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. |
Chemical reactions are invariably associated with the transfter of energy either in the form of heat or light. In the laboratory, heat changes in physical and chemical processes are measured with an instrument called calorimeter. Heat change in the process is calculated as {:(q = ms DeltaT,,s ="Specific heat"),(=cDeltaT,,c ="Heat capacity"):} Heat of reaction at constant volume is measured using bomb calorimeter. q_(V) = DeltaU = Internal energy change Heat of reaction at constant pressure is measured using simple or water calorimeter. q_(p)= DeltaH q_(p) = q_(V) +P DeltaV DeltaH = DeltaU +DeltanRT What value of DeltaT should be used for the calorimetry experiment that gives the following graphical results? |
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Answer» `30^(@)C` `:. DeltaH = DeltaU` |
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| 2. |
Chemical oxygen demand is determined by using |
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Answer» Methylorgane |
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| 4. |
Chemical equilibrium is dynamic. Justify. |
| Answer» SOLUTION :Chemical equilibrium is dynamic : because the velocities of forward and backward reactions are equal, as ALSO the concentration of reactant and PRODUCT remain unchanged at this equilibrium state. (But may not be equal). The equilibrium can be made to shift in either direction by CHANGING the concentration, Pressure, or Volume. Hence it is dynamic in nature. | |
| 5. |
Chemical equilibrium is attained in a reversible reaction carried in a close container and is of dynamic nature. The value of equilibrium constant may be expressed either as K_(p)" and " K_(c) and the two are related to each other as : K_(p) =K_(c)(RT)^(Deltang) Free energy change (DeltaG) at equilibrium point is zero. The value of equilibrium constant gives the extent to which a particular reation has proceeded to attain the equilibrium . Its value gets reversed if the reaction is reversed and becomes the square root of the initial value if the reaction is divided by 2. For the reaction PCI_(3)(g)+ CI_(2) (s) hArr PCI_(5) (g) The value of k_(c) " at " 250^(@)C " is " " mol"^(-1) L^(-1). The value of k_(p) at the same temperaturewill be : |
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Answer» <P>`0.61 ATM^(-1)` `x (0.0821 L atm K^(-1) mol^(-1) XX 523 K^(-1))` `= 0.61 K` |
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| 6. |
Chemical equilibrium is attained in a reversible reaction carried in a close container and is of dynamic nature. The value of equilibrium constant may be expressed either as K_(p)" and " K_(c) and the two are related to each other as : K_(p) =K_(c)(RT)^(Deltang) Free energy change (DeltaG) at equilibrium point is zero. The value of equilibrium constant gives the extent to which a particular reation has proceeded to attain the equilibrium . Its value gets reversed if the reaction is reversed and becomes the square root of the initial value if the reaction is divided by 2. A reaction attains equilibrium when the free energy change accompanying the reaction is : |
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Answer» POSITIVE and large |
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| 7. |
Chemical equilibrium, explain by example of 'chemical reaction'. |
Answer» Solution :(A) Example of Synthesis of ammonia : The chemical equilibrium can be demonstrated in the synthesis of ammonia by Haber.s process. In a series of experiments , Haber started with known amounts of dinitrogen and dihydrogen maintained at high temperature and PRESSURE and at REGULAR intervals determined the amount of ammoniapresent . He was successful in determining also the concentration of unreacted dihydrogen and dinitrogen .It is shown in following figure. (B) Example of decomposition of hydrogen iodide (HI) : If we start with equal initial concentration of `H_2` and `I_2` the reaction proceeds in the forward direction and the concentration of `H_2` and `I_2` decreases while that of HI increases until. After some time concentration of all (of `H_2 ,I_2`, HI) become constant and it reach to equilibrium . `H_(2(g)) + I_(2(g)) hArr 2HI_((g))` ...(iii) At the same temperature we can take only HI and make the reaction to proceed in the reverse direction, the concentration of HI will decrease and of `H_2` and `I_2` will INCREASE. After some time concentration of all `(H_2, I_2, H_1)` will reach constant and equilibrium SITUATION formed. `2HI_((g)) hArr H_(2(g)) + I_(2(g))`...(iv) "Equilibrium can attain from any of the side while reaction." ![]() In starting, `H_2` and `I_2` are reactant not products. At equilibrium all are components. If in forward and backward reaction at definite temperature he no. of atoms H and I in constant volume be same then both will give equilibrium mixture. |
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| 8. |
Chemical equation is balanced according to the law ofmultiple proportionsreciprocal proportionsconservation of massdefinite proportions. |
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Answer» MULTIPLE PROPORTIONS |
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| 9. |
Composition of PAN is |
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Answer» `CH_(2)=CH-CHO` |
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| 10. |
Chemical analysis of a carbon compound gave the following percentage composition by weight of the elements present, carbon = 10.06%, hydrogen = 0.84%, chlorine = 89.10%. Calculate the empirical formula of the compound. |
| Answer» SOLUTION :`(CHCl_(3))` | |
| 11. |
Chemical A is used for water softening to remove temporary hardness. A reacts with sodium carbonate to generate caustic soda. When CO_(2) is bubbled through a solution of A, it turns cloudy. What is the chemical formula of A ? |
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Answer» `CaCO_(3)` `underset((A))(Ca(OH)_(2)) +Na_(2) CO_(3) to underset("Caustic SODA")(CaCO_(3) darr )+ CaCO_(3)` `underset(A)(Ca(OH)_(2)) + CO_(2) to underset("Milkiness")(CaCO_(3)) darr +H_(2)O` |
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| 12. |
CH=C-CH_(3)+H_(2)overset(pt)rarr?overset(H_(2)rarr |
| Answer» Solution :`CH=C-underset("PROPYNE")(CH )_(3)+H_(2) OVERSET(2T)(to)CH_(2)= underset("Propene") (CH ) = CH_(3)overset(H_(2)pt)(to) CH_(3)= CH_(3) CH_(3)` | |
| 13. |
Charle's law may be expressed as |
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Answer» <P>`(dV//dT)_(P) = K` `V PROP T` or `V=KT` `(dV//dT)_(P)=K` |
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| 14. |
Charge of electron is |
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Answer» `1,602xx10^(-10)" COULOMB"` |
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| 15. |
Charcoal possesses adsorption property, because |
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Answer» it is a non-CONDUCTOR of electricity. |
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| 16. |
Characteristic reaction of alkenes is |
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Answer» ELECTROPHILIC ADDITION |
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| 17. |
Char Coal is a ___________ element. |
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Answer» ACID -making |
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| 18. |
Changes in a system from an initial state to the final state were made by a different manner that DeltaH remains same but q changes because |
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Answer» `DELTAH` is a path function but q is astate function |
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| 19. |
Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is givenout by the system is____ |
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Answer» Solution :`DELTAU=q+w` `DeltaU`=-1 kJ+ 4 kJ `DeltaU` =+3 kJ |
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| 20. |
Changein Gibbs free energy is given by _____ |
| Answer» SOLUTION :`DeltaG=DeltaH-TDeltaS` | |
| 21. |
Change in heat at constant volume (q_(v))=…...... |
| Answer» Answer :A | