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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. |
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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` |
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| 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`. |
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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` |
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| 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. |
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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. |
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| 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` |
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Answer» Correct Answer - C::D |
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| 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)` |
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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)` |
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| 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.` |
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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` |
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| 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. |
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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` |
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| 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` |
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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` |
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| 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`. |
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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)` |
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| 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)` ? |
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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` |
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| 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 |
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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)` |
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| 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: |
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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)` |
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| 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`. |
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Answer» Correct Answer - `-41.104kcal;` |
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| 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. |
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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` |
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| 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. |
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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. |
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| 117. |
Heat of neutralisation of `HF` is `:`A. `57.32kJ`B. `gt57.32kJ`C. `lt57.32kJ`D. none of these |
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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. |
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| 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)` |
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Answer» Correct Answer - B `A_((s))rarr A_((g)),DeltaH=(x+y)`. |
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| 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. |
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Answer» Correct Answer - `-150.0kJ mol^(-1);` |
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| 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)` |
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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)` |
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| 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. |
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Answer» Correct Answer - `(i)40, (ii) ` Gas is monoatomic. `(iii) 6.023xx10^(23)at oms.` |
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