52671 + Interview Questions in Chemistry in Class 11

1.	H_2O_2 is obtained by adding dil H_2SO_4 to
Answer» `PbO_2` `MnO_2` `BaO_(2) 8H_2O` `BaCO_3` ANSWER :C

Discussion

2.	H_(2)O_(2) is an unstable liquid . On standing or on heating it decomposes to H_(2)O and O_(2) . H_(2)O_(2) can acts as oxidising agent and reducing agent . The concentration of H_(2)O_(2) is expressed differently with volume strength and the concentration of H_(2)O_(2) at a particular time is measured by titrating it with acidified KMnO_(4) or by titrating liberated I_(2) from acidified KI and H_(2)O_(2)with hypo solution . A sample of H_(2)O_(2)has 3.4 g of H_(2)O_(2)in 100 mL solution . The bottle containing this sample was kept at 25^(0)C for 15 days then 20mL of this sample is treated with excess KI and the liberated iodine requires 50 mL , 0.2 M Na_(2)S_(2)O_(3) solution . Assume the volume of solution remains unchanged . The volume strength of H_(2)O_(2)in the begining and after 15 days are
Answer» `5.6 , 3.4` `11.2 , 2.8` `5.6 , 4.6` `11.2 , 5.6` ANSWER :B

Discussion

3.	H_(2)O_(2) acts both as oxidant and reductant. H_(2)O and O_(2) are products when H_(2)O_(2) acts as oxidant and reductant respectively. The strength of H_(2)O_(2) is expressed in terms of molarity, normality, % strength and volume strength. H_(2)O_(2) decomposes as H_(2)O_(2)rarrH_(2)+1//2O_(2)(g) i.e., one mole O_(2) is released from 2 mole H_(2)O_(2) .x. .volume. strength of H_(2)O_(2) means 1 volume (mL or litre) of H_(2)O_(2) sample released x volume (mL or litre) O_(2) gas at NTP on its decomposition. Hence molarity = x//11.2 moles per litre, i.e., normality of H_(2)O_(2)=x//5.6 Thus volume strength, i.e., x=5.6xx Normality. Weigth of H_(2)O_(2) (in gm) present in 100 mL H_(2)O_(2) solution is called percentage strength of H_(2)O_(2) How much volume of H_(2)O_(2) solution of 22.4 .vol. strength is required to oxidise 6.3 gm oxalic acid
Answer» 10mL 11.2mL 25 ML 30 mL Solution :eq. `H_(2)O_(2)` = eq. `H_(2)C_(2)O_(4)` `(22.4)/(5.6)xx(V)/(1000)=(6.3)/(126)xx2impliesV_(mL)=25mL`

Discussion

4.	H_2O_2 is always stored in plastic bottles ? Why ?
Answer» Solution :The aqueous solution of hydrogen PEROXIDE is enontanamelu dienogen peroxide is spontaneously disproportionate to give OXYGEN. The reaction is slow but it is explosive when it is catalyzed by METAL or alkali DISSOLVED from glass. For this reason, its solution are stored in plastic bottles. `H_(2)O_(2(aq))toH_(2)O_((l))+1//2O_(2(G))`

Discussion

5.	H_(2)O_(2) is a powerful oxidisingagent. It is an electron acceptor in acidic as well as in alkaline medium. It can also act as reducing agent towards oxidising agents. In alkaline medium, the reducing nature of H_(2)O_(2) is even more effective Decolourisation of acidified KMnO_(4) occurs when H_(2)O_(2) is added to it. This is due to
Answer» Oxisation of `KMnO_(4)` REDUCING of `KMnO_(4)` Both oxidation and reduction of `KMnO_(4)`0 none of these Solution :(b) In the ACIDIC medium `KMnO_(4)` (violet ) is REDUCED to `K_(2)SO_(4) and MnSO_(4)` which are both colourless.

Discussion

6.	H_(2)O_(2) is a powerful oxidisingagent. It is an electron acceptor in acidic as well as in alkaline medium. It can also act as reducing agent towards oxidising agents. In alkaline medium, the reducing nature of H_(2)O_(2) is even more effective The bleaching properties of H_(2)O_(2) are due to its:
Answer» Unstable NATURE Acidic nature Reducing nature OXIDISING nature. Solution :(d) Bleaching ACTION of `H_(2)O_(2)` is due to its oxidising nature.

