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401.

Combustion of sucrose is used by aerobic organisms for providing energy for the life sustaining process. If all the capturing of energy from the reaction is done through electrical process (non P-V work), then calculate, maximum available energy which can be captured by combustion of 34.2 g of sucrose : (Given : `DeltaH_("combustion")("sucrose")=-6000kJmol^(-1)` `DeltaS_("combustion")=180j//K-mol` and bodyntemperature is 300 K)A. 600 kJB. 594.6 kJC. 5.4 kJD. 605.4 kJ

Answer» Correct Answer - D
Discussion
402.

Combustion of sucrose is used by aerobic organisms for providing energy for the life sustaining process. If all the capturing of energy from the reaction is done through electrical process (non P-V work), then calculate, maximum available energy which can be captured by combustion of 34.2 g of sucrose : (Given : `DeltaH_("combustion")("sucrose")=-6000kJmol^(-1)` `DeltaS_("combustion")=180j//K-mol` and bodyntemperature is 300 K)A. 600 kJB. `594.6` kJC. `5.4` kJD. 605.4 kJ

Answer» Correct Answer - D
No. of moles of surcose `=(34.2)/(342)=0.1`
`-(DeltaG)_(T,P)` = useful work done by the system
`DeltaG=-DeltaH+T.DeltaS`
`=+(6000xx0.1)+(180xx0.1xx300)/(1000)`
= 605.4 kJ
Discussion
403.

Boron can undergo the following reactions with the given enthalpy changes : `2B(s)+(3)/(2)O_(2)(g)rarrB_(2)O_(3)(s)," "DeltaH=-1260 " kJ"` `2B(s)+3H_(2)(g)rarrB_(2)H_(6)(g)," "DeltaH=30 " kJ"` Assume no other reactions are occuring. If in a container (operating at constant pressure) which is isolated from the surrounding, mixture of `H_(2)` (gas) and `O_(2)` (gas) are passed over excess of B(s), then calculate the molar ratio `(O_(2) : H_(2))` so that temperature of the container do not change :A. `15 : 3`B. `42 : 1`C. `1 : 42`D. `1 : 84`

Answer» Correct Answer - D
No. of moles of `O_(2)` required to supplied 30 kJ heat to second reaction
`=(30)/(1260)xx(3)/(2)=(1)/(28)`
`Son_(O_(2)):n_(H_(2))=(1)/(28):3 " or "1 : 84`
Discussion
404.

Determine the enthalpy of formation of `B_(2)H_(6)`(g) in kJ/mol of the following reaction : `B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(3)(s)+3H_(2)O(g)`, Given : `Delta_(r )H^(@)=-1941 " kJ"//"mol", " "DeltaH_(f)^(@)(B_(2)O_(3),s)=-1273" kJ"//"mol,"` `DeltaH_(f)^(@)(H_(2)O,g)=-241.8 " kJ"//"mol"`A. `-75.6`B. `+75.6`C. `-57.4`D. `-28.4`

Answer» Correct Answer - C
`-1941 =-1273 + 3 (-241.8)-(Delta_(f)H)B_(2)H_(6)`
`(Delta_(f)H)_(B_(2)H_(6))=-1273-725.4 + 1941 =-57.4`
Discussion
405.

The relation between change in internal energy`(DeltaE)`change in enthalpy`(DeltaH)` and work done`(W)`is represented asA. 4 kJB. 5 kJC. 3 kJD. zero

Answer» Correct Answer - D
`Delta H _(35) - DeltaH_(25) = 0`
Discussion
406.

Assertion. For the reaction ,`2 NH_(3)(g) rarr N_(2)(g) + 3H_(2)(g) , DeltaH gt DeltaE` Reason . Enthalpy change is always greater than internal energy change.A. If both A and R are true,andR is the true explanation of A.B. If both A and R are true,but R is not the true explanation of A.C. If A is true, but R is falseD. If both A and R are false.

Answer» Correct Answer - c
Discussion
407.

The volume of gas is reduced to half from its original volume. The specific heat will beA. reduce to halfB. be doubledC. remains constantD. increases four times

Answer» Correct Answer - C
Discussion
408.

