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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 651. |
The value of `DeltaH-DeltaU` when 2 moles of solid benzoic acid undergoes combustion at 300 K is given by :A. `-1.247 kJ`B. `-2.494 kJ`C. `+2.494 kJ`D. `+1.247 kJ` |
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Answer» Correct Answer - B |
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| 652. |
At `298 K, K_(p)` for reaction `N_(2)O_(4) (g) hArr 2NO_(2) (g)` is 0.98. Predict whether the reaction is spontaneous or not. |
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Answer» For the reaction, `N_(2)O_(4) (g) hArr 2NO_(2) (g), L_(P) = 0.98` As we know that `Delta_(r) G^(Θ) = - 2.303 RT log K_(P)` Here, `K_(P) = 0.98` i.e., `K_(P) lt 1` therefore, `Delta_(r) G^(@)` is positive, hence the reaction is non-spontaneous. |
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| 653. |
A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in figure. What will be the value of `Delta H` for the cycle as a whole ? |
| Answer» The net enthalpy change, `Delta H` for a cyclic process is zero as enthalpy change is a state function, i.e., `Delta H` (cycle) `= 0` | |
| 654. |
Calculate heat capacity of a diatomic ideal gas (Molar mass=`11 gm//mol`) if it is subJected to a process such that pressure exerted is directly proportional to cube of the volume.[R=`2cal//mol K`]A. `5cal//mol K`B. `11cal//gm K`C. `0.5cal//gm K`D. `0.5cal//mol K` |
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Answer» Correct Answer - C |
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| 655. |
`2` moles of ideal gas is expanded isothermally & reversibly from `1` litre to `10` litre. Find the enthalpy changes in `KJ mol^(-1)`. |
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Answer» Correct Answer - A ` because H=E+PV` and `Delta H=Delta E+PDelta V` `:. Delta H=Delta E +nRDelta T` For isotermal and reversible process `Delta T=0` `:. Delta H=Delta E+0` ,brgt `because Delta E=0` `because Delta H` is also equal to zero. |
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| 656. |
If the bond dissociation energies of XY , `X_(2)` and `Y_(2)` ( all diatomic molecules )are in the ratio `1:1 : 0.5` and `DeltaH _(f) ` for the formation of XY of `- 200 kJ mol^(-1)` , the bond dissociation energy of `X_(2)` will beA. `100 kJ mol^(-1)`B. `200 kJ mol^(-1)`C. ` 400 kJ mol^(-1)`D. ` 800 kJmol^(-1)` |
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Answer» Correct Answer - D Suppose bond dissociation energy of XY `=akJ mol^(-1) , i.e., `BE ( XY )= a . Then BE `9X_(2)0= a, BE(Y_(2))= 0.5 a` Aim `: (1)/(2) X_(2)+ (1)/(2) Y_(2) rarr XY` `Delta_(r)H = BE` ( Reactants ) - BE ( Products ) `= [ (1)/(2) BE (X_(2)) + (1)/(2) BE(Y_(2)) ] = BE(XY)` `:. - 200 = ((a)/(2) + (0.5a)/( 2)) -a` or ` - 200 = - 0.25 a` or `a= 800 kJ mol^(-1)` |
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| 657. |
Three moles of an ideal gas being initially at a temperature `T_i=273K` were isothermally expanded 5 times its initial volume and then isochorically heated so that the pressure in the final state becomes equal to that in the initial state. The total heat supplied in the process is 80kJ. Find `gamma(=(C_p)/(C_V))` of the gas. |