Discussion

7.	H_(2)O_(2) is a powerful oxidisingagent. It is an electron acceptor in acidic as well as in alkaline medium. It can also act as reducing agent towards oxidising agents. In alkaline medium, the reducing nature of H_(2)O_(2) is even more effective In which of the following reactions H_(2)O_(2) acts as a reducing agent?
Answer» `PbO_(2)+H_(2)O_(2) to PBO+H_(2)O+O_(2)` `Na_(2)SO_(3)+H_(2)O_(2) to Na_(2)SO_(4)+H_(2)O` `2KI+H_(2)O_(2) to 2KOH+I_(2)` `KNO_(2)+H_(2)O_(2) to KNO_(3)+H_(2)O` SOLUTION :(a) `H_(2)O_(2)` REDUCES `PbO_(2) "to" PbO`.

Discussion

8.	H_(2)O_(2) is a powerful oxidisingagent. It is an electron acceptor in acidic as well as in alkaline medium. It can also act as reducing agent towards oxidising agents. In alkaline medium, the reducing nature of H_(2)O_(2) is even more effective In which of the following reactions, H_(2)O_(2) acts as an oxidising agent?
Answer» `IO_(4)^(-)+H_(2)O_(2) to IO_(3)^(-)+H_(2)O+O_(2)` `2I^(-)+H_(2)O_(2)+2H^(+) to I_(2)+2H_(2)O` `Ag_(2)O+H_(2)O_(2) to 2Ag+H_(2)O+O_(2)` `2MnO_(4)^(-)+6H^(-)+5H_(2)O_(2) to 2Mn^(2+)+8H_(2)O+5O_(2)` Solution :(b) `H_(2)O_(2)` oxidises `I^(-)` ions to iodine.

Discussion

9.	H_(2)O_(2) is a disbasic acid. The salts formed by it are
Answer» OXIDES and peroxides Peroxides and SUPER oxides Peroxides and sesqioxides Peroxides and Hydroperoxides Answer :D

Discussion

10.	H_2O_2 is a ____________ acid.
Answer» MONO basic di basic tri basic none Answer :B

Discussion

11.	H_(2)O_(2) is a
Answer» MONOBASIC acid dibasic acid neutral compound weak alkali Answer :B

Discussion

12.	H_(2)O_(2) is ''5.6 volume then
Answer» It is 1.7% WEIGHT by volume It is 1 N It is 1M It is 5.6M Solution :`5.6 VOL to 1N to 0.5 M to 1.7 (W/V)%`

Discussion

13.	H_(2)O_(2) is "5.6 volume" then
Answer» It is 1.7%WEIGHT by vollume It is 1 N It is 1 M It is 5.6 M Answer :A::B

Discussion

14.	H_(2)O_(2) is ''5.6 volume '' then
Answer» It is 1.7% WEIGHT by volume It is 1N It is 1 M It is `5.6M` SOLUTION :`5.6 Vol to 1N to 0.5 M to 1.7 ((W)/(V))%`

Discussion

15.	H_2O_2 + H_2O to H_2O^(+) toHO_2^-This reaction indicates
Answer» `H_2O_2` is more ACIDIC than `H_2O` `H_2O` is more acidic than `H_2O_2` both `H_2O` and `H_2O_2` are acidic `H_2O_2` is a BLEACHING agent Answer :A

Discussion

16.	H_2O_2 forms prismatic crystal at
Answer» `-0.4^@C` `-2^@C` `4^@C` `11.6^@C` ANSWER :B

Discussion

17.	H_(2)O_(2) exists as ……….. In alkaline medium.
Answer» `HO_(2)^(-)` `HO_(2)^(OPLUS)` `O_(2)^(2-)` Both 1 &3 Answer :D

Discussion

18.	H_(2)O_(2) does not act as
Answer» Reducing agent OXIDISING agent Dehydrating agent BLEACHING agent ANSWER :C

Discussion

19.	H_2O_2 decomposes slowly on exposure to light.
Answer» ANSWER :1

Discussion

20.	H_(2)O_(2) converts potassium ferrocyanide to ferricyanide. The change observed in the oxidation state of iron is
Answer» `FE^(2+)RARRFE^(2+)` `FerarrFe^(2+)` `Fe^(3+)rarrFe^(2+)` `Fe^(2+)rarrFe^(+)` ANSWER :A

