4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 4 InterviewSolution

151.	The pressue of a fluid is a linear function of volume (P=a+bV)and the internal energy of the fluide is `U=34+ 3PV`(S.I.units). Find a,b,w, `DeltaE` and q for change is state from `( 100 Pa, 3m^(3))` to `(400 Pa, 6m^(3))`
Answer» `W = - underset(V_(1))overset(V_(2))intPdV" "(100 = a + 3b, 400= a +6b, a= - 200 & b = 100)` `W = - underset(V_(1))overset(V_(2))int(a + bV)dV" "- {[aV]+[(bV^(2))/(2)]}_(V_(1))^(V_(2))` ` =- [-600+13.5 xx 100] =- 750` `DeltaU = 6300`

Discussion

152.	The molar heat capacity for the process shown in fig. is A. `C=C_(v)`B. `C=C_(p)`C. `C gt C_(v)`D. `C lt C_(v)`
Answer» Correct Answer - D For polytropic process `PV^(x)=k`, `C=C_(v)+(R )/(1-x) rArr As " "PV^(2)=K(given)" "rArr"put" x=2` `C=C_(v)+(R )/(1-2)=C_(v)-R :. C lt C_(v)`

Discussion

153.	In a cyclic process shown in the figure an ideal gas is adiabatically taken from `B`to`A`, the work done on the gas during the process `B`to`A` is `30,J` when the gas is taken from `A` to `B` the heat absorbed by the gas is `20 J` Then change in internal energy of the gas in the process `A` to `B` is : A. `20 J`B. `-30 J`C. `50 J`D. `-10 J`
Answer» Correct Answer - B

Discussion

154.	Calculate charge in interanl energy for a gas under going from state-I `(300 K, xx 2 xx10^(-2) m^(3))` to state -II `(400 K, 4 xx 10^(-2) m^(3))` for one mol .of vanderwaal gas. `[C_(V)=121J//K//"mole"]` `{Given : ((delU)/(delV))_(T)=T((delP)/(delT))_(V)-P` `C_(V)=12 J//K//mol` `a=2 J .m.//mol^(2)}`
Answer» `DeltaU= C_(V)(T_(2)-T_(1))+a((1)/(V_(1))-(1)/(V_(2)))=C_(V)(100)+a((1)/(4))xx10^(2)=12xx100+2((1)/(4))xx100=125`

Discussion

155.	Calculate work for the expansion of a subsatnce from `3m^(3)` to `5m^(3)` against. Constant pressure `=10^(5)Pa`
Answer» `W=- 10^(5)xx2=-2xx10^(5)J`

Discussion

156.	Find the work done when 18 ml of water is getting vapourised at 373 K is open vessel (Assume the ideal behaviour of water vapour).
Answer» `PV=nRT" "["V in liter and T in kelvin"]` `PV=1xx0.0821xx373` `PV=30 "litre"` `V=30 "litre"` `W=-PDeltaV` `=-1(30-V_(gas))" "[18"ml is negligilble as compared to 30 litre"]` ` =-1xx30=-30 "litre atm".`

Discussion

157.	Show that pressure of a fixed amount of a ideal gas is a state function`V=(nRT)/(p)`
Answer» `((delV)/(delT))_(p)=(nR)/(P)` `(del)/(delP)[((delV)/(delT))_(p)]_(T)=(nR)/(P^(2)) rArr((delV)/(delp))_(T)=(nRT)/(P^(2))` `(del)/(delT)[((delV)/(delT))_(p)]_(T)=(-nR)/(P^(2)) `

Discussion

158.	When 100mL 0.2 MKOH is mixed with 100mL0.2 M HCl in a rigid adiaabatic container ,temperature of solution increase by `t_(1)^(@)C` while when 300mL 0.1 M Koh is mixed with 3000mL 0.1 M HCl then increase in temperature is `t_(2)^(@)`C then which one is correct? (Assume density as well as specfic heat capacity o final solution are same.)A. `t_(1)=t_(2)`B. `t_(1)gtt_(2)`C. `t_(1)ltt_(2)`D. none of these
Answer» Correct Answer - b

Discussion

159.	The decomposition of limestone `CaCO_(3) = CaO_(s) + CO_(2) (g)` is nonspontaneous are `176 kJ` and `160 Jk^(-1)`, respectively. At what temperature, the decomposition becomes spontaneous?A. At `10000 K`B. Below `500^(@)C`C. At `500^(@)C`D. Above `827^(@)C`
Answer» Both `Delta H^(@)` and `Delta S^(@)` are positive. Thus, `Detla G` will be negative only when the `T Delta S` term is greater in magnitude than `Delta H`. This condition is met when `T` is large relative to the equilibrium temperature `(T_(eq))`. At equilibrium, `Delta G - 0`. Thus, `0 = Delta H^(@) = T Delta S^(@)` or `T = (Delta H^(@))/(Delta S^(@)) = (176 xx 10^(3) J mol^(-1))/(160 J K mol^(-1))` `= 1100 K = 827^(@)C` Thus, above `827^(@)C, T Delta S^(@) gt Delta H^(@)` and `Delta G^(@)` is negative.