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Answer» In the isothermal process, heat transfer to the gas is given by `Q_1 = v RT_0 1n (V_2)/(V_1) = v RT_0 1 n eta (For eta = (V_2)/(V_1) = (p_1)/(p_2))` In the isochoric process, `A = 0` Thus heat transfer to the gas is given by `Q_2 = Delta U = v C_V Delta T = (v R)/(gamma - 1) Delta T(for C_V = (R)/(gamma - 1))` But `(p_2)/(p_1) = (T_0)/(T)`, or `T = T_0 (p_1)/(p_2) = eta T_0 (for eta = (p_1)/(p_2))` or, `Delta T = eta T_0 - T_0 = (eta - 1) T_0` so, `Q_2 = (vR)/(gamma - 1).(eta - 1) T_0` Thus, net heat transfer to the gas `Q = vRT_0 1n eta + (vR)/(gamma-1).(eta - 1) T_0` or, `(Q)/(vRT_0) = 1n eta + (eta - 1)/(gamma - 1)`, or, `(Q)/(vRT_0) - 1n eta = (eta - 1)/(gamma -1)` or `gamma = 1 + (eta - 1)/((Q)/(vRT_0) - 1n eta) = 1+ (6.1)/(((80 xx 10^3)/(3 xx 8.314 xx 273)) - 1n 6) = 1.4`. |
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| 658. |
Two mole of an ideal gas is heated at constant pressure of one atmosphere from `27^(@)C` to `127^(@)C`. If `C_(v,m)=20+10^(-2)"T JK"^(-1).mol^(-1)`, then q and `DeltaU` for the process are respectively:A. `6362.8J,4700J`B. `3037.2J,4700J`C. `7062.8J,5400J`D. None of these |
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Answer» Correct Answer - A `w=-nRDeltaT=-2xx8.314xx100` `=-1662.8J` `DeltaU=n intC_(v,m)dT` `=2xxint(20+10^(-2)T)dT` `=2xx20xx(T_(2)-T_(1))+2xx10^(-2)xx((T_(2)^(2)-T_(1)^(2)))/(2)` = 4700 J `4700=q-1662.8` `therefore" "q=6362.8 J` |
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| 659. |
The enthalpy of neutralisation of a weak acid in `1 M` solution with a strong base is `-5.6 kJ mol^(-1)`. Enthalpy of ionization of the acid is `1.5 kJ mol^(-1)` and enthalpy of neutralization of the strong acid with a strong base is `-57.3 kJ"equiv"^(-1)`, what is the % ionization of the weak acid in molar solution (assume the acid to be monobasic) ? |
| Answer» Correct Answer - 0.2 | |
| 660. |
One mole of an ideal gas (`C_(V) = 20 JK^(-1) mol^(-1))` initially at STP is heated at constant volume to twice the initial temeprature. For the process W and q will beA. W = 0, q = 5.46 kJB. W = 0,q=0C. W = -5.46kJ, q = 5.46 kJD. W=5.46kJ, q = 5.46 kJ |
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Answer» Correct Answer - A `W = P Delta V = P xx 0 = 0` `Q = C_(V) (Delta T) = 20 xx 273 = 5460 J = 5.46 kJ` |
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| 661. |
A diatomic ideal gas initially at 273 K is given 100 cal heat due to which system did 209 J work. Molar heat capacity ` (C_(m))`n of gas for the process is :A. `(3)/(2) R`B. `(5)/(2) R`C. `(5)/(4)R`D. 5 R |
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Answer» Correct Answer - D `DeltaU = 100 -(209)/(4.18) = 50 cal rArr DeltaU = nC_(V)dT rArr C_(v,m) = (5)/(2)R rArr ndT = (20)/(R) rArr q = nC_(m) dT` `C_(m)= (100R)/(20) R` |
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| 662. |
Calculate the amount of heat evolved during the complete combustion of `100 ml` of liquid benzene from the following data. Predict your answer as `(Delta H)/(100)` ( in KJ/mol). (i) `18 gm` of graphite on complete combustion evolve 585 KJ heat (ii) 15540 KJ heat is required to dissociate all the molecules of 1 litre water into `H_(2)` and `O_(2)`. (iii) The heat of formation of liquid benzene is 48 kJ/mol (iv) Density of `C_(6)H_(6)(l)=0.87 gm//ml` |
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Answer» Correct Answer - 36 KJ `(i)" "C(s) + O_(2)(g) rarr CO_(2)(g) , DeltaH_(f)^(0) = -390KJ//mol, " " DeltaH_(f)^(0) = (585)/(18)xx12` `(ii) " " H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)(l) , DeltaH_(f)^(0) =-280 KJ//mol ," " Delta H_(f)^(0) = (15540)/(55.5)` `(iii)" " C_(6)H_(6)(l) + (15)/(2)O_(2)(g) rarr 6CO_(2)(g) + 3H_(2)O(l) ," " DeltaH_(f)^(0) = -48` `therefore" "DeltaH^(0) = [6(-390) + 3(-280)] - 48 = -3228 KJ//mol` `" "` Mass of benzene is `= 0.87 xx 100 = 87g` `therefore" "` Heat evolved from `87` gm benzene `=3600KJ` `" "` Hence , `(DeltaH)/(100)=36KJ.` |