The relation between change in internal energy`(DeltaE)`change in enthalpy`(DeltaH)` and work done`(W)`is represented asA. 30.5 kJB. 33 kJC. `-28 kJ`D. `-30.5 kJ`

Answer» Correct Answer - D
`DeltaG_(30) = Delta H_(30 + T. (d)/(dT) (Delta G)_(p)`
Discussion
409.

In the reaction `AB_(2)(l)+3X_(2)(g)hArrAX_(2)(g)+2BX_(2)(g) DeltaH=-2700` kcal per mole. Of `AB_(2)(l)`. The entalpies of formation of `AX_(2)(g)` and `BX_(2)(g)` are in the ratio of `4 : 3` and have opposite sign. The value of `DeltaH_(f)^(@)(AB_(2)(l))=30 kcal//mol`. Then :A. `DeltaH_(f)^(@)(AX_(2))=-96kcal//mol`B. `DeltaH_(f)^(@)(BX_(2))=+480kcal//mol`C. `K_(p)=K_(c)` and `DeltaH_(f)^(@)(AX_(2))=+480kcal//mol`D. `K_(p)=K_(c) " RT"` and `DeltaH_(f)^(@)(AX_(2))+DeltaH_(f)^(@)(BX_(2)=-240kcal//mol`

Answer» Correct Answer - C
Discussion
410.

The volume of gas is reduced to half from its original volume. The specific heat will beA. reduce to halfB. be doubledC. remain constantD. increase four times

Answer» Correct Answer - C
The specific het of a substance is the heat required to raise the temperature of 1 gram of a substance by one degree (1K or `1^(@)C`). It is an intensive property and is independent of the volume of the substance
Discussion
411.

Under which of the following conditionis the relation `DeltaH = DeltaE+ P DeltaH` valid for a closed system?A. constant pressureB. constant temperatureC. constant temperature and pressureD. constant temperature, pressure and composition

Answer» Correct Answer - A
`H=E+PV :. DeltaH = DeltaE + P DeltaV + V DeltaP `. If P is constant, i.e., `DeltaP =0`,only then `DeltaH = DeltaE + P DeltaV`
Discussion
412.

For the auto-ionization of water at `25^(@)C, H_(2)O(l)iff H^(+)(aq)+OH^(-)` (aq) equilibrium constant is `10^(-14)`. What is `DeltaG^(@)` for the process?A. `~=8xx10^(4)J`B. `~=3.5xx10^(4)J`C. `~=10^(4)J`D. None of these

Answer» Correct Answer - A
`DeltaG^(@)=-RTln K_(w)`
`=-8.314 xx 298 xx 2.303 (-14)`
`~=80000J//"mol"`
Discussion
413.

In the reaction `AB_(2)(I) + 2X_(2) (g) hArr AX_(2) (g) + BX_(2) (g) Delta H = -270` k cal per mol. Of `AB_(2) (I)` the enthalpies of formation of `AX_(2) (g)` & `BX_(2) (g)` are in the ratio of `4 : 3` and have opposite sign. The value of `Delta H_(f)^(0) (AB_(2)(I)) = + 30` k cal/mol. ThenA. `Delta H_(f)^(0) (AX_(2)) = - 96` kcal/molB. `Delta H_(f)^(0) (BX_(2)) = + 480` k cal/molC. `K_(P) = K_(c) " & " DeltaH_(f)^(0) (AX_(2)) = + 480 kcal//mol`D. `K_(P) = K_(c) RT " & " DeltaH_(f)^(0) (AX_(2)) + Delta H_(f)^(0) (BX_(2)) = -240 kcal//mol`

Answer» Correct Answer - C
`AB_(2) (l) + 4X_(2) (g) hArr AX_(2) (g) + 2BX_(2) (g)`
`Delta H = -270, -270 = - 2(3x) + 4x -30`
Discussion
414.