Discussion

160.	For a reaction, `CaCO_(3(s)) rarr CaO_((s))+CO_(2(g))` `Delta_(f)H^(@)(CaO)=-"631.1 kJ mol"^(-1)` `Delta_(f)H^(@)(CO_(2))=-"393.5 kJ mol"^(-1) and` `Delta_(f)H^(@)(CaCO_(3))=-"1206.9 kJ mol"^(-1)` Which of the following is a correct statement?A. A large amount of heat is evolved during the decomposition of `CaCO_(3)`.B. Decomposition of `CaCO_(3)` is an endothermic process and heat is provided for decompositionC. The amount of heat evolved cannot be calculated from the data provided .D. `Delta_(f)H^(@)=SigmaDelta_(f)H^(@)" (reactants)"-Sigma Delta_(f)H^(@)" (products)"`
Answer» Correct Answer - B `Delta_(r)H=Sigma_(f)H_(p)^(@)-SigmaDelta_(f)H_(R)^(@)` `=[-635.1+(-393.5)]-[-1206.9]=+178.4kJ`

Discussion

161.	Assertion : The enthalpy change for the reaction `CaO_((s))+CO_(2(g)) rarr CaCO_(3(s))` is calld enthalpy of formation of calcium carbonate. Reason : The reaction involves formation of 1 mole of `CaCO_(3)` from its constituent elements.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - D The enthalpy change for the given reaction is not an enthalpy of formation of calcium carbonate, since `CaCO_(3)` has been formed from other compounds, and not from its constituent elements.

Discussion

162.	In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burtn in excess oxygen at 298.0 K. The temperaure of the calorimeter was found to increase from 298.0 Kto298.45 K due to the combustion process.Given that the heat energy capacity of the calorimeter is `2.5 kJ K^(-1)`, the numerical value for the enthalpy of combustion of the gas in kJ `mol^(-1)`is
Answer» Correct Answer - 9

Discussion

163.	The latent heat of vapourisation of a liquid at `500K` and `atm` pressure is `30kcal mol^(-1)`. What will be change in internal energy of `3mol` of liquid at same temperature?A. `13.0 kcal`B. `-13.0 kcal`C. `27.0 kcal`D. `-27.0 kcal`
Answer» `:. DeltaH = DeltaU + DeltanRT` `:. DeltaH = 300 kcal` `3H_(2)O(l) rarr 3H_(2)O(v)` `Deltan = 3 - 0 = 3` `rArr 30 = DeltaU +3 xx 0.0821 xx 500 xx 10^(-3)` `DeltaU = 27 kcal`

Discussion

164.	A certain engine which operates in a Carnot cycle absorbe `3.0 kJ` at `400^(@)C` in a cycle. If it rejects heat at `100^(@)C`, how much work is done on the engine per cycle and how much heat is evolved at `100^(@)C` in each cycle?
Answer» Given, Heat absorbs `(q_(2)) = 3.0 kJ` `T_(2) = 400 + 273.15 K = 673.15 K` `T_(1) = 100 +273.15 K = 373.15 K` Efficiency of the Carnot cycle is given by `eta = (T_(2)-T_(1))/(T_(2)) = (q_(2)+q_(1))/(q_(2))` Thus, `(-T_(1))/(T_(2)) = (q_(1))/(q_(2))` Hence, `q_(1) = ((-T_(1))/(T_(2))) (q_(2))` Thus, heat evolved `q_(1) =- ((373.15K)/(673.15K)) (3.0kJ) =- 1.662 kJ` The work done on the engine is `w =- (q_(2)+q_(1)) = - (3.0 +1.662) kJ =- 4.662kJ`

Discussion

165.	Calculate the work done by `0.1` mole of a gas at `27^(@)C` to double its volume at constant pressure `(R = 2 cal mol^(-1)K^(-1))`
Answer» `(V_(1))/(V_(2)) = (T_(1))/(T_(2)) rArr (V)/(2V) = (300)/(T_(2)) rArr T_(2) = 600 rArr Delta T = 300` `W = -P Delta V = nRDelta T` `rArr W = -0.1 xx 2 xx 300 = -60 cal`.