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| 663. |
1 gram graphite on combustion in oxygen liberates 7.82 K.cal of heat. The amount of heat evolved when 1 mole of oxygen is used in this oxidation process isA. `7.82 K.cal`B. `78.2 K.cal`C. `9.384 K.cal`D. `93.84 K.cal` |
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Answer» Correct Answer - D `C+O_(2) rarr CO_(2)` |
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| 664. |
For the equations `C("diamond") +2H_(2)(g) rarr CH_(4)(g) DeltaH_(1)` `C(g)+4H(g) rarr CH_(4)(g) DeltaH_(2)` Predict whtherA. `DeltaH_(1) = DeltaH_(2)`B. `DeltaH_(1) gt DeltaH_(2)`C. `DeltaH_(1) lt DeltaH_(2)`D. `DeltaH_(1) = DeltaH_(2) + Delta_(vap)H(C )+ Delta_(diss)H(H_(2))` |
| Answer» `DeltaH_(1) gt DeltaH_(2)` | |
| 665. |
For the equations `C("diamond")+2H_(2)(g)toCH_(4)(g)" "DeltaH_(1),` `C(g)+4H(g)toCH_(4)(g)" "DeltaH_(2)` Predict whetherA. `C(g)+4H(g)toCH_(4)(g)" "DeltaH_(2),`B. `DeltaH_(1)gtDeltaH_(2)`C. `DeltaH_(1)ltDeltaH_(2)`D. `DeltaH_(1)=DeltaH_(2)+Delta_(vap)H(C)=Delta_(diss)H(H_(2))` |
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Answer» Correct Answer - B |
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| 666. |
Calculate the amount of heat evolved during the complete combustion of `100 ml` of liquid benzene from the following data. (i) `18 g` of graphite on complete combustion evolve `590 KJ` heat (ii) `15889 KJ` heat is required to dissociate all the molecules of `1` litre water into `H_(2)` and `O_(2)` (iii) The heat of formation of liquid benzene is `50 KJ//mol` (iv) Density of `C_(6)H_(6)(l)=0.87 g//ml` |
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Answer» (i) `C(s)+O_(2)(g)rarrCO_(2)(g), DeltaH_(f)^(@)= -393.33 KJ//mol` (ii) `H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l), Delta H_(f)^(@)= -286 KJ//mol` (iii) `C_(6)H_(6)(l)+(15)/(2)O_(2)(g)rarr6CO_(2)(g)+3H_(2)O(l)` `:. DeltaH^(@)=[6(-393.33)+3(-286)]-50=-3268 KJ//"mole"` `:.` Heat evolved from `87 g` benzene `3645 KJ` |
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| 667. |
For the equation `C("diamond")+2H_(2)(g)rarrCH_(4)(g) DeltaH_(1) , C(g)+5H(g)rarrCH_(4)(g) DeltaH_(2)`A. `Delta H_(1)=DeltaH_(2)`B. `Delta H_(1) gt Delta H_(2)`C. `Delta H_(1) lt Delta H_(2)`D. `DeltaH_(1)=DeltaH_(2)+Delta_(vap)H(C )+Delta_(diss)H(H_(2))` |
| Answer» Correct Answer - B | |
| 668. |
A gas mixture of `4` litres of ethylene and methane on complete combustion at `25^(@)C` produces `6` litres of `CO_(2)`. Find out the amount of heat evolved on burning one litre of the gas mixture. The heats of combustion of ethylene and methane are`-1464` and `-976 KJ mol^(-1)` at `25^(@)C` |
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Answer» gt `X "moles"" " (4-x)"moles" " "6"moles"` Applying `POAC` for `C` atoms, `2xxx+1xx(4-x)=1xx6, x=2 "lit"` Thus, the volume of ` C_(2)H_(4)=2"lit"`, and volume of `CH_(2)=2"lit"`. `:.` volume of `C_(2)H_(4)` in a `1` litre mixture `=2//4=0.5 "lit"`. and volume of `CH_(4)` in a `1` litre mixture `=1-0.5=0.5"lit"` Now, thermochemical reactions for `C_(2)H_(4)` and `CH_(4)` are `{:(C_(2)H_(4)+3O_(2)rarr2CO_(2)+2H_(2)O, Delta H= -1464KJ,,,,),(CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O, Delta H= -976 KJ,,,,):}` As `Delta H` values given at `25^(@)C`. let us first calculate the volume occupied by one mole of any gas at `25^(@)C` (supposiong pressures as `1 atm`) Volume per mole at `25^(@)C=(298)/(273)xx22.4=24.4"lit"` Thus, heat evolved in the combustion of `0.5` lit of `C_(2)H_(4)=(1464)/(24.4)xx0.5= -30KJ` and heat evolved in the combustion of `0.5` lit of `CH_(4)=(976)/(24.4)xx0.5=-20KJ`. `:.` total heat evolved in the combustion of `1` litre of the mixture `= -30+(-20)= -50 KJ` |
