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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Calculate `DeltaU and DeltaH` when `10dm^(3)` of helium at NTP is heated in a cylinder to `100^(@)C`, assuming the the gas behaves ideally. `(C_(V)=3//2R)` |
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Answer» Correct Answer - `DeltaU=556.74J` `DeltaH=927.9J` `DeltaU=n*C_(V)*DeltaT and DeltaH=n*C_P*DeltaT` `=(10)/(22.4)xx(3)/(2)xx8.314xx100=556.74J` `=(10)/(22.4)xx(5)/(2)xx8.314xx100` `=927.9J` |
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| 2. |
Calculate q,w, `DeltaU` and `DeltaH` for the reversible isothermal expansion of one mole of an ideal gas at `127^(@)C` from a volume of `10dm^(3)` to `20dm^(3)`. |
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Answer» Since, the process is isothermal, `DeltaU=DeltaH=0` From first law of thermodynamics, `DeltaU=q+w=0` `q=-w` `w=-2.30nRTlog((V_(2))/(V_(1)))` `=-2.303xx1xx8.314xx400log((20)/(10))` `=-2.303xx1xx8.314xx400xx0.3010` `=-2305.3J` (Work is one by the system) `q=-w=2305.3J` (Heat is absorbed by the system). |
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| 3. |
The standard Gibbs free energies for the reaction at `1773K` are given below: `C(s) +O_(2)(g) rarr CO_(2)(g), DeltaG^(Theta) =- 380 kJ mol^(-1)` `2C(s) +O_(2)(g) hArr 2CO(g),DeltaG^(Theta) =- 500 kJ mol^(-1)` Discuss the possibility of reducing `AI_(2)O_(3)` and `PbO` with carbon at this temperature, `4AI +3O_(2)(g) rarr 2AI_(2)O_(3)(s),DeltaG^(Theta) =- 22500 kJ mol^(-1)` `2Pb +O_(2)(g) rarr 2PbO(s),DeltaG^(Theta) =- 120 kJ mol^(-1)` |
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Answer» Let us consider the reduction of `Al_(2)O_(3)` by carbon: `2Al_(2)O_(3)+3C(s)to4Al(s)+3CO_(2)(g)`, `DeltaG^(@)=-380xx3+22500=+21360kJ` `2Al_(2)O_(3)+6C(s)to4Al(s)+6CO(g)," "DeltaG^(@)=-500xx3+22500=+21000kJ` Positive values of `DeltaG^(@)` show that the redcution of `Al_(2)O_(3)` is not possible by any of the above methods `2PbO(s)+Cto2Pb+CO_(2),DeltaG^(@)=+120(-380)=-260kJ` `2PbO(s)+2Cto2Pb+2CO,DeltaG^(@)=+120+(-500)=-380kJ` Negative value of `DeltaG^(@)` shows that the process is spontaneous and PbO can be reduced by carbon. |
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| 4. |
Calculate the heat of formation of ammonia from the following data: `N_(2)(g)+3H_(2)(g)to2NH_(3)(g)` The bond energies of `N-=N,H-H and N-H ` bonds are 226,104 and 92 kcal respectively. |
| Answer» Correct Answer - `-10kcal` | |
| 5. |
The bond energies of `C=C` and `C-C` at `298 K` are `590` and `331 kJ mol^(-1)`, respectively. The enthalpy of polymerisation per mole of ethaylene isA. `-70kJ`B. `-72kJ`C. `72kJ`D. `-68kJ` |
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Answer» Correct Answer - B The polymeridation of ethene may be represented as `nCH_(2)=CH_(2)to[-CH_(2)-CH_(2)-]_(n)` One mole of C=C bond is decomposed and two moles of C-C bonds are formed per mole of ethane `thereforeDeltaH=590-2xx331=-72kJ` per mol of ethylene. |
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| 6. |
In `C_(2)H_(4)` energies of formation of `(C=C) and (C-C)` are -145 kJ/mol and -80kJ/mol respectively. What is the enthalpy change w hen ethylene polymerises to form polythene?A. `+650kJ//mol`B. `+65kJ//mol`C. `-650kJ" "mol^(-1)`D. `-65kJ" "mol^(-1)` |
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Answer» Correct Answer - B Polymerisation of ethane may be given as: `nCH_(2)=CH_(2)to(-CH_(2)-CH_(2)-)_(n)` `DeltaH=sum(BE)_("reactants")-sum(BE)_("Products")` `=(+145)-(+80)=+65kJ//mol` |
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| 7. |
For the reaction: `2H_(2)(g) +O_(2)(g) rarr 2H_(2)O(g), DeltaH =- 571 kJ` bond enegry of `(H-H) = 435 kJ` and of `(O=O) = 498 kJ`. Then, calculate the average bond enegry of `(O-H)` bond using the above data.A. 484 kJB. `-484kJ`C. `271kJ`D. `-271kJ` |
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Answer» Correct Answer - A `2(H-H)+O=Oto2(H-O-H)` `DeltaH=sum(BE)_("reactants")-sum(BE)_("products")` `=571=[2xx435+498]-4xx(BE)_(O-H)` `(BE)_(O-H)=(2xx435+498+571)/(4)=~~484kJ` |
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| 8. |
Heat of formation of `2 mol` of `NH_(3)(g)` is `= -90 kJ`, bond energies of `H-H` and `N-H` bonds are `435kJ` and `390 kJ mol^(-1)`, respectively. The value of the bond enegry of `N-=N` will beA. `-472.5kJ`B. `-945kJ`C. `472.5kJ`D. `945kJ" "mol^(-1)` |
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Answer» Correct Answer - D `N-=+3(H-H)to2underset(H)underset(|)overset(H)overset(|)(N)-H," "DeltaH=-90kJ` `DeltaH_("reaction")=sum(BE)_("reactants")-sum(BE)_("products")` `-90=[(BE)_(N-=N)+3(BE)_(H-H)]-[6(BE)_(N-H)]` `-90=x+3xx435-6xx390` `x=945" kJ "mol^(-1)`. |
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