Discussion

21.	H_(2)O_(2) Changes black lead sulphide to white
Answer» PB `PbO_(2)` PBO `PbSO_(4)` ANSWER :D

Discussion

22.	H_(2)O_(2) cannot oxidise:
Answer» `Na_(2)SO_(4)` KI PbS `O_(3)` Answer :C

Discussion

23.	H_(2)O_(2) cannot oxide
Answer» `Na_(2)SO_(3)` `KI` PbS `O_(3)` SOLUTION :`H_(2)O_(2)` OXIDISES `Na_(2)SO_(4)`, KI to `I_(2),` PbSto `PbSO_(4)` but reduces `O_(3)` to `O_(2)`

Discussion

24.	H_(2)O_(2) cannot act as
Answer» OXIDISING AGENT REDUCING agent Antichlor DEHYDRATING agent Answer :D

Discussion

25.	In auto oxidation preparation of H_(2)O_(2) the oxidizing agent is
Answer» 2-ethyl anthraquinone 2-ethyl anthraquinol p-benzoquinone N-methyl aniline Answer :B

Discussion

26.	H_(2)O_(2) acts both as oxidant and reductant. H_(2)O and O_(2) are products when H_(2)O_(2) acts as oxidant and reductant respectively. The strength of H_(2)O_(2) is expressed in terms of molarity, normality, % strength and volume strength. H_(2)O_(2) decomposes as H_(2)O_(2)rarrH_(2)+1//2O_(2)(g) i.e., one mole O_(2) is released from 2 mole H_(2)O_(2) .x. .volume. strength of H_(2)O_(2) means 1 volume (mL or litre) of H_(2)O_(2) sample released x volume (mL or litre) O_(2) gas at NTP on its decomposition. Hence molarity = x//11.2 moles per litre, i.e., normality of H_(2)O_(2)=x//5.6 Thus volume strength, i.e., x=5.6xx Normality. Weigth of H_(2)O_(2) (in gm) present in 100 mL H_(2)O_(2) solution is called percentage strength of H_(2)O_(2) 50 mL of H_(2)O_(2) solution was diluted to 200 mL and 10 mL of this diluted H_(2)O_(2) solution reduced 10 mL of 0.1 M KMnO_(4) acidic solution. The volume strength of H_(2)O_(2) is
Answer» 2.8 Vol 5.6 Vol 11.2 Vol 22.4 Vol Solution :`M_(1)V_(1)=M_(2)V_(2)` `M_(2)=(50xxM_(1))/(200)=(M_(1))/(4)` EQ. `H_(2)O_(2)` = eq. `KMnO_(4)` `(M_(1))/(4)xx10xx2=10xx0.1xx5implies10xx0.1xx5` `M_(1)=1impliesV.S.=1xx11.2=11.2" Vol".`

Discussion

27.	H_(2)O_(2) can be estimated by
Answer» Tiration with acidified `K_(2)Cr_(2)O_(7)` TITRATION with NaOH solution Titration with acidified `KMnO_(4)` solution Titration with `Na_(2)CO_(3)` solution Solution :`H_(2)O_(2)` estmated by titration with acidifed `KMnO_(4)` solution.

Discussion

28.	H_(2)O_(2) acts as strong oxidising agent in
Answer» ACIDIC MEDIUM In the PRESENCE of Glycerol Alkaline medium Neutral medium Answer :A

Discussion

29.	H_(2)O_(2) acts as on oxidizing agent as well as reducing agent Why ?
Answer» Solution :In `H_(2)O_(2),` oxygen has - 1 oxidation state which lies between maximum `(0"or"+2"in"OF_(2))` and minimum -2 Therefore, oxygen can be oxidized to `O_(2)` (zero oxidation state) acting as REDUCING agent or can be reduced to `H_(2)O"or"OH^(-)` (-2 oxidation state) acting as an oxidizing agent. `UNDERSET("Reducing agent")(O_(2)^(-1))toO_(2)^(0)+2e,orunderset("OXIDISING agent")(O_(2)^(-1))+2e^(-)to2O^(2-)`

Discussion

30.	H_(2)O_(2) acts as a bleaching agent because of
Answer» reducing nature of `H_(2)O_(2)` OXIDISING nature of `H_(2)O_(2)` ACIDIC nature of `H_(2)O_(2)` basic nature of `H_(2)O_(2)`. Solution :Due to oxidising nature of `H_(2)O_(2)`, it acts as a BLEACHING agent.