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| 669. |
At absolute zero, the entropy of a pure crystal is zero. This isA. First lawB. Second lawC. Third lawD. None |
| Answer» Correct Answer - Third law | |
| 670. |
An ideal gas undergoes four different processes from the same initial state (figure). Four process are adiabatic, isothermal, isobaric and isochloric. Out of 1, 2, 3, and 4 which one is idabatic. A. 4B. 3C. 2D. 1 |
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Answer» Correct Answer - C As is clear from (figure). In curve 1, `V` is constant. It represents Isochoric process. In curve 4, `P` is constant. It represents Isobaric process. Out of curves 2 and 3, one is isothermal and other is adiabatic, As slope of curve 2 is more than the slope of curve 3, therefore curve 2 represent aidabatic process. |
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| 671. |
What work is to be done on `2mol` of a perfect gas at `27^(@)C` it is compressed reversibly and isothermally from a pressure of `1.01 xx 10^(5) Nm^(-2) to 5.05 xx 10^(6)Nm^(-2)`? |
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Answer» For reversible process: `w_(rev) = - 2.303 nRT log10 .(P_(1))/(P_(2))` `=- 2.303 xx2xx 8.314 xx 300 log_(10).(1.01 xx 10^(5))/(5.05 xx 10^(6))` `=+ 1.9518 xx 10^(4)J` |
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| 672. |
What work is to be done on `2mol` of a perfect gas at `27^(@)C` it is compressed reversibly and isothermally from a pressure of `1.01 xx 10^(5) Nm^(-2) to 5.05 xx 10^(6)Nm^(-2)`?A. work done on the gas is `1.9518 xx 10^(4)J`B. Free energy change for the process is `-1.9518 xx 10^(4)J`C. work done on the gas is `1.19 xx 10^(7)J`D. Free energy change for the process is `-1.19 xx 10^(7)J` |
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Answer» Correct Answer - A::B reversible process: `W_(rev) = -2.303 nRT log_(10).(p_(1))/(p_(2))` `= -2.303 xx 2 xx 8.314 xx 300 log_(10).(1.01 xx 10^(5))/(5.05 xx 10^(5))` `+ 1.9518 xx 10^(4) "joule"` Since W reversible is a measure of free change `-Delta G = -W_(rev) = -W_(max) = 1.9518 xx 10^(4)J` |
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| 673. |
The following curve represent adiabatic expansions of gases `He,O_(2)`,and `O_(3)` not necessarily in order. Which curve represnets for `O_(3)`? A. `I`B. `II`C. `III`D. Any one of these |
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Answer» `pV^(gamma) =` constant or `log P =- gamma log V` `gamma` is slope for `P-V` plots for `He, O_(2)` and `O_(3)`, i.e., `(5)/(3),(7)/(5)`and `(4)/(3)`, respectively. |
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| 674. |
Which of the following statement is incorrect regarding adiabatic and isothermal processes for an ideal gas, starting from same initial state to same final volume?A. In expansion, more work is done by the gas in isothermal processs.B. In compression, less work will be done on the gas in isothermal process.C. The slope of adiabatic `P-V` graph is negative.D. In expansion, final temperature of adiabatic will be mor as compared to isothermal. |
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Answer» Correct Answer - D |
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| 675. |
What work is to be done on `2` mole of a perfect gas at `27^@C` if it is compressed reversibly and isothermally from a pressure of `1.01 xx 10^5 Nm^(-2)` to `5.05 xx 10^6 Nm^(-2)`? Also calculate the free energy change. |