Discussion

166.	`(Delta H - Delta U)` for the formation of carbon monoxide `(CO)` from its elements at `298K` is `(R = 8.314 K^(-1)mol^(-1))`A. `-1238.78 J mol^(-1)`B. `1238.78 J mol^(-1)`C. `-2477.57J mol^(-1)`D. `2477.57J mol^(-1)`
Answer» `C (s) +(1)/(2)O_(2) (g) rarr CO(g)` `DeltaH - DeltaU = DeltanRT` `= (1)/(2) xx 8.314 xx 298 = 1238.78 mol^(-1)`

Discussion

167.	Heat supplied to a Carnot engine is `37.3 kJ`. How much useful work in `kJ` can be done by the engine that operates between `0^(@)C` and `100^(@)C`?
Answer» `eta = (w)/(q_("supplied")) =1 -(T_(2))/(T_(1)) rArr (w)/(37.3) =1- (273)/(373) rArr w = 10 kJ`

Discussion

168.	Heat supplied to a Carnot egine is `453.6 kcal`. How much useful work can be done by the engine that works between `10^(@)C`and `100^(@)C`?
Answer» `T_(2) = 100 +273 = 373 K , T_(1) = 10 + 273 = 283K,` `q_(2) = 453.6 xx 4.184 = 1897.86 kJ` We know that `w = q_(2).(T_(2)-T_(1))/(T_(2))` `= 1897.86 xx ((373-283))/(373)` `= (1897.86 xx 90)/(373) = 457.92 kJ`

Discussion

169.	The efficiency of the Carnot engine is 25% when the temperature of sink is 27C. Then temperature of source is:
Answer» \(\eta = 1 - \frac{T_2}{T_1}\) \(\frac{25}{100} = 1 - \frac{300}{T_1}\) \(\frac14 = 1- \frac{300}{T_1}\) \(\frac{300}{T_1} = \frac34\) \(T_2 = \frac{300\times4}{3}\) \(T_1 = 400 K° \) \(T_1 = 127°C\)

169.

The efficiency of the Carnot engine is 25% when the temperature of sink is 27C. Then temperature of source is:

Answer»

\(\eta = 1 - \frac{T_2}{T_1}\)

\(\frac{25}{100} = 1 - \frac{300}{T_1}\)

\(\frac14 = 1- \frac{300}{T_1}\)

\(\frac{300}{T_1} = \frac34\)

\(T_2 = \frac{300\times4}{3}\)

\(T_1 = 400 K° \)

\(T_1 = 127°C\)

Discussion

170.	The following sequence of reaction occurs in commercial producatio of aqueous nitric acid. `4NH_(3)(g)+5O_(2)(g) to 4NO(g)+6H_(2)O(l),` `DeltaH=-904Kj...(i)``2NO(g)+O_(2)(g)to 2NO_(2)(g), DeltaH=-112KJ...(ii)` `2NO_(2)(g)+2H_(2)O(l)to 2HNO_(3)(aq)+H_(2)(g),` `DeltaH=-140KJ...(iii)` Determine the total heat liberated (in KJ/mol) at mole of aqueous nitric acid from `NH_(3)` by this process:A. `-352`B. `-405`C. `246.5`D. none of these
Answer» Correct Answer - a

Discussion

171.	The enthalpy of combustion of `H_(2)` , cyclohexene `(C_(6)H_(10))` and cyclohexane `(C_(6)H_(12))` are `-241` , `-3800` and `-3920KJ` per mol respectively. Heat of hydrogenation of cyclohexene isA. `-121 "KJ mol"^(-1)`B. `121 "KJ mol"^(-1)`C. `-242 "KJ mol"^(-1)`D. `242 "KJ mol^(-1)`
Answer» Correct Answer - A

Discussion

172.	Heat of neutralization between HCl and NaOH is `-13.7 k.cal`. If heat of neutralization between `CH_(3)COOH` and `NaOH` is `-11.7 k.cal`. Calculate heat of ionization of `CH_(3)COOH`.
Answer» Correct Answer - 8 Joule `Q = 13.7 -11.7 = 2`