Discussion

31.	H_2O_2 act as a antiseptic then it is known as ....... in market.
Answer» SOLUTION :PERHYDROL

Discussion

32.	H_2O_2 acts a ____________ agent.
Answer» oxidizing REDUCING both reducing and oxidizing neither reducing nor oxidizing Answer :C

Discussion

33.	H_2O_2 acts as
Answer» OXIDISING agent reducing agent bleaching agent all of these Answer :D

Discussion

34.	H_2O_2+ 2KI overset("40% yield")to I_2 + 2KOHH_2O_2 + 2KMnO_4 + 3H_2SO_4 overset("50% yield ")to K_2SO_4 + 2MnSO_4 + 3O_2 + 4H_2O 150 ml of H_2O_2sample was divided into two parts. First part was treated with KI and Formed KOH required 200 ml. of M//2 H_2SO_4 for neutralisation.Other part was trated with KMnO_4 yielding 6.74 litre of O_2 at STP.Using % yieldindicated find volume stregth of H_2O_2 sample used.
Answer» 5.04 1.08 3.36 11.33 Solution :No. of GEW .s of `H_2SO_4` = No.of GEW.s of KOH = No. of GEW.s `H_2O_2` = 0.2 No. of GEW.s `H_2O_2` = No. of GEW.s of `O_2 = 6.74/22.4 XX 2 = 0.6` in part I 0.4 X = 0.2 X = 0.5 in part II 0.5 X = 0.6 X =1.2 (where X is No. of GEW.s `H_2O_2`) TOTAL No. of GEW.s of `H_2O_2` `=1.7 N = 1.7 xx 100/15` volume stregth = 11.33

Discussion

35.	H_(2)O is liquid whereas H_(2)S is gas.
Answer» Solution :In `H_(2)O`, hydrogen is combines with highly electrnegative ELEMENT and also forms INTER molecular hydrogen bond. But S is less ELECTRONEGATIVE and in `H_(2)S` there is no hydrogen bonding.

Discussion

36.	H_(2)O is polar , whereas BeF_(2) is not It is because
Answer» The ELECTRONEGATIVITY of F is GREATER than that of O . `H_(2)O` INVOLVES hydrogen bonding whereas `BeF_(2)` is a discrete molecule `H_(2)O` is linear and `BeF_(2)` is angular . `H_(2)O` is angular and `BeF_(2)` is linear ANSWER :c

Discussion

37.	H_(2)O has a higher boiling point than that of H_(2)O due to
Answer» `H_(2)S` molecular weight is more than `H_(2)O` Dispersion forces between WATER MOLECULES are higher than `H_(2)S` Strong ionic bonds between water molecules Hydrogen bonds between water molecules Answer :B

Discussion

38.	H_2O + H_2O iff H_3O^(+) + OH^- ,In this reaction water acts asI) Bronsted Acid II) Bronsted Base III) Amphoteric oxide 1)
Answer» I only I, II only II, III only I, II, III ANSWER :D

Discussion

39.	H_2O and Cl_2O have different bond angles due to
Answer» Number of lone PAIRS on cental atom in `H_2O and Cl_2O` are DIFFERENT HYBRIDISATION is different REPULSIONS are more among BULKY chlorine atoms in `Cl_2O` DEN betweencentral atom and bonded atoms is different . Answer :C

Discussion

40.	H_2O and H_2O_2 resemble in
Answer» BOND ANGLE HYBRIDIZATION of oxygen structure Oxidation STATE of oxygen Answer :B

Discussion

41.	H_(2)NCONH_(2) overset(Delta)rarr?
Answer» `NH_(3) + CO_(2)` Urethane BIURET Urotropin Solution :`2H_(2) N - underset("UREA")overset(O)overset(\|\|)(C) - NH_(2) underset(-NH_(3))overset(Delta)rarr H_(2) N - overset(O)overset(\|\|)(C) - underset("Biuret")(NH ) - overset(O)overset(\|\|)(C) - NH_(2)`

