Explore topic-wise InterviewSolutions in .

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.

101.

Calculate the enthalpy change when infinitely solution of `CaCl_(2)` and `Na_(2)CO_(3)` are mixed `Delta_(f)H^(@)` for `Ca^(2+)(aq, CO_(3)^(2-) (aq)` and `CaCO_(3) (s)` are `-129.80, -161.65, -288.5" kCal mol"^(-1)` respectively.

Answer» Correct Answer - 2.95 kCal
`Ca^(+2) + CO_(3)^(-2) rarr CaCO_(3)`
`Delta H = - 288.5 - [-129.8 - 161.65] = 2.95 Kcal`
102.

A swimmer coming out from a pool is covered with a film of water weighing about `80g`. How much heat must be supplied to evaporate this waateer ? If latent heat of evaporation for `H_(2)O` is `40.79 kJ mol^(-1)` at `100^(@)C`.

Answer» Amount of water present on the body of swimmer
`=80g=(80)/(18)mol e`
Thus, heat required to evaporate `(80)/(18) mol e`
`H_(2)O=40.79xx(80)/(18)=181.2kJ`
103.

When `0.1025 g` of benzoic acid was burnt in a bomb calorimeter the temperature of the calorimeter increased by `2.165^(@)C`. For benzoic acid `DeltaH_("comb")^(@)= -3227 kJ mol^(-1)`. Calculate the heat capacity of the calorimeter.

Answer» The equation for the combustion is ,
`C_(7)H_(6)O_(2)(s) + 7.5 O_(2)(g) rarr 7CO_(2)(g) + 3H_(2)O(l), " " Delta H^(@) = 3227 kJ`
Since `7.5` moles of `O_(2)` gas is needed, and 7 moles of `CO_(2)` is produced, some pressure-volume work is done, to the calorimeter:
`PV = Delta n_(g) R T`, where `Delta n =(7-7.5) = -0.5 mol`
`Delta E = Delta H - Delta Hn_(g) RT`
`= -3227 - (0.5 xx 8.314298 xx 298)`
`= -3226kJ//mol` (a small correction)
The amount of heat produced by 0.1025 g benzoic acid is
`q = 0.1025//122.13 xx 3226 = 2.680 kJ`
Thus, the heat capacity is
`C = q_(v)//Delta T = 2.680//2.165 = 1.238 kJ//K`.
After the heat capacity is determined, the calorimeter is ready to be used to measure the enthalpy of combustion of other substances.
104.

The heat of combustion of ethanol was determined in a bomb calorimeter and was found to be `-670.48kcal mol^(-1)` at `25^(@)C`. What will be `DeltaU` for the same reaction at `298K` ?A. `-2802.6kJ`B. `-669.28cal`C. `-670.48kcal`D. `-280.26xx10^(4)kJ`

Answer» Correct Answer - C::D
105.

Assuming that water vapour is an ideal gas, the internal energy change `(DeltaU)` when 1 mole of water is vaporised at `1 ba r` pressure and` 100^(@)C, (` given`:` molar enthalpy of vaporization of water `41kJ mol^(-1)` at `1 ba r` and `373 K ` and `R=8.3 J mol^(-1)K^(-1))` will be `:`A. `4.100kJ mol^(-1)`B. `3.7904kJ mol^(-1)`C. `37.904kJ mol^(-1)`D. `41.00kJ mol^(-1)`

Answer» Correct Answer - C
`H_(2)O_((l))overset("vaporisation")rarr H_(2(g))`
`Deltan_(g)=1-0=1`
`DeltaH+Deltan_(g)RT`
`DeltaU=DeltaH-Deltan_(g)RT=41.8.3xx10^(-3)xx373`
`=37.9kJ mol^(-1)`
106.

How much heat is necessary to convert `100g` ice at `0^(@)C` to water vapour `(` steam`)` at `100^(@)C` ? Given `DeltaH_(fusion)=80kcal//kg, DeltaH_(vap)=540kcal//kg,`heat capacity `=1.0 kcal//kg.`

Answer» `underset(0^(@)C)(Ice) underset(Fusion)overset(DeltaH_(1))rarrunderset(0^(@)C)(Water)overset(DeltaH_(2))rarr underset(100^(@)C)(Water)underset(V a p o risation) overset(DeltaH_(3))rarr underset(100^(@)C)(Steam)`
Thus, total heat needed for desired change
`=DeltaH_(1)+DeltaH_(2)+DeltaH_(3)`
`:. DeltaH_(Tot al)=0.1xx80+0.1xx1.0xx100+0.1xx540`
`=8.0+10.0+54.0=72.0kcal`
107.