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Answer» For reversible process: `W_(rev)=-2.303 nRT"log"_(10)(P_(1))/(P_(2))` `=-2.303xx2xx8.314xx300 "log"_(10)(1.01xx10^(5))/(5.05xx10^(6))` `=+1.9518xx10^(4)` joule Since, `W_(rev)` is a measure of free energy change `:. -DeltaG=-W_(rev)=-W_(max)` or `DeltaG=1.9518xx10^(4)` joule |
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| 676. |
If a certain mass of gas is made to undergo separately adiabatic and isothermal expansions to the same pressure, starting form the same initial conditions of temperature and pressure, then, as compared to that of isothermal expansion, in the case of adiabatic expansion, the finalA. Volume and temperature will be higherB. Volume and temperature will be lowerC. Temperature will be lower but the final volume will be higherD. Volume will be lower but the final temperature will be higher |
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Answer» Correct Answer - B Adiabatic `P,V,T P_(1),V_(a),T` Isothermal `P,V,T, P_(1),V_(i),T_(i)` `T_(i) lt T` Also `PV^(gamma)=P_(1)V_(a)^(gamma)` `PV=P_(1).V_(i)` `:. (V)/(V^(gamma))=(V_(i))/(V_(a)^(gamma))` `:. V_(i) gt V_(a)` |
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| 677. |
If separate samples of argon, methane, nitrogen and ammonia, all at the same initial temperature and pressure and expanded adiabatically to double their original volumes, then which one of these gases will require the greatest quantity of heat to restore the original temperature?A. NitrogenB. ArgonC. MethaneD. Ammonia |
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Answer» Correct Answer - B `gamma`is maximum for `Ar`. |
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| 678. |
`Delta_(f)H^(Theta)` per mole of `NH_(3)(g). NO(g)`, and `H_(2)O(l)` are `-11.04 + 21.60` and `-68.32 kcal`, respectively. Calculate the standard heat of reaction at constant pressure and at a constant volume for the reaction: `4NH_(3)(g) +5O_(2)(g) rarr 4NO(g) +6H_(2)O(l)` |
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Answer» `Deltan_(g) = 4 - 9 = 5` `Delta_(f)H^(Theta)O_(2) = 0` `DeltaH = [4Delta_(f)H^(Theta)(NO)+6Delta_(f)H^(Theta) (H_(2)O)] -4 Delta_(f)H^(Theta)(NH_(3))` `=4 xx 21.6 +6 xx -(68.32-(-11.04) =- 279.36 kcal` `DeltaU = DeltaH - Deltan_(g)RT` `=- 279.36 -(-5) xx2xx 10^(-3) xx 298` `=- 276.38 kcal` |
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| 679. |
Work done on a system when one mole of an an ideal gas at 500 K is compressed isothermally and reversibly to 1/10th of its original volume. (R = 2cal)A. `1Kcal`B. `2.303 Kcal`C. `4.606 Kcal`D. `2.303 cal` |
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Answer» Correct Answer - B `W = 2.303 nRT log.(V_(2))/(V_(1))` |
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| 680. |
Calculate `Deltah` for the eaction `BaCO_(3)(s)+2HCI(aq) rarr BaCI_(2)(aq)+CO_(2)(g)+H_(2)O(l)` `Delta_(f)H^(Theta) (BaCO_(3)) =- 290.8 kcal mol^(-1), Delta_(f)H^(Theta) (H^(o+)) =0` `Delta_(f)H^(Theta) (Ba^(++)) = - 128.67 kcal mol^(-1)`, `Delta_(f)H^(Theta)(CO_(2)) =- 94.05 kcal mol^(-1)`, `Delta_(f)H^(Theta) (H_(2)O) =- 68.32 kcal mol^(-1)` |
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Answer» `{:(underset(darr)BaCO_(3)(s)+2H^(o+)+2CI^(Theta)),(Ba^(2)+2CI^(Theta)+CO_(2)+H_(2)O):}` `DeltaH = (Delta_(f)H^(Theta)Ba^(2+) +Delta_(f)H^(Theta)CO_(2)+H_(2)O)-(Delta_(f)H^(Theta)BaCO_(3) +Delta_(f)H^(Theta)H^(o+))` `=(-128.67 - 94.05 -68.32) -[-290.8 xx2xx0)` `=-- 0.24 kcal` |