During complete combustion of one mole of butane, 2658kJ of heat is released. The thermochemical reaction for above changeisA. `2C_(4)H_(10)(g)+13O_(2)(g) rarr 8 CO_(2)(g) +10 H_(2)O(l) Delta_(c) H=- 2658.0 kJ mol^(-1)`B. `C_(4)H_(10)(g)+ (13)/(2) O_(2)(g) rarr4CO_(2)(g)+5H_(2)O(g) Delta_(c)H=- 1329.0 kJ mol^(-1)`C. `C_(4)H_(10)(g)+ (13)/(2) O_(2)(g) rarr4CO_(2)(g)+5H_(2)O(l) Delta_(c)H=- 2658.0 kJ mol^(-1)`D. `C_(4)H_(10)(g)+ (13)/(2) O_(2)(g) rarr4CO_(2)(g)+5H_(2)O(l) Delta_(c)H=+ 2658.0 kJ mol^(-1)`

Answer» Correct Answer - c
We have to take the combustion of one mole of `C_(4)H_(10)` and `Delta_(c) H` should be `-ve` and have a value of 2658 kJ mol`^(-1)`
Discussion
415.

During complete combustion of one mole of butane ,2658Kj of heat is released. The thermochemical reaction for above change isA. `2C_(4)H_(10(g))+130_(2(g)) rarr 8CO_(2(g))+10H_(2)O_((l)) Delta H_(c) = -2658.0 kJ mol^(-1)`B. `C_(4)H_(10(g))+(13)/(2)O_(2(g)) rarr4CO_(2(g))+5H_(2)O_(l) Delta H_(c) = -1329.0 kJ mol^(-1)`C. `C_(4)H_(10(g))+(13)/(2)O_(2(g)) rarr 4CO_(2(g))+5H_(2)O_((l)) DeltaH_(c) = -2659.0kJ mol^(-1)`D. `C_(4)H_(10(g))+(13)/(2)O_(2(g)) rarr4CO_(2(g))+5H_(2)O_((l)) Delta H_(c) = +2658.0 kJ mol^(-1)`

Answer» Correct Answer - C
Exothermic reaction for combustion of one mole of butane is represented as :
`C_(4)H_(10(g))+(13)/(2)O_(2(g)) rarr 4CO_(2(g))+5H_(2)O_((l))`
`Delta H_(c) = -2658.0 kJ mol^(-1)`
Discussion
416.

During complete combustion of one mole of butane ,2658Kj of heat is released. The thermochemical reaction for above change isA. `2C_(4) H_(10) (g) + 13 O_(2) (g) rarr 8 CO_(2)(g) + 10 H_(2) O (l), Delta_(c) H = -2658.0 kJ mol^(-1)`B. `C_(4)H_(10) (g) + (13)/(2) O_(2) (g) rarr 4CO_(2) (g) + 5H_(2)O (l) , Delta_(c) H = - 1329.0 kJ mol^(-1)`C. `C_(4)H_(10) (g) + (13)/(2) O_(2)(g) rarr 4CO_(2) (g) + 5H_(2) O (l), Delta_(c) H = - 2658.0 kJ mol^(-1)`D. `C_(4) H_(10) (g) + (13)/(2) O_(2) (g) rarr 4CO_(2) (g) + 5H_(2) O (l) , Delta_(c) H = + 2658.0 kJ mol^(-1)`

Answer» Correct Answer - C
Give that, the complete combustion of one mole of butane is represented by thermochemical reaction as
`C_(4) H_(10) (g) + (13)/(2) O_(2) (g) rarr 4CO_(2) (g) + 5H_(2) O (l)`
We have to take the combustion of one mole of `C_(4)H_(10)` and `Delta_(c)H` should be negative and have a value of `2658 kJ mol^(-1)`
Discussion
417.