Discussion

42.	H_(2(g)) rarr 2H_((g)) , DeltaH = 400 KJ , then DeltaS_("system") at 2000 K is
Answer» `0.2 KJ//mol//K` ` -0.2 KJ//mol//K` `200 KJ//g//K` `-400 KJ//mol//K` ANSWER :A

Discussion

43.	H_(2(g)) +I_(2(g)) hArr 2HI_((g)) in this equilibrium process what is the relation between K_c and K'_c.
Answer» SOLUTION :Both are INVERSE to each other `K_c=1/(K._c)`

Discussion

44.	H_(2(g)) + I_(2(g)) hArr 2HI_((g)) for this reaction K_c=[HI]^2/([H_2][I_2])=9 what is the equilibriumconstant for reverse reaction ?
Answer» Solution :Reverse reaction `2HI_((G)) HARR H_(2(g)) + I_(2(g))` `K._c=([H_2][I_2])/[HI]^2=1/K_c=1/9`

Discussion

45.	H_(2(g)) +I_(2(g)) hArr 2HI_((g)) and 2HI_((g)) hArr H_(2(g)) +I_(2(g))what is indicates ?
Answer» Solution :It INDICATE EQUILIBRIUM can be established by FORWARD or REVERSE REACTION.

Discussion

46.	H_(2(g)) +I_(2(g)) hArr 2HI_((g)) and 2HI_((g)) hArr H_(2(g)) +I_(2(g))If the volume of vessel is same them what can be predicted for equilibrium mixture ?
Answer» SOLUTION :All the components will be in same PROPORTION as TEMPERATURE, pressure, VOLUME is same.

Discussion

47.	H_(2(g)) + I_(2(g)) hArr 2HI in this equilibrium if H_2 is added than state the direction of reaction and explain it with help of Q_c.
Answer» SOLUTION :`Q_c=[HI]^2/([H_2][I_2])` and `K_c=[HI]^2/([H_2][I_2])` As `H_2` is added it react with `I_2` So in EQUATION of `Q_c, H_2` and `I_2` increase so `Q_c` decreases. So the value of `Q_c lt K_c` Hence, reaction PROCEED in FORWARD direction and new equilibrium is attain.

Discussion

48.	H_2(g) + Cl_2(g) to 2HCl(g)……."rate" = r_1D_2(g) + Cl_2(g) + 2DCl(g) ….."rate" = r_2.The value of r_1 gt r_2 . Then the correct statement among the following is
Answer» H - H and D - D bond length are same H - H has greater bond energy than D-D The ACTIVATION energy for the first reaction is LESS than the of the SECOND reaction The activation energy for the first reaction is greater than the of the second reaction Answer :C

Discussion

49.	H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(l) BE (H-H) = x_(1), BE (O=O)=x_(2) BE(O=H)=x_(3) Latent heat of vaporisation of water liquid into water vapour =x_(4), then Delta_(f)H (heat of formation of liquid water) is
Answer» `x_(1) +(x_(2))/(2) -x_(3) +x_(4)` `2x_(3)-x_(1) -(x_(2))/(2)-x_(4)` `x_(1)+(x_(2))/(2)-2x_(3)-x_(4)` `x_(1)+(x_(2))/(2)-2x_(3)+x_(4)` Solution :`DeltaH = (BE)_("REACTANT") -(BE)_("products")` [But all the species must be in gaseous state. In product, `[H_(2)O(L) rarr H_(2)O(g)] DeltaH` must be added. Hence, `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(l)` `DeltaH = [(BE)_(H-H)+(1)/(2)(BE)_(O=O)]` `= [(DeltaH)_(vap) +2(BE)_(O-H)]` `= x_(1) +(x_(2))/(2) -[x_(4)+2x_(3)]` `= x_(1) +(x_(2))/(2) -x_(4) - 2x_(3)`