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| 681. |
`Delta_(f)H(H_(2)O) =- 68 kcal mol^(-1)` and `DeltaH` of neutralisation is `-13.7 kcal mol^(-1)`, then the heat of formation of `overset(Theta)OH` isA. `-68 kcal mol^(-1)`B. `-54.3 kcal mol^(-1)`C. `54.3 kcal mol^(-1)`D. `-71.7 kcal mol^(-1)` |
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Answer» `H^(o+) +overset(Theta)OH hArr H_(2)O` `DeltaH =- 13.7 kcal` `DeltaH^(Theta) = Delta_(f)H^(Theta) (H_(2)O) -[Delta_(f)H^(Theta)(H^(o+))+Delta_(f)H^(Theta) (overset(Theta)OH))]` `=- 68 - (0-13.7) =- 54.3 kcal mol^(-1)` `[Delta_(f)H^(Theta) (H^(o+)) =0 (convention)]` |
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| 682. |
The `C-H` bond of the side chain in toluene, `C_(6)H_(5)-CH_(3)`, has a dissociation energy of `77.5 kcal mol^(-1)`. Calculate `Delta_(f)H^(Theta)` of benzy`1` radical and the strength of the central bond in dibenzy`1 C_(6)H_(5)-CH_(2)-CH_(2)-C_(6)H_(5)` given that `Delta_(f)H^(Theta)` to toluene vapour in `12 kcal mol^(-1)` and that of dibenzy`1` vapour is `27.8 kcal mol^(-1). BE` of `H_(2) = 104 kcal mol^(-1)`. |
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Answer» `BE of H -H = 104 kcal mol^(-1)` `H -H =2H` `C_(6)H_(5)CH_(3)(g) rarr C_(6)H_(5)CH_(2)+H DeltaH = 77.5 ….(i)` `DeltaH_(1) = [Delta_(f)C_(6)H_(5) +Delta_(f)H H] -[ Delta_(f)H(Toluene)]` `2(C_(6)H_(5)CH_(2)) rarr underset(("Dibenzyl"))(C_(6)H_(5)-CH_(2) -CH_(2)-PH)` ....(ii) `DeltaH_(2) = 2_(f)H ("Dibenzyl") -2Delta_(f)H^(Theta) (C_(6)H_(5)CH_(2))` `Delta_(f)H^(Theta)H =(104)/(2) = 52` `[(1)/(2)H_(2) rarr H]` From reaction (i), `77.5 = (Delta_(f)H(C_(6)H_(5)CH_(2)) +52) -12` `rArr Delta_(f)H^(Theta)C_(6)H_(5)CH_(2) = 37.5 kcal mol^(-1)` From reaction (ii), `DeltaH_(2) = (27.8 -2 xx 37.5) = - 47.2 kcal mol^(-1)` The bond strength of `PhCH_(2) -CH_(2)Ph = 47.2 kcal mol^(-1)` |
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| 683. |
What is the bond enthalpy of `Xe-F` bond ? `XeF_(4)(g)rarrXe^(+)(g)+F^(-)(g)+F_(2)(g)+F(g)," "Delta_(r)H=292"kcal/mol"` Given : Ionization energy of `Xe=279"kcal/mol"` `B.E.(F-F) = 30"kcal/mol"`, Electron affinity of F = 85 kcal/molA. 24 kcal/molB. 34 kcal/molC. `8.5` kcal/molD. None of these |
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Answer» Correct Answer - B `Delta_(r)H` = [Heat supplied] - [Heat evolved] `292=[4x + 279] - [38 + 85]` implies x = 34 kcal/mol |
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| 684. |
If `C_(6)H_(12)O_(6)(s)+9O_(2)(g)rarr6CO_(2)(g)+6H_(2)O(g) , Delta H=-680` Kcal The weight of `CO_(2)(g)` produced when 170 Kcal of heat is evolved in the combustion of glucose is :-A. 265 gmB. 66 gmC. 11 gmD. 64 gm |
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Answer» Correct Answer - B `C_(6)H_(12)O_(6(s))+9O_(2(g))rarr 5CO_(2(g))+6H_(2)O_((g)) , Delta H=-680 KCal` . 680 Kcal heat is released when 6 mole `CO_(2)` is produced so, 170 KCal heat is released when `((6)/(680)xx170)` mole `CO_(2)` is produced. `therefore` Mass of `CO_(2)` produced `=(6)/(680)xx170xx44 g` `=66 g CO_(2)` |
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| 685. |
For an isothermal processA. `Delta H = 0`B. `Delta U = 0`C. `Delta T = 0`D. All of these |
| Answer» Correct Answer - D | |
| 686. |
From the following data, calculate the standard enthalpy of formation of propane `Delta_(f)H^(Theta) CH_(4) = - 17 kcal mol^(-1)` `Delta_(f)H^(Theta)C_(2)H_(6) =- 24 kcal mol^(-1), BE (C-H) = 99 kcal mol^(-1)` `(C- C) = 84 kcal mol^(-1)`. |