During complete combustion of one mole of butane ,2658Kj of heat is released. The thermochemical reaction for above change isA. `2C_(4)H_(10)(g)+130_(2)(g)to8CO_(2)(g)+10H_(2)O(l)` `Delta_(c)H=-2658.0KJmol^(-1)`B. `C_(4)H_(10)(g)+(13)/(2)O_(2)(g)to4CO_(2)(g)+5H_(2)O(g)` `Delta_(c)H=-1329.0KJmol^(-1)`C. `C_(4)H_(10)(g)+(13)/(2)O_(2)(g)to4CO_(2)(g)+5H_(2)O(l)` `Delta_(c)H=-2658.0KJmol^(-1)`D. `C_(4)H_(10)(g)+(13)/(2)O_(2)(g)to4CO_(2)(g)+5H_(2)O(l)` `Delta_(c)H=-2658.0KJmol^(-1)`

Answer» Correct Answer - c
Discussion
418.

On the basis of thermochemical equations (i), (ii) and(iii) , find out which of the algebric relationships given in options (a) to (d) is correct. (i) C(graphite) `+O_(2)(g) rarr CO_(2)(g), Delta_(r)H = xkJ mol^(-1)` (ii) C(graphite) `+ (1)/(2) O_(2)(g) rarr CO(g), Delta_(r)H= y kJ mol^(-1)` (iii) `CO(g)+(1)/(2)O_(2)(g) rarrCO_(2)(g), Delta_(r)H = z kJ mol^(-1)`A. `z= x+y`B. `x= y-z`C. `x=y+z`D. `y= 2z-x`

Answer» Correct Answer - c
Eqn. (i) - Eqn. (ii)=Eqn. (iii) ,.Hence, `x-y = z`or `x= y + z`
Discussion
419.

If the entropy of vaporisation of a liquid is 110 `JK^(-1)mol^(-1)` and its enthalpy of vaporisation is 50 kJ `mol^(-1)`. The boiling point of the liquid isA. 354.5 KB. 454.5 KC. 554.5 KD. 445.5 K

Answer» Correct Answer - B
Discussion
420.

Consider the reactions given below .On the basis of these reactions ,Find out which of the algebraic relations given in options (a) to(d) is correct? `(p)C(g)+4H(g)toCH_(4)(g), Delta_(r)H=xKJ mol^(-1)` `(Q)C("graphite")+2H_(2)(g)to CH_(4)(g),` `Delta_(r)H=yKJmol^(-1)`A. `x=y`B. `x=2y`C. `x gt y`D. `x lt y`

Answer» Correct Answer - C
`x gr y` because same bonds are formed in reactions (i) and (ii) but bonds between reactant molecules are broken, only in reactions (ii). As energy is absorbed when bonds are broken energy released in reaction (i) is greater than in reaction (ii)
Discussion
421.

A substance is carried through the following transformations: The equation `Delta S_(1) + Delta S_(2) = Delta S_(3) + Delta S_(4) + Delta D_(5)`A. Is true only if the steps are carried out reversiblyB. Is always true because entropy is a state functionC. May be true but need more information on the processesD. Is incorrect

Answer» Correct Answer - B
State functions do not depend on path followed.
Discussion
422.

At `27^(@)` C a gas `(gamma = (5)/(3))` is compressed reversible adiabatically so that its pressure becomes 1/8 of original pressure. Final temperature of gas would beA. 420 KB. 300 KC. `-142^(@)C`D. `327^(@)C`

Answer» Correct Answer - C
In adiabatic process
`P_(1)^(1 - gamma) . T_(1)^(gamma) = P_(2)^(1 - gamma)=` constant
Discussion
423.

The entropy change can be calculated by using the expression `DeltaS-(q_(rev))/(T)`. When water freezes in a glass beaker, choose the correct statement amongst the following:A. `Delta S` (system) decreases but `Delta S` (surroundings) remains the sameB. `Delta S` (system) increases but `Delta S` (surroundings) decreasesC. `Delta S` (system) decreases but `Delta S` (surroundings) increasesD. `Delta S` (system) decreases but `Delta S` (surroundings) also decreases

Answer» Correct Answer - C
The entropy change can be calculated by using the expression
`Delta S = (q_("rev"))/(T)`
When water freezes in a glass beaker, `Delta S` (system) decreases because molecules in solid ice are less random than in liquid water, However, when water freezes to ice, heat. Hence, entropy of the surrounding increases
Discussion
424.