Discussion

173.	Born- Hyber cycle below respesents the energy changes occurring as 298 K, jwhen MO(s) is formed from its elements, wherx x, y,z, a, b,c and d are enthalpy change elements, for corresponding pro0cesses respectively `DeltaH_("sub")"of" M=180kJ//mol` `I.E_(1)(M)=218kJ//mol` `I.E_(2)(M)=384kJ//mol` `DeltaH_("atomisation")"of" O_(2)=640kJ//mol``DeltaH(OrarrO^(-))=-142kJ//mol` `DeltaH(OrarrO^(-2))=-844kJ//mol` In terms of x,y,z,a,b, c and d enthalpy change for the reaction, `2M(s)+O_(2)(g)rarr2MO(s)`,is:A. `2x+2y+2z+2a+2b+2c+2d`B. `x+y+z+a+b+c+d`C. `2x+2y+z+2a+2b+2c+d`D. `2x+2y+2z+2a+2b+2c-2d`
Answer» Correct Answer - a

Discussion

174.	Enthalpy of neutralization of strong acid with strong base is 13.7 kcal. When an unknown acid (1 eq) is neutralizedwith (1 eq.) strong base, the enthalpy change is 10.7 kcal which of the following statements is/are correct regarding unknown acid?A. Unknown acid is strong acidB. Unknown acid is weak acidC. 3.0 kcal heat utilised to dissociate the unknown acidD. 10.7 kcal heat utilised to dissociate unknown acid
Answer» Correct Answer - B::C

Discussion

175.	Enthalpy of neutralization of `CH_(3)COOH` by NaOH is less than that of HCl by NaOH. Enthalpy of neutralization of `CH_(3)COOH` is less because of the absorption of heat in the ionization process.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-15B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-15C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE
Answer» Correct Answer - A

Discussion

176.	Enthalpy of hydrogenation of one mole benzene to cyclohexane is : [Given : Resonance energy of benzene=-70KJ/mol, Enthalpy of hydrogenation of cyclohexene =-100KJ/mol]A. `-170KJ`B. `-30KJ`C. `-370KJ`D. `230KJ`
Answer» Correct Answer - d

Discussion

177.	magnitude of enthalpy of neutralization is minimum for:A. `HCN+KOH`B. `HCl+KOH`C. `HCl+NH_(4)OH`D. `HCN+NH_(4)OH`
Answer» Correct Answer - d

Discussion

178.	Heat of neuralisation is amount of heat evolved or absorbed when `1g-` equivalent of an acid reacts with `1g-` equivalent of a base in dilute solution . If weak acid or weak bae are neutralised, the heat released during neutralisation is somewhat lesser than `-13.7kcal` or `-57.27 kJ` . Het of neutralisation is also referred as heat of formation of water from `H^(+)` and `OH^(-)` ions `i.e., H^(+)+OH^(-) rarr H_(2)O, DeltaH=-13.7kcal. ` `200mL` of `0.1M NaOH` is mixed with `100mL` of `0.1M H_(2)SO_(4)` in 1 experiment. In `II` experiment `100mL` of `0.1M NaOH` is mixed with `50mL` of `0.1M H_(2)SO_(4)`. Select the correct statements: `(1)` heat liberated in each of the two reactions is `274cal`. `(2)` heat liberated in `I` is `274 cal` and in `II` is `137 cal`. ,brgt `(3)` temperature rise is `I` reaction is equal to the temperature rise in `II`. `(4)` temperature rise in `I` reaction is equal to the temperaure rise in `II`A. 1,3B. 2,4C. 2,3D. 1,4
Answer» Correct Answer - b

Discussion

179.	A piston filled with 0.04 mol of ideal gas expands reversibly from 50.0mL to 375 mLat a constant temperature of `37.0^(@)C`. As it does so, it absorbs 208J of heat. The value of q and w for the process will be `:` `( R = 8.314J // mol K , ln7.5 =2.01 )`A. `q= + 208 J , w= + 208 J`B. `q = + 208 J , w = - 208 J`C. ` q = - 208 J , w= - 208 J`D. `q = - 208 J, w= +208 J`
Answer» Correct Answer - b For isothermal expansion of an ideal gas, `DeltaU = 0`. Hence,from first law of thermodynamics, `DeltaU = q+w` , we have `q= -w` . As process involves adsorption of heat, i.e., it is endothermic `q= + 208 J :. W= - 208 J`