`100 ml 0.25 M H_(2)SO_(4)` (strong acid) is neutralised with `200` ml `0.2M NH_(4)OH` in a constant pressure Calorimeter which results in temperature rise of `1.4^(@)C`. If heat capacity of Calorimeter constent is `1.5 kJ//^(@)C`. Which statement is/are correct `{:("Given :",HCl+NaOH rarr NaCl+H_(2)O+57 kJ),(,CH_(3)COOH+NH_(4)OH rarr CH_(3)COONH_(4)+H_(2)O+48.1 kJ):}`A. Enthalpy of neutralisation of `HCl` v/s `NH_(4)OH` is `-52.5` kJ/molB. Enthalpy of dissociation (ionization) of `NH_(4)OH` is `4.5` kJ/molC. Enthalpy of dissociation of `CH_(3)COOH` is `4.6` kJ/molD. `DeltaH` for `2H_(2)O(l) rarr 2H^(+) (aq.)+2OH^(-) (aq.)` is `114` kJ

Answer» Correct Answer - A::B::D
108.

Two solutions initially at `25^(@)C` were mixed in an adiabatic constant pressure calorimeter. One contains 400 ml of 0.2 M weak monoprotic acid solution. The other contain 100 ml of 0.80 M NaOH. After mixing temperature increased to `26.2^(@)C`. How much heat is evolved in the neutralization of 1 mole of acid? Assume density of solution `1.0 g//cm^(3)`, and specific heat of solution 4.2 J/g-K. Neglect heat capacity of the Calorimeter.

Answer» Correct Answer - `-31.5` kJ/mole
Mass= density `xx` vol. `=1xx500=500` gram
`DeltaT=1.2`
`q=mc.DeltaT=2520`
Heat released by 1 mole of acid `=2520/0.08=-31500 J=-31.5 kJ`
109.

The standard enthalpies of formation of `CO_(2)(g), H_(2)O(l)` and glucose (s) at `25^(@)C` are `-400` kJ/mol, `-300` kJ/mol and `-1300` kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at `25^(@)C` is :A. `+2900 kJ`B. `-2900 kJ`C. `-16.11 kJ`D. `+16.11 kJ`

Answer» Correct Answer - C
`C_(6)H_(12)O_(6) (s)+6O_(2)(g) rarr 6CO_(2)(g) +6H_(2)O(l)`
`DeltaH_("comb. "C_(6)H_(12)O_(6))^(@)=6 DeltaH_(f CO_(2))^(@)+6DeltaH_(f H_(2)O(l))^(@)-DeltaH_(f C_(6)H_(12)O_(6)(s))^(@)`
`=6xx(-400)+6 xx(-300) -(-1300)`
`=-2900" kJ/mol"`
`:.` Stadard enthalpy of combustion per gram of glucose `=(-2900)/180=-16.11 kJ`
110.

Specific heat of `Li(s),Na(s),K(s),Rb(s)` and `Cs(s)` at `398K` are `3.57,1.23,0.756,0.363` and `0.242 J g^(-1)K^(-1)` respectively. Compute the molar heat capacity of these elements and identify any periodic trend. If there is trend, use it to predict the molar heat capacity of `Fr`.

Answer» Molar heat capacity `=at.wt. xx ` specific heat.
Molar heat capacity `( i n J mol^(-1) K^(-1))` of `:`
`{:(Li(s),=,3.57xx6.94,=,24.78),(Na(s),=,1.23xx22.99,=,28.28),(K(s),=,0.756xx39.10,=,29.56),(Rb(s),=,0.363xx85.47,=,31.03),(Cs(s),=,0.242xx132.91,=,32.16):}` ltbr. There is a trend plotting these values with atomic number, the extrapolation of graph gives the value of `Fr(s0`
`=33.5J mol^(-1)K^(-1)`
111.