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Answer» `C(s) +2H_(2)(g) rarr CH_(4)(g), DeltaH_(1) = - 17 kcal …(i)` `2C(s)+3H_(2)(g) rarr C_(2)H_(6)(g), DeltaH_(2) =- 24 kcal …(ii)` ltbr. `3C(s) +4H_(2)(g) rarr C_(3)H_(8)(g), Delta_(f)H = ?....(iii)` From equations (i) and (ii), we get Let `x kcal mol^(-1)` is the energy of `C(s) rarr C(g)` Let `y kcal mol^(-1)` is the `BE` energy of `(H -H)` From equation (i), we get `-17 = x +2y - 4 xx 99` From equation (ii), we get `-24 = 2x +3y -(84 +6 xx 99)` SOlve for `x` and `y`, `x = 171 kcal, y = 104 kcal` From equation (iii), we get `Delta_(f)H = 3x +4y - [2 xx 84 +8 xx 99]` `=- 31 kcal mol^(-1)` |
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| 687. |
Calculate the C-C bond enthalpy from the following data: (a) `C(s) rarr C(g) , Delta H=170 kcal` (b) `(1)/(2)H_(2)(g)rarrH(g), Delta H= 52 kcal` (c) Heat of combustion of ethane `= -20 kcal` (d) C-H bond enthalpy `= 99 kcal`. |
| Answer» Correct Answer - 78 Kcal | |
| 688. |
Under which of the following condition is the relation `DeltaH = DeltaU +P DeltaV` valid for a closed system atA. Constant temperature, pressrure , and compositionB. Constant temperature and pressureC. Constant pressureD. Constant temperature |
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Answer» Correct Answer - C According to thermodynamics, `H = U + PV` `:. Delta H = Delta U + Delta (pV)` `= Delta U + p Delta V + V Delta p + Delta p Delta V` If `Delta p = 0` (constant pressure), than `Delta H = Delta U + p Delta V` |
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| 689. |
Which one of the following equations does not correctly respresents the first law of thermodynamics for the given process?A. Isothermal process, q = -WB. Expansion of a gas in vacuum, W = 0C. Adiabatic process, `Delta U` = qD. Process at constant volume, `Delta U` = q |
| Answer» Correct Answer - C | |
| 690. |
Which one of the following equations does not correctly respresents the first law of thermodynamics for the given process?A. Isothermal process: `q= -w`B. Cyclic process: q= -wC. Adiabatic process: `Delta E=q`D. Expansion of a gas into vaccum: `Delta E=q` |
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Answer» Correct Answer - C Accoring to 1 st law of thermodynamics, `Delta E=q+w`. For isothermal process, `Delta E=0`. Hence, `q= -w` For cyclic process, `Delta E=0` Hence, For expansion into vaccum, `w=0`. Hence `Delta E=q`. |
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| 691. |
Given that standard heat enthalpy of `CH_(4), C_(2)H_(4)` and `C_(3)H_(8)` are -17.9, 12.5, -24.8 Kcal/mol. The `Delta H` for `CH_(4)+C_(2)Hrarr C_(3)H_(8)` is :A. `-55.2` KcalB. `-30.2` KcalC. 55.2 KcalD. `-19.4` Kcal |
| Answer» Correct Answer - D | |
| 692. |
Which of the following equations respresents standard heat of formation of `CH_(4)` ?A. `C_(("diamond"))+2H_(2(g))rarr CH_(4(g))`B. `(C_(("graphite"))+2H_(2(g))rarr CH_(4(g))`C. `(C_(("diamond"))+4H_((g))rarr CH_(4(g))`D. `C_(("graphite"))+4H_((g))rarr CH_(4(g))` |
| Answer» Correct Answer - B | |
| 693. |
The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :A. `46.0 KJ mol^(-1)`B. `92.0 KJ mol^(-1)`C. `-23.0 KJ mol^(-1)`D. `-92.0 KJ mol^(-1)` |
| Answer» Correct Answer - B | |
| 694. |
The enthalpy of formation of ammonia is `- 46.2 mol^(-1)`. The enthalpy change for the reaction `2NH_(3) rarr N_(2) + 3 H_(2)` isA. `42 kJ`B. `64 kJ`C. `80 kJ`D. `92 kJ` |
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Answer» Correct Answer - D The enthalpy of formation of ammonia describes the following thermochemical equation: `(1)/(2) N_(2) (g) + (3)/(2) H_(2) (g) rarr 2 NH_(3)` To get the enthalpy change for the given reaction, we need to reverse it and then multiply it by 2. Thus, `Delta H = - (2) (46.2) = 92.4 kJ` |