Discussion

180.	A gas expands from a volume of 1`m^(3)` to a volume of`2m^(3)` against an external pressure of`10^(5) Nm^(-2)`. The work done by the gas will beA. `10^(5) kJ `B. ` 10^(2) kJ`C. ` 10^(2) J`D. `10^(3)J`
Answer» Correct Answer - b `w = - PDeltaV = - 10^(5) Nm^(-2) ( 2-1) m^(3) = - 10^(5) Nm = - 10^(5) J = -10^(2) kJ `

Discussion

181.	The heat of combustion of sucrose `(C_(12)H_(22)O_(11))` is `1350 kcal//mol`. How much of heat will be liberated when `17.1 g` of sucrose is burnt ?A. `67.5 kcal`B. `13.5 kcal`C. `40.5 kcal`D. `25.5 kcal`
Answer» Correct Answer - A

Discussion

182.	The lattice energy of solid `NACl` is `180K. Cal mol^(-1)`. The dissolution of the solid in water in the form of ions is endothermic to the extent of `1 K.cal mol^(-1)`. If the hydration energies of `NA^(+)` and `Cl^(-)` are in ratio `6 : 5`, what is the enthalpy of hydration of `NA^(+)` ionA. `-85.6 K. cal mol^(-1)`B. `-97.5 K.cal mol^(-1)`C. `-182.6 K.cal mol^(-1)`D. `+100K. Cal mol^(-1)`
Answer» Correct Answer - B `DeltaH_(sol^(n))=DeltaH_(L.E)+DeltaH_(H.E)`

Discussion

183.	`AB,A_(2)` and `B_(2)` are diatomic molecules. If the bond enthalpies of `A_(2), AB` and `B_(2)` are in the ratio `1:1:0.5` and the enthalpy of formation of `AB` from `A_(2)` and `B_(2)` is `-100kJ mol^(-1)` , what is the bond enthalpy of `A_(2)` ?A. `200 kJ mol^(-1)`B. `100 kJ mol^(-1)`C. `300 kJ mol^(-1)`D. `400 kJ mol^(-1)`
Answer» Correct Answer - D `DeltaH=H_(R)-H_(P)`

Discussion

184.	An athelet takes 20 breaths per minute at room temperature. The air inhaled in each breath is `200mL` which contains `20%` oxygen by volume, while exhaled air contains `10%` oxygen by volume. Assuming that all the oxygen consumed if used for converting glucose into `CO_(2)` and `H_(2)O_((l))` , how much glucose will be burnt in the body in one hour and what is the heat produced ? `(` Room temperature `-27^(@)C` and enthalpy of combustion of glucose is `-2822.5kJ mol^(-1) at 0^(@)C)`
Answer» Inhaled `O_(2)` in one breath `= (200 xx 20)/(100) mL = 40 mL` Exhaled `O_(2)` in one breath `= (200 xx 10)/(100) mL = 20 mL` Used `O_(2)` in one breath `= 40 - 20 = 20 mL` Now `20` breaths are taken in `1min` Breath taken in `1hour = 20 xx 60 = 120 0 breath` Volume of `O_(2)` used in `1h` `= 20 xx 1200 = 24000 mL at 27^(@)C` `= (24000)/(300) xx 273 mL at 0^(@)C` `= 21840 mL = 21.84 L at 0^(@)C` For glucose: `C_(2)H_(12)O_(6) +6O_(2) rarr 6CO_(2) +6H_(2)O DeltaH =- 2822.5 kJ` `6 xx 22.4 L of O_(2)` is used for `180` g glucose `21.84 L of O_(2)` is used for `= (180 xx 21.4)/(6 xx 22.4) = 29.25 g` Also `6 xx 22.4 L of O_(2)` with glucose gives `= 2822.5 kJ 21.84 L O_(2)` with glucose give `=(2822 xx 21.84)/(6 xx 22.4) = 458.66 kJ`

Discussion

185.	In which of the following reactions do you except to have a decrease in entropy?A. `Fe(s) to Fe(l)`B. `Fe(s)+(3)/(2)O_(2)(g)to Fe_(2)O_(3)(s)`C. `Fe(l) to Fe(g)`D. `2H_(2)O_(2)(l) to 2H_(2)(l)+O_(2)(g)`
Answer» Correct Answer - B