`(a)` Calculate the energy needed to raise the temperature of `10.0g` of iron from `25^(@)C` to `500^(@)C` if specific heat capacity of iron if `0.45J( .^(@)C)^(-1)g^(-1)` `(b)` What mass of gold `(` of specific heat capacity `0.13J ( .^(@)C)^(-1)g^(-1)` can be heated can be heated through the same temperature difference when supplied with the same amount of energy as in `(a)` ?

Answer» `(a)q=m.s.DeltaT=10xx0.45xx(773-298)J`
`=2137.5J`
`q=m.s.DeltaT`
`[`Given `=2137.5,T=(773-2980=475,s=0.13]`
`2137.5=mxx0.13xx475`
`m=34.61g`
112.

Find `Delta_(r)U^(@)` for the reaction `4HCl (g) + O_(2) (g) rArr 2Cl_(2) (g) + 2H_(2)O (g)` at 300 K. Assume all gases are ideal `{:("Given",H_(2) (g) + Cl_(2) (g) rarr 2HCl (g),Delta_(r) H_(300)^(@) - 184.5 kJ//"mole",),(,2H_(2) (g) + O_(2) (g) rarr 2H_(2)O (g),Delta_(r) H_(300)^(@) = - 483 kJ//"mole" ("Use" R = 8.3 J//"mole"),):}`A. 111.5 kJ/moleB. `-109.01 kJ//"mole"`C. `-111.5 kJ//"mole"`D. None

Answer» Correct Answer - C
`4HCl+O_(2) square hArr 2Cl_(2)+2H_(2)O" "DeltaH=-114`
`DeltaU=DeltaH-Deltan_(g)RT=-114+(1xx8.314xx10^(-3) xx300)`
`=-111.5" kJ mol"^(-1)`
113.

Diborane is a potential rocket fuel that undergoes combustion according to the reaction, `B_(2)H_(6)(g) +3O_(2)(g) rarr B_(2)O_(3)(g) +3H_(2)O(g)` From the following data, calculate the enthalpy change for the combustion of diborane:

Answer» `B_(2)H_(6(g))+3O_(2(g))rarr B_(2)O_(3(s))+3H_(2)O_((g)), DeltaH=?`
Multiply `eq.(2)` and `(3)` by 3, add `eqs.(1),(2) and (3)` and then subtract `eq.(4)` from that
`2B_((s))+(3)/(2)O_(2(g))rarr B_(2)O_(3(s)), DeltaH=-1273kJmol^(-1).....(1)`
`3H_(2(g))+(3)/(2)O_(2(g)) rarr 3H_(2)O_((l)), DeltaH=-286xx3 ....(2)`
`ul (3H_(2)O_((l)) rarr 3H_(2)O_((g)), DeltaH=44xx3 .....(3))`
`2B_((s))+3H_(2(g))+3O_(2(g)) rarr B_(2)O_(3(s))+3H_(2)O_((g)),DeltaH=-1999`
`ul ({:(2B_((s)),+3H_(2(g)),rarr,B_(2)H_(6(g)),,DeltaH=36),(-,-,,-," -"):})`
`B_(2)H_(6(g))+3O_(2(g))rarr B_(2)O_(3(s))+3H_(2)O_((g)), DeltaH=-2035kJ mol^(-1)`
114.

Given the following standard heats of reactions: (a) heat of formation of water `= -68.3 kcal`, (b) heat of combustion of `C_(2)H_(2) =- 310.6 kcal`, (c ) heat of combustion of ethylene `=- 337.2 kcal`. Calculate the heat of reaction for the hydrogenation of acetylene at constant volume and at `25^(@)C`.

Answer» Correct Answer - `-41.104kcal;`
115.

The heats of combustion of `C_(2)H_(4(g)), C_(2)H_(6(g))` and `H_(2(g))` are `-1405,-1558.3` and `-285.6 kJ` respectively. Calculate heat of hydrogenation of ethylene.