The entropy change can be calculated by using the expression, `DeltaS=(q_("rev"))/(T)`. When water freezes in a glass beaker, choose the correct statement amongst the following:A. `DeltaS_("system")` decreases but `DeltaS_("surroundings")` remains the same.B. `DeltaS_("system")` increases but `DeltaS_("surroundings")` decreases.C. `DeltaS_("system")` decrease but `DeltaS_("surroundings")` increases.D. `DeltaS_("system")` decreases and `DeltaS_("surroundings")` also decreases.

Answer» Correct Answer - C
As water freezes from liquid to solid, randomness decreases, i.e., `DeltaS_("system")` decreases. Heat released during the process, is absorbed by the surroundings hence, `DeltaS_("surroundings")` increases.
Discussion
425.

The entropy change can be calculate by using the expression `DeltaS= ( q_("rev"))/( T)`. When water freezes in a glass beaker, choose the correct statement amongst the following `:`A. `DeltaS`( system ) decreases but`DeltaS`( surroundings) remainsthe same.B. `DeltaS ` ( system ) increases but `DeltaS` ( surroundings) decreasesC. `DeltaS` ( system) decreases but `DeltaS` ( surroundings) increases.D. `DeltaS` ( system ) decreases and`DeltaS` ( surroundings) also decreases.

Answer» Correct Answer - c
`DeltaS` ( system)decrease because moleculesin solid ice are less random than in liquid water. However, when water freezes to ice, heat is released which is absorbed by the surroundings. Hence, entropy of the surroundings increases.
Discussion
426.

The energies of the bonds broken in a certain eaction arae greater than the energies of the bonds formed. Which one of the following statements about this reaction must be true?A. The reaction is endothermic.B. The reaction is exothermic.C. The reaction is spontaneous.D. The reaction is non- spontaneous.

Answer» Correct Answer - a
Discussion
427.

How much heat is reuired to convert 5.0 g of ice at `-10.0^(@)` C to liquid water at `15.0^(@)` C ? (Assume heart capacities are indendent o ftemperature.) A. `4.2xx10^(2)j`B. `2.1xx10^(3)J`C. `9.3xx10^(3)J`D. `3.8xx10^(4)J`

Answer» Correct Answer - b
Discussion
428.

The entropy change can be calculated by using the expression `DeltaS-(q_(rev))/(T)`. When water freezes in a glass beaker, choose the correct statement amongst the following:A. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` remains the sameB. `Delta S_(("system"))` increases but `Delta S_(("surroundings"))` decreasesC. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` increasesD. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` also decrease

Answer» Correct Answer - C
During the process of freezing energy is released, which is absorbed by the surroundings
`:. Delta S_("system") = (-q_(rev))/(T)`,
`Delta S_("surroundings") = (q_(rev))/(T)`
Therefore the entropy of the system decrease and that of surroundings increases
Discussion
429.

Two reactions are given below : `C_("(graphite)")+O_(2(g))rarrCO_(2(g)),DeltaH=-393.7kJ` `C_("(diamond)")rarrC_("(graphite)"),DeltaH=-2.1kJ` What quantity of diamond will give 800 kJ of heat on burning ?A. 24.25 gB. 15.24 gC. 2 gD. 12.12 g

Answer» Correct Answer - A
`C_("(diamond)")+O_(2(g))rarrCO_(2(g))`
`=-393.7+(-2.1)=-"395.8 kJ"`
`"800 kJ will b given by "(12)/(395.8)xx800=24.25g`
Discussion
430.

In conversation of lime-stone ti lime, `CaCO_(3)(s)toCaO(s)+CO_(2)(g) 7`the value of `DeltaH^(@)and DeltaS^(@)` are `+179.1KJ mol^(-1)`and `160.2J//K` respectively at 298K and 1 bar. Assuming that `DeltaH and DeltaS` do not change with temperature, temperature above which coversation of lime-stone to lime will be just spontaneous is:

Answer» Correct Answer - B
Discussion
431.