Discussion

186.	Calculate the heat produced when `3.785 L` of octabe reacts with oxygeb to form carbon mono oxide and water vapour at `25^(@)C`. (Density of octane is `0.75025 g mL^(-1)). Delta_(comb)H^(Theta)(C_(8)H_(18)) =- 1302.7 kcal, Delta_(f)H^(Theta) (CO_(2)) =- 94.05 kcal, Delta_(f)H^(Theta) (H_(2)O) =- 64.32 kcal, Delta_(f)H^(Theta) (CO) =- 26.41 kcal mo^(-1)`
Answer» Number of moles of octane `=(3785 xx 0.7025)/(114.2) = 23.38` `C_(8)H_(18) +(17)/(2)O_(2) rarr 8CO +9H_(2)O DeltaH = ?` `C_(8)H_(18) +(25)/(2)O_(2) rarr 8CO_(2) +9H_(2)O DeltaH =- 1302.7 kcal` ..(i) For equation (i) `DeltaH = 8 Delta_(f)H (CO_(2)) +9Delta_(f)H(H_(2)O) -Delta_(f)H(C_(8)H_(18))` `Delta_(f)H^(Theta) (C_(8)H_(18)) = 1302.7 +8 (-94.05) +9(-68.32)` `=- 64.6 kcal` `DeltaH = 8 Delta_(f)H^(Theta) (CO) +9Delta_(f)H^(Theta) (H_(2)O) -Delta_(f)H^(Theta) (C_(8)H_(18))` `=- 761.56` `DeltaH` for `23.38 mol` `=- 761.56 xx 23.8 =- 17729 kcal`

Discussion

187.	The heat of formation of `C_(2)H_(5)OH(l)` is `-66 " kcal"//"mole"`. The heat of combustion of `CH_(3)OCH_(3)(g)` is `-348 " kcal"//"mole",DeltaH_(f)` for `H_(2)O(l)` and `CO_(2)(g)` are `-68 kcal//"mole"` and `-94 kcal//"mole"` respectively,. Then the `DeltaH` for the isomerisation reaction, `C_(2)H_(5)OH(l)toCH_(3)OCH(g)`, and `DeltaE` for the same are : (Take: `T_("surr")=298 K`)A. `DeltaH=18 kcal//"mole",DeltaE=17.301 kcal//"mole"`B. `DeltaH=22 kcal//"mole",DeltaE=21.404 kcal//"mole"`C. `DeltaH=26 kcal//"mole",DeltaE=25.709 kcal//"mole"`D. `DeltaH=30 kcal//"mole",DeltaE=28.522 kcal//"mole"`
Answer» Correct Answer - B

Discussion

188.	The heat of formation of `C_(2)H_(5)OH(l)` is `-66 kcal//"mole"`. The heat of combustion of `CH_(3)OCH_(3) (g)` is -348 kcal/mole. `DeltaH_(f)` for `H_(2)O` and `CO_(2)` are -68 kcal/mole respectively. Then, the `Delta H` for the isomerisation reaction `C_(2)H_(5)OH(l)rarrCH_(3)OCH_(3)(g)`,and `Delta E` for the same are at `T=25^(@)C`A. `Delta H=18` kcal/mole,`Delta E= 17.301`kcal/moleB. `Delta H=22` kcal/mole,`Delta E=21.408` kcal/moleC. `DeltaH=26` kcal/mole, `Delta E=25.709` kcal/moleD. `Delta H=30` kcal/mole, `Delta E= 28.522` kcal/mole
Answer» Correct Answer - B `DeltaH_(F)^(o)C_(2)OH_(5)(l)= -66` Kcal/mole `2C+3H_(2)+(1)/(2)O_(2)rarrC_(2)H_(5)OH " " DeltaH= -66 kcal//"mole"`.....(1) `CH_(3)-O-CH_(3)+3O_(2)rarr2CO_(2)+3H_(2)O " "Delta H= -348 kcal//"mole"`.....(2) `[H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O " " DeltaH_(3)-68 kcal//"mole"`.....(3) `[C+O_(2)rarrCO" "Delta H_(4)= -94 kcal//"mole"`......(4) Target equation `= - eq-eq2+3eq3+2 eq 2` `DeltaH=+66 + 348 -3xx68-2xx94=+66+348-204-108 " "implies DeltaH=22 K `cal/mole `DeltaH=DeltaE+DeltangRT" "22=DeltaE+1xx 2xx298 xx 10^(-3) " " implies DeltaE=21.4` Kcal/mole

Discussion

189.	`AB, A_(2)` and `B_(2)` are diatomic molecules.If the bond enthalpy of `A_(2,)` AB from `A_(2)` and `B_(2)` od `-100 kJ//"mol"`, what is the bond enthalpy of `A_(2)" in" kJ//mol`.` ?
Answer» Correct Answer - 400

