2+)hArrZn^(2+)+Cd, If E_(Cd^(2+)//Cd)^(@)=-0.403V and E_(Zn^(2+)//Zn)^(@)=-0.763V |
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| 44. |
Calculate the equilibrium constant K_(sp) for the reaction AgCl(s) hArr Ag^(+)(aq)+Cl^(-)(aq) Using the data : DeltaG_(f)^(0)(AgCl)=-109.7 kJ, DeltaG_(f)^(0)(Ag^(+))=77.1 kJ and DeltaG_(f)^(0)(Cl^(-))=-131.2 kJ. |
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Answer» SOLUTION :`DELTAG^(@)` for the reaction is calculated as `DeltaG^(@)=77.1+(-131.2)-(-109.7)=55.6kJ` We have, `DeltaG^(@)=-2.303RTlogK` `logK=(DeltaG^(@))/(2.303RT)=(55.6xx10^(3))/(2.303xx8.314xx298)=-9.75` `:.K=K_(SP)=1.8xx10^(-10)` |
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| 45. |
Calculate the equilibrium constant K for the reaction at 298 K Zn(s) + Cu^(2+)(aq) |
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Answer» Solution :`DeltaG^(@) =-nFE^(@)` `n=2, F = 96487 C MOL^(-1), E_("cell")^(@) = 1.1 V` Therefore, `DeltaG^(@) =-2 xx 96487 xx 1.1 =-21227 J mol^(-1)` Equilibrium constant K can be OBTAINED from the relation `DeltaG^(@) =-RT ln K` or `-21227 =-8.314 xx 298 xx ln K` or ln `K = 21227/(8.314 xx 298) = 8.5677` or `K = 3.696 xx 10^(8)` |
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| 46. |
Calculate the equilibrium constant for the reaction, Zn+Cu^(2+)hArrCu+Zn^(2+) Given: E^(@) for Zn^(2+)//Zn=-0.763V and for Cu^(2+)//Cu=+0.34V,R=8.314JK^(-1)mol^(-1),F=96500" C "mol^(-1) |
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| 47. |
Calculate the equilibrium constant for the reaction given E_("cell")^(@) = +0.46 V Cu(s) + 2Ag^(+) (aq) |
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Answer» Solution :We know `K = "Antilog" [NE^(@)]/[0.059]` =Antilog `[(2 xx 0.46)/(0.059)] = 3.92 xx 10^(15)` |
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| 48. |
Calculate the equilibrium constant for the reaction Fe^(3+) + 3I iff 2Fe^(2+) + I_3^(-). The standard reduction potentials in acidic-medium conditions are 0.77 and 0.54 V respectively for Fe^(3+)//Fe^(2+) and I_3^(-) //I^-couples. |
| Answer» SOLUTION :`6.07 XX 10^7` | |
| 49. |
Calculate the equilibrium constant for the reaction Cu(s)+2Ag^(+)(aq)hArrCu^(2+)(aq)+2Ag(s) Given that E_(Ag^(+)//Ag)^(@)=0.80V and E_(Cu^(2+)//Cu)^(@)=0.34V |
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Answer» Solution :The CELL MAY be represented as: `Cu|Cu^(2+)(aq)||AG^(+)(aq)|Ag` `E_(cell)^(@)=E_(RHS)^(@)-E_(LHS)^(@)=0.80V-(0.34V)=046V` `E_(cell)^(@)=(0.0591)/(n)" LOG "K_(c)` For the given reaction, `n=2,E_(cell)^(@)=0.46V` `therefore0.46=(0.0591)/(2)logK_(c)" or "logK_(c)=(0.46xx2)/(0.0591)=15.5668""thereforeK_(c)"Antilog "15.5668=3.6xx10^(15)`. |
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