Entropy change for reversible phase transition at constant pressure P and temperature T is calculated by the formula `DeltaS=(DeltaH)/(T),` where `DeltaH` is the enthalpy change for phase transition. For irreversible phase transition `DeltaSgt(DeltaH)/(T).` Consider a phase transition. `Sn("white",s)iffSn("grey,s")` `DeltaH^(@) `at 1 atm and` 300K=-2 kJmol^(-1)` The equilibrium temperature at 1 atm is 400 K. Assume `C_(p,m)` of Sn (white,s) and Sn(grey,s) are equal. `DeltaG^(@)` for above phase transition at 1 atm and 300K is :A. `-500J mol^(-1)`B. `-500 kJ mol^(-1)`C. 0D. `-100J mol^(-1)`

Answer» Correct Answer - a
Discussion
432.

Entropy change for reversible phase transition at constant pressure P and temperature T is calculated by the formula `DeltaS=(DeltaH)/(T),` where `DeltaH` is the enthalpy change for phase transition. For irreversible phase transition `DeltaSgt(DeltaH)/(T).` Consider a phase transition. `Sn("white",s)iffSn("grey,s")` `DeltaH^(@) `at 1 atm and` 300K=-2 kJmol^(-1)` The equilibrium temperature at 1 atm is 400 K. Assume `C_(p,m)` of Sn (white,s) and Sn(grey,s) are equal. `DeltaS^(@)` for above phase transition at 1 atm and 300K is :A. `-5JK^(-1)mol^(-1)`B. `-(20)/(3)JK^(-1)mol^(-1)`C. `-0.0055JK^(-1)mol^(-1)`D. `-(2000)/(3)JK^(-1)mol^(-1)`

Answer» Correct Answer - a
Discussion
433.

Calculate the entropy changes of fustion and vaporisation for chlorine from the following data: `Delta_(fus)H = 6.40 kJ mol^(-1)`, melting point `=- 100^(@)C` `Delta_(vap)H = 20.4 kJ mol^(-1)`, boiling point `=- 30^(@)C`

Answer» `Delta_(fus)S = (6400)/(173K) = 36.9 J K^(-1) mol^(-1)`
`Delta_(vap)S = (20400)/(243) = 83.9 J K^(-1)mol^(-1)`
Discussion
434.

Benzene burns according to the following equation: `2C_(6)H_(6)(l)+15O_(2)(g)to12CO_(2)(g)+6H_(2)O(l)` `DeltaH^(@)=-6542 KJ//mol` what is the `DeltaE^(@)` for the combustion of 1.5 mol of benzene?A. `-3271 KJ`B. `-9813KJ`C. `-4906 KJ`D. none of these

Answer» Correct Answer - d
Discussion
435.

Ethyl chloride `(C_(2)H_(5)Cl)` , is prepared by reaction of ethylene with hydrogen chloride: `C_(2)H_(4)(g) + HCl(g) rarr C_(2)H_(5)Cl(g) " " DeltaH=-72.3 KJ//"mol"` What is the value of `DeltaE ("in KJ")`, if 98 g fo ethylene and 109.5 g if HCl are allowed to react at 300KA. `-64.81`B. `-190.71`C. `-209.41`D. `-229.38`

Answer» Correct Answer - C
`Deltan_(g)=1-2=-1`
`DeltaE=DeltaH-Deltan_(g)RT -DeltaH-RT =-72.3 + 8.314 xx 300 xx 10^(-3)=-69.806 KJ//"mole"`
so for 3 mole we will get `DeltaE=-69.806 xx 3 KJ//"mol" =209.42 KJ//"mole"`
Discussion
436.

A large positive value of `DeltaG^(ɵ)` corresponds to which of these?A. small positive KB. small positive JC. large positive KD. large negative K

Answer» Correct Answer - A
Discussion
437.