Discussion

190.	Calculate the bond energy of `Cl-Cl` bond from the following data: `CH_(4)(g)+Cl_(2)(g)rarrCH_(3)Cl(g)+HCl(g), Delta H= -100.3 kJ`. Also the bond enthalpies of `C-H, C-Cl,H-Cl` bonds are `413,326` and `431 kJ mol^(-1)` respectively.
Answer» Correct Answer - `243.7 kJ mol^(-1)` `Delta H= 4 Delta H_(C-H)+DeltaH_(Cl-Cl)-3 Delta H_(C-H)-Delta H_(H-Cl)` `= -100.3=(4xx413)+x-(3xx413)-326-431` `x= 243.7 KJ mol^(-1)`

Discussion

191.	Calculate the enthalpy of formation of ammonia from the following bond enegry data: `(N-H) bond = 389 kJ mol^(-1), (H-H) bond = 435 kJ mol^(-1)`, and `(N-=N)bond = 945.36kJ mol^(-1)`.
Answer» `N-= N +3 (H-H) rarr underset(H)underset(\|)overset(H)overset(\|)(2N)-H,DeltaH^(Theta) =`? `DeltaH^(Theta) = [DeltaH^(Theta)underset((N-=N))+3xxDeltaH_((H-H))^(Theta)]-[6xxDeltaH_((N-H))^(Theta)]` `= 945.36 +3 xx 435.0 - 6 xx 389.0 =- 83.64 kJ` Heat of formation of `NH_(3) = (DeltaH^(Theta))/(2) =- (83.64)/(2)` `=- 41.82kJ mol^(-1)`

Discussion

192.	Calculate the ΔH° of the reaction\(H-\overset{\overset{H}{\|}}{\underset{\underset{H}{\|}}{C}}-Cl\) (g) → C(g) + 2H(g) + 2Cl(g)(Use table for standard enthalpy of formation).
Answer» The given reaction represents a process in which chemical bonds are broken and constituent atoms are produced. Clearly this is a case of atomization of CH₂Cl₂(g) as in this case no new bond is formed. Thus Bond energy in (to break bonds) 2ΔH°(C-H) = 2 x 413 kJ mol^-1 2ΔH°(C-Cl) Bond energy out (to form bond) No bond is formed and ΔH° (out) is zero. = \(\frac{2 \times 328 kJ\,mol^{-1}}{1482\, kJ\,mol^{-1}}\) ΔH°(reaction) = Bond energy in - Bond energy out = 1482 kJ mol^-1 - 0 = 1482 kJ mol^-1

192.

Calculate the ΔH° of the reaction\(H-\overset{\overset{H}{|}}{\underset{\underset{H}{|}}{C}}-Cl\) (g) → C(g) + 2H(g) + 2Cl(g)(Use table for standard enthalpy of formation).

Answer»

The given reaction represents a process in which chemical bonds are broken and constituent atoms are produced. Clearly this is a case of atomization of CH₂Cl₂(g) as in this case no new bond is formed. Thus

Bond energy in (to break bonds)

2ΔH°(C-H) = 2 x 413 kJ mol^-1

2ΔH°(C-Cl)

Bond energy out (to form bond)

No bond is formed and ΔH° (out) is zero.

= \(\frac{2 \times 328 kJ\,mol^{-1}}{1482\, kJ\,mol^{-1}}\)

ΔH°(reaction) = Bond energy in - Bond energy out

= 1482 kJ mol^-1 - 0

= 1482 kJ mol^-1

Discussion

193.	Use the following data to calculate the molar enthalpy of combustion of ethane ,`C_(2)H_(6)`: `2C_(2)H_(2)(g)+5O_(2)(g)to4CO_(2)(g)+2H_(2)O(l),` `DeltaH=-2511KJ//mol``C_(2)H_(2)(g)+2H_(2)(G)to C_(2)H^(6)(g), DeltaH=-311KJ//mol` `2H_(2)(g)+O_(2)(g)to 2H_(2)O(l), DeltaH=-484KJ//mol`A. `-1428KJ//mol`B. `-2684KJ//mol`C. `-2856KJ//mol`D. `-3306KJ//mol`
Answer» Correct Answer - a

Discussion

194.	The heats of combustion of `Ch_(4)` and `C_(4)H_(10)` are `-890.3` and `-2878.7 kJ mol^(-1)`, respectively. Which of the two has greater efficiency as fuel per gram?
Answer» Molar mass of methane `= 16` Heat produced per gram of methane `=- (890.3)/(16) =- 55.64 kJ` Molar mass of butane `= 58` Heat produced per gram of butane `=- (2878.7)/(58) =- 49.63 kJ` Thus, methane has greater fuel efficiency than butane.