Answer» `C_(2)H_(4(g))+H_(2(g)) rarrC_(2)H_(6(g)),DeltaH=?`
Given,
`C_(2)H_(4(g))+3O_(2(g))rarr 2CO_(2(g))+2H_(2)O_((g)), DeltaH=-1409.5kJ ...(1)`
`C_(2)H_(6)+(7)/(2)O_(2(g))rarr 2CO_(2(g))+3H_(2)O_((g)), DeltaH=-1558.3kJ ..(2)`
`H_(2)+(1)/(2)O_(2(g)) rarrH_(2)O_((g)),DeltaH=-285kJ ......(3)`
Adding `eqs. (1)` and `(3)`
`C_(2)H_(4(g))+(7)/(2)O_(2(g))+H_(2(g)) rarr 2CO_(2(g))+3H_(2)O_((g)), DeltaH=-1695.1kJ ....(4)`
Subtracting `eq. (2)` from `(4)`
`ul( {:(C_(2)H_(6(g)),+(7)/(2)O_(2(g)),rarr,2CO_(2(g)),+3H_(2)O_((g)),,DeltaH=-1558.3kJ .....(5)),(-,-,,-,-," +"):} ) `
`C_(2)H_(4(g))+H_(2(g))rarr C_(2)H_(6(g)), DeltaH=-136.8kJ`
`:.` Heat of hydrogenation of `C_(2)H_(4)=-136.8kJ`
116.

The heat released on neutralisation of a strong alkali say `NaOH` with strong acid is `13.7kcal//mol.` The heat released on neutralisation of `NaOH` with `HF` if `-16.4 kcal//mol.` Calculate `DeltaH^(@)` of ionisation of `HF` in water.

Answer» `OH_((aq.))^(-)+H_((aq.))^(+)rarr H_(2)O_((aq.)), ....(1)`
`DeltaH^(@)=-13.7kcal`
`OH_((aq.))^(-)+HF_((l)) rarr H_(2)O+F_((aq.))^(-), ....(2)`
`DeltaH^(@)=-16.4kcal`
By `eq. (2)-(1) HF rarr H_((aq.))^(+)+F_((aq.))^(-),`
`DeltaH^(@)=-2.7kcal`
The negative value is due to high heat of hydration of `F^(-)` ion on account of smaller size.
117.

Heat of neutralisation of `HF` is `:`A. `57.32kJ`B. `gt57.32kJ`C. `lt57.32kJ`D. none of these

Answer» Correct Answer - B
Due to extensive solvation of `F^(-)` ion on account of smaller size, observed value of heat of neutralisation of `HF` appears more.
118.

Given that `,` `A(s) rarr A(l)DeltaH=x` `A(l) rarr A(g), DeltaH=y` The heat of sublimation of `A` will be `:`A. `x-y`B. `x+y`C. `x or y`D. `-(x+y)`

Answer» Correct Answer - B
`A_((s))rarr A_((g)),DeltaH=(x+y)`.
119.

Calculate the resonance energy of `C_(6)H_(6)` using Kerkule formula for `C_(6)H_(6)` from the following data. `(i)` `DeltaH_(f)^(@) fo r C_(6)H_(6)=-358.5kJ mol^(-1)` `(ii)` Heat of atomisation of `C=716.8kJ mol^(-1)` `(iii)` Bond energy of `C-H,C-C,C=C` and `H-H` are `490,340,620,436.9kJ mol^(-1)` respectively.

Answer» Correct Answer - `-150.0kJ mol^(-1);`
120.

Given that `:` `2CO_((g))+O_(2(g))rarr2CO_(2(g)),DeltaH^(@)=-P kJ` `C_((s))+O_(2(g))rarr CO_(2(g)), DeltaH^(@)=-Q kJ` the enthalpy of formation of carbon monoxide is `:`A. `(P-Q)/(2)`B. `2Q-P`C. `P-2Q`D. `(P-2Q)/(2)`

Answer» Correct Answer - D
`2CO_((g))+O_(2(g))rarr2CO_(2(g)),DeltaH_(1)=-PkJ`
`C_((s))+O_(2(g))rarr CO_(2(g)),DeltaH_(2)=-QkJ`
`C_((s))+``^(1/2)O_(2(g))rarrCO_((g)),DeltaH_(3)=?`
`DeltaH_(3)=DeltaH_(2)-1//2DeltaH_(1)`
`=-Q-(-(P)/(2))=-Q+(P)/(2)`
`=(P-2Q)/(2)`
121.

The specific heat at constant volume for a gas `0.075cal//g` and at constant pressure is `0.125cal//g` Calculate `:` `(i)` The molecular weight of gas, `(ii)` Atomicity of gas, `(iii)` No. of atoms of gas in its 1 mole.

Answer» Correct Answer - `(i)40, (ii) ` Gas is monoatomic.
`(iii) 6.023xx10^(23)at oms.`