Consider the values of `DeltaH("in" kJ mol^(-))` and for `DeltaS("in" mol^(-)K^(-1))` given for four different reactions. For which reaction will `DeltaG` increases the most (becoming more positive) when the temperature is increased form `0^(@)C "to" 25^(@)C`?A. `DeltaH^(@)C=50, DeltaS^(@)=50`B. `DeltaH^(@)C=90, DeltaS^(@)=20`C. `DeltaH^(@)C=-90, DeltaS^(@)=-50`D. `DeltaH^(@)C=-90, DeltaS^(@)=-20`

Answer» Correct Answer - C
Discussion
438.

An ideal gas expand against a constant external pressure at 2.0 atmosphere from 20 litre to 40 litre and absorb `10kJ` of energy from surrounding . What is the change in internal energy of the system ?A. 4052 JB. 5948 JC. 14052 JD. 9940 J

Answer» Correct Answer - B
`DeltaU=q+w`
`=10 xx 1000 -2 xx (20)xx101.3 = 5948 J`
Discussion
439.

Heat of combustion of ethanol at constant pressure and at temperature T K(=298 K) is found to be -qJ `"mol"^(-1)`. Hence , heat of combustion (in J `"mol"^(-1)`) of ehtanol at the same temperature at constant volume will be:A. RT-qB. `-(q+RT)`C. q-RTD. q+RT

Answer» Correct Answer - A
`C_(2)H_(5)OH(l) + 3O_(2)(g) rarr 3H_(2)O(r)`
`Deltan_(3)=2-3=-1`
so `DeltaU=DeltaH-Deltan_(g)RT`
`=-q+RT`
Discussion
440.

Standard enthalpy of fusion `(Delta_(fus) H^(@))` is the amount of heat required to melt ____ of a solid at its melting point and at a standard pressure of 1 bar.A. gramB. kilogramC. moleD. moelcule

Answer» Correct Answer - C
Therefore, it is also called molar enthaply of fusion.
Discussion
441.

Benzene burns according to the following equations at 300K (R=8.314 J `"mole"^(-1)K^(-1))` `2C_(6)H_(6)(l) + 15O_(2)(g) rarr 12 CO_(2)(g) + 6H_(2)O(l) " " DeltaH^(@) =- 6542 KJ` What is the `DeltaE^(@)` for the combustion of 1.5 mol of benzeneA. `-3271 KJ`B. `-9812 KJ`C. `-4906.5 `KJD. None of these

Answer» Correct Answer - D
From given reaction `Deltan_(g) =12 -15=-3`
so `DeltaE^(@) =DeltaH^(@)-Deltan_(g)RT =-6542 + 3RT`
for 1.5 mole , `Delta E^(@) =(1.5)/(2) {-6542 + 3RT} = 4900 KJ`
Discussion
442.

A carnot engine operates between temperatures `T` and `400K(Tgt400K)`. If efficiency of engine is `25%`, the temperature `T` is :A. `400 K`B. `500 K`C. `533.3 K`D. `600 K`

Answer» Correct Answer - C
`eta_(("efficiency")) =(T_(2)-T_(1))/(T_(2))`
or `0.25=(T-400)/T`
`:. T=533.3 K`
Discussion
443.

One mole of an ideal gas at `25^(@)C` expands in volume from 1.0 L to 4.0 L at constant temperature. What work (in J) is done if the gas expands against vacuum `(P_("external") = 0)`?A. `-4.0 xx 10^(2)`B. `-3.0 xx 10^(2)`C. `-1.0 xx 10^(2)`D. Zero

Answer» Correct Answer - D
`w=-intPdV=0`
Discussion
444.

The molar heat capacity of water at constant pressure, C, is `75 JK^(-1) mol^(-1)`. When 1.0 kJ of heat is supplied to 100 g water which is free to expand, the increase in temperature of water is :A. 0.24 KB. 2.4 KC. 1.3 KD. 0.13 K

Answer» Correct Answer - B
`DeltaH=nc_(P)DeltaT`
`DeltaT=(DeltaH)/(nc_(P))`
Discussion
445.