Discussion

195.	Calculate `DeltaH` of the reaction `H - underset(Cl) underset(\|) overset(H) overset(\|) (C ) - Cl(g) rarr C(g) +2H(g) + 2Cl(g)` The average bond energies of C-H bond and C-Cl bond are 416kJ and 325 kJ `mol^(-1)` respectively.
Answer» In the given reaction, two moles of C-H bond and two molesof C-Cl bond are broken. Hence energy absorbed `= 2 xx 416 + 2 xx 325` `= 1485kJ`

Discussion

196.	Calculate `DeltaH^(@)` `("in" kJmol^(-1))` for the reaction `CH_(2)Cl_(2)(g)rarrC(g)+2H(g)+2Cl(g).` The average bond enthalpie of `C-H` and `C-C1` bonds are `414 kJ mol^(-1)`.
Answer» Correct Answer - 1488

Discussion

197.	Calculate the enthalpy change `(DeltaH)` in `KJ mol^(-1)`, of the follwing reaction `2C_(2)H_(2)(g)+50_(2)(g)rarr4CO_(2)(g)+2H_(2)O(g)` givin average bond enthalpies of various bonds i.e., `C-H,C=C,O=O,O-H as 414,814,499,724` and `640 kJ mol^(-1) respectively. Express magnitude only.
Answer» Correct Answer - 2573

Discussion

198.	One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is `P_0`. Choose the correct option (s) from the following A. Internal energies at A and B are the sameB. Work done by the gas process AB is `P_(0)V_(0)` in 4C. Pressure at C is `(P_(0))/4`D. Temperature at C is `(T_(0))/4`
Answer» Correct Answer - A::B Process AB is isothermal process, Therfore, internal energies at A and B are the same. Work done by AB `W= RT_(0) "log"_(e)(4V_(0))/(V_(0))= P_(0)V_(0)"log"_(e)4` Since the coordinates of C are not given, hence we cannot predict the pressure and temperature at C.

Discussion

199.	A closed and isolated cylinder contains ideal gas. An adiabatic separator of mass m, cross sectional area `A` divides the cylinder into two equal parts each with volume `V_(0)` and pressure `P_(0)` in equilibrium Assume the sepatator to move without friction. Identify the correct statementA. The process is adiabatic only when the piston is displaced suddenlyB. The process is isothermal when the piston is moved slowlyC. The motion is periodic for any displacement of the pistonD. The motion is `SHM` for any displacement of the piston
Answer» Correct Answer - A,B,D `DeltaQ=0` for system `P_(0)V_(0)^(gamma)=P_(1)V_(1)^(gamma)=P_(2)V_(2)^(gamma)` `impliesP_(1)=P_(0)[1-(gammaAx)/(V_(0))]impliesP_(2)=P_(0)[1+(gammaAx)/(V_(0))]` ` implies F = (2P_(0)gammaA^(2)x)/(V_(0)`

Discussion

200.	One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is `P_0`. Choose the correct option (s) from the following A. internal energies at `A and B` are the sameB. work done by the gas in process `AB` is `P_(0)V_(0)ln4`C. pressure at `C` is `(P_(0))/(4)`D. temperature at `C` is `(T_(0))/(4)`
Answer» Correct Answer - A::B::C::D Internal energy of in ideal gas depends on temperature `W_(BC)=nRT "ln"(V_(2))/(V_(1))=(1) (R)(P_(0)V_(0))/(R) "ln"(4V_(0))/(V_(0))` `=P_(0)V_(0) ln 4` For `CA (P)/(T)=` constant P at `C =(P_(0))/(4)` T at `C=(T_(0))/(4)` Hence all option are correct.

Discussion

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Calculate `DeltaH^(@)` `("in" kJmol^(-1))` for the reaction `CH_(2)Cl_(2)(g)rarrC(g)+2H(g)+2Cl(g).` The average bond enthalpie of `C-H` and `C-C1` bonds are `414 kJ mol^(-1)`.

Calculate the enthalpy change `(DeltaH)` in `KJ mol^(-1)`, of the follwing reaction `2C_(2)H_(2)(g)+50_(2)(g)rarr4CO_(2)(g)+2H_(2)O(g)` givin average bond enthalpies of various bonds i.e., `C-H,C=C,O=O,O-H as 414,814,499,724` and `640 kJ mol^(-1) respectively. Express magnitude only.