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| 695. |
The enthalpy change `(DeltaH)` for the reaction, `N_(2(g))+3H_(2(g)) rarr 2NH_(3g)` is `-92.38kJ` at `298K` What is `DeltaU` at `298K ?` |
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Answer» `DeltaH` and `DeltaU` are related as `DeltaH = DeltaU +Deltan_(g)RT` For the reaction, `N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g)` `Deltan_(g) = 2 -(1+3) =- 2mol,T = 298K` `DeltaH =- 90.00 kJ, R = 8.314 J K^(-1) mol^(-1)` `-90000 = DeltaU +(-2mol) xx (8.314 J mol^(-1)K^(-1)) xx (298K)` `-9000 = DeltaU - 4955` `DeltaU =- 90000 + 4955 = 85045J =- 85.045 kJ` |
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| 696. |
The enthalpy and entropy changes for the reaction C(s) diamond `+ O_(2) rarr CO_(2)(g)` at `25^(@)C` and 1 atm. Are -393.4 kJ `mol^(-1)` and 0.006 `kJ mol^(-1)` respectively . Is the conversion of diamond to `CO_(2)` at room temperature a spontaneous process ? |
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Answer» Correct Answer - Yes `DeltaG = DeltaH - T DeltaS = -393.4kJ mol^(-1) - 298 K ( 0.006kJ K^(-1)mol^(-1) )= -ve.` |
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| 697. |
For a given reaction, `DeltaH = 35.5kJ mol^(-1)` and `DeltaS = 83.6 JK^(-1) mol^(-1)` .The reaction s spontaneous at `:` ( Assume at `DeltaH ` and `DeltaS` do not vary with temperature )A. `T lt425 K`B. `T gt 425 K`C. All temperaturesD. `T gt 298 K` |
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Answer» Correct Answer - B `DeltaS= DeltaH - T DeltaS` For the reaction to be spontaneous , `DeltaG = - ve` As`DeltaH` and`DeltaS` both are positive,`DeltaG` can be `-ve` only if`T DeltaS gt DeltaH` or`T gt ( DeltaH)/( DeltaS)` `i.e,T gt ( 35.5xx 1000J)/( 83.6JK^(-1)) ` or `T gt 425 K` |
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| 698. |
Enthalpy of neutralisation of acetic acid by `NaOH` is `-50.6 kJ mol^(-1)`. Calculate `DeltaH` for ionisation of `CH_(3)COOH`. Given. The heat of neutralisation of a strong acid with a strong base is `-55.9 kJ mol^(-1)`. |
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Answer» The neutrlisation of a strong acid by a strong base is represented by: `H^(o+) (aq) +overset(Theta)OH(aq) rarr H_(2)O(l)` `DeltaH =- 55.9 kJ ….(i)` We have to calculate: `CH_(3)COOH rarr CH_(3)COO^(Theta) +H^(o+), DeltaH = ?` Given: `CH_(3)COOH +overset(Theta)OH rarr CH_(3)COO^(Theta) +H_(2)O` `DeltaH_(2) = - 50.6` Operate: `:. (ii) -(i)` `:. DeltaH = DeltaH_(2) - DeltaH_(1)` `=- 50.6 -(-55.9) = 5.3 kJ mol^(-1)` |
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| 699. |
Calculate the enthalpy of reaction for`CO(g) + (1)/(2) O_(2)(g) rarr CO_(2)(g)` Given `C(s) + O_(2) (g) rarr CO_(2)(g) , DeltaH = -393.5 kJ mol^(-1)` `C(s) + (1)/(2) O_(2)(g) rarr CO(g) Delta H = - 110 .5 kJ mol^(-1)` |
| Answer» Correct Answer - `-283kJ mol^(-1)` | |
| 700. |
The enthalpy of neutralisation of HCl by NaOH IS `-55.9 kJ ` and that of HCN by NaOH is `-12.1 kJ mol^(-1)`. The enthalpy of ionisation of HCN isA. `-43.8 KJ`B. `43.8 KJ`C. 68 KJD. `-68` KJ |
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Answer» Correct Answer - B `HCl+NaOH rarr NaCl + H_(2)O` OR `H^(+)+OH^(-)rarr H_(2)O , Delta H = -55.9 KJ mol^(-1)` …(i) `HCN+NaOHrarr NaCN+H_(2)O` OR `HCN+OH^(-)rarr CN^(-)+H_(2)O , Delta H=-121.1 KJ mol^(-1)` …(ii) `HCN rarr H^(+)+CN^(-) , Delta H = ?` eq(iii) = eq(ii) - eq(i) `Delta H_(3)=-12.1-(-55.9)=43.8 KJ` |
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