Discussion

50.	H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(g) Delta H = -242 kJ mol^(-1). Bond energy of H_(2) and O_(2) are 436 mol^(-1) and 500 mol^(-1) respectively. What is the bond energy of the O-H bond ?
Answer» Solution :`H_(2)(g) + (1)/(2)O_(2)(g) RARR H_(2)O(g) , Delta_(F)H^(@) = -242 kJ mol^(-1)` H = Bond energy of reaction - Bond energy of products `Delta H = B_(H-H) + (1)/(2)B_(O-O) - 2B_(O-H) , -242 kJ = 436 + (1)/(2)(500) - 2B_(O-H) hArr -928 kJ = -2B_(O-H), B_(O-H) = (928)/(2) = 464 kJ mol^(-1)`.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

H_2O_2 is obtained by adding dil H_2SO_4 to

H_2O_2 is always stored in plastic bottles ? Why ?

H_(2)O_(2) is a disbasic acid. The salts formed by it are

H_2O_2 is a ____________ acid.

H_(2)O_(2) is a

H_(2)O_(2) is ''5.6 volume then

H_(2)O_(2) is "5.6 volume" then

H_(2)O_(2) is ''5.6 volume '' then

H_2O_2 + H_2O to H_2O^(+) toHO_2^-This reaction indicates

H_2O_2 forms prismatic crystal at

H_(2)O_(2) exists as ……….. In alkaline medium.

H_(2)O_(2) does not act as

H_2O_2 decomposes slowly on exposure to light.

H_(2)O_(2) converts potassium ferrocyanide to ferricyanide. The change observed in the oxidation state of iron is

H_(2)O_(2) Changes black lead sulphide to white

H_(2)O_(2) cannot oxidise:

H_(2)O_(2) cannot oxide

H_(2)O_(2) cannot act as

In auto oxidation preparation of H_(2)O_(2) the oxidizing agent is

H_(2)O_(2) can be estimated by

H_(2)O_(2) acts as strong oxidising agent in

H_(2)O_(2) acts as on oxidizing agent as well as reducing agent Why ?

H_(2)O_(2) acts as a bleaching agent because of

H_2O_2 act as a antiseptic then it is known as ....... in market.

H_2O_2 acts a ____________ agent.

H_2O_2 acts as

H_(2)O is liquid whereas H_(2)S is gas.

H_(2)O is polar , whereas BeF_(2) is not It is because

H_(2)O has a higher boiling point than that of H_(2)O due to

H_2O + H_2O iff H_3O^(+) + OH^- ,In this reaction water acts asI) Bronsted Acid II) Bronsted Base III) Amphoteric oxide 1)

H_2O and Cl_2O have different bond angles due to

H_2O and H_2O_2 resemble in

H_(2)NCONH_(2) overset(Delta)rarr?

H_(2(g)) rarr 2H_((g)) , DeltaH = 400 KJ , then DeltaS_("system") at 2000 K is

H_(2(g)) +I_(2(g)) hArr 2HI_((g)) in this equilibrium process what is the relation between K_c and K'_c.

H_(2(g)) + I_(2(g)) hArr 2HI_((g)) for this reaction K_c=[HI]^2/([H_2][I_2])=9 what is the equilibriumconstant for reverse reaction ?

H_(2(g)) +I_(2(g)) hArr 2HI_((g)) and 2HI_((g)) hArr H_(2(g)) +I_(2(g))what is indicates ?

H_(2(g)) +I_(2(g)) hArr 2HI_((g)) and 2HI_((g)) hArr H_(2(g)) +I_(2(g))If the volume of vessel is same them what can be predicted for equilibrium mixture ?

H_(2(g)) + I_(2(g)) hArr 2HI in this equilibrium if H_2 is added than state the direction of reaction and explain it with help of Q_c.

H_2(g) + Cl_2(g) to 2HCl(g)……."rate" = r_1D_2(g) + Cl_2(g) + 2DCl(g) ….."rate" = r_2.The value of r_1 gt r_2 . Then the correct statement among the following is

H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(l) BE (H-H) = x_(1), BE (O=O)=x_(2) BE(O=H)=x_(3) Latent heat of vaporisation of water liquid into water vapour =x_(4), then Delta_(f)H (heat of formation of liquid water) is

H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(g) Delta H = -242 kJ mol^(-1). Bond energy of H_(2) and O_(2) are 436 mol^(-1) and 500 mol^(-1) respectively. What is the bond energy of the O-H bond ?