For the reaction, `X_(2)O_(4)(l)rarr2XO_(2)(g),DeltaE=2.1Kcal` , `DeltaS=20cal//K` at `300K` . Hence `DeltaG` isA. `2.7 kcal`B. `-2.7 kcal`C. `9.3 kcal`D. `-9.3 kcal`

Answer» `DeltaH = DeltaU +Deltan_(g)RT`
`= 2.1 xx2xx 0.02 xx 300 = 3.3 kcal`
`DeltaG = DeltaH - T DeltaS`
`=3.3 xx -300 xx (0.02) =- 2.7 kcal`
Discussion
446.

When 1.0 g of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose capacity is 8.75 KJ/K, the enthalpy of combustion of oxalic acid at`27^(@)C` is :A. `-245.7 KJ//mol`B. `-244.452 KJ//mol`C. `-246.947 KJ//mol`D. None of these

Answer» Correct Answer - C
`H_(2)C_(2)O(l) + (1)/(2)O_(2)(g) rarr H_(2)O(l) + 2O_(2)(g),`
`Deltan_(g)=3//2`
`DeltaU_(0) =-(0.312 xx8.75)/(1) xx 90 =-245.7 KJ//"mol"`
`DeltaH=DeltaU+ Deltan_(g)RT =- 245.7 KJ//"mol"`
`DeltaH=DeltaU + Deltan_(g)RT =-245.7 +(3)/(2) xx (8.314xx300)/(1000)=-241.947 KJ//"mol"`
Discussion
447.

When 1.0 g of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose capacity is 8.75 KJ/K, the enthalpy of combustion of oxalic acid at`27^(@)C` is :A. `-245.7kJ//mol`B. `-244.452kJ//mol`C. `-241.5kJ//mol`D. None of these

Answer» Correct Answer - D
`H_(2)C_(2)O_(4)(l)+(1)/(2)O_(2)(g)rarrH_(2)O(l)+2CO_(2)(g)`,
`Deltan_(g)=3//2`
`DeltaU_(c )=-(0.312xx8.75)/(1)xx90`
=-245.7 kJ/mol
`DeltaH=DeltaU+Deltan_(g)RT`
`=-245.7 +(3)/(2)xx(8.314xx300)/(1000)`
=-241.95 kJ/mol
Discussion
448.

When 1.0 g of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose capacity is 8.75 KJ/K, the enthalpy of combustion of oxalic acid at`27^(@)C` is :A. `-245.7KJ//mol`B. `-244.452 KJ//mol`C. `-246.947 KJ//mol`D. none of these

Answer» Correct Answer - d
Discussion
449.

The molar heat capacity of water at constant pressure, C, is `75 JK^(-1) mol^(-1)`. When 1.0 kJ of heat is supplied to 100 g water which is free to expand, the increase in temperature of water is :A. 4.8 KB. 6.6 KC. 1.2 KD. 2.4 K

Answer» Correct Answer - D
According to heat capacity rule capcaity rule.
`q=mcDeltaT, c=(q)/(m(T_(2)-T_(1))`
Given that , `c=75 JK^(-1) "mol"^(-1)`
`q=10 KJ = 1000 J`
Mass= 100 g water
Molar mass of water =18 g
`75=(1000)/(5.55xxDeltaT)" "` (Number of moles `=(100)/(18)=5.55)`
`therefore DeltaT=(1000)/(5.55xx75)=2.4 K`
Discussion
450.

When `1.0g` of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose heat capacity is `8.75 kJ//K`, the temperature increases by 0.312 K. The enthalpy of combustion of oxalic acid at `27^(@)C` isA. `-245.7 kJ//mol`B. `-2.43.45 kJ//mol`C. `-246.95 kJ//mol`D. `-241.95 kJ//mol`

Answer» Correct Answer - D
`H_(2)C_(2)O_(4)(l)+(1)/(2)O_(2)(g) rarr H_(2)O(l) + 2CO_(2)(g)`,
`Delta n_(g) = 3//2, Delta U_(c) = -(0.312 xx 8.75)/(1) xx 90`
`= -245.7 kJ//mol`,
`Delta H = Delta U + Delta n_(g)RT`
`= -245.7+(3)/(2) xx (8.314 xx 300)/(1000) = -241.95 KJ//mol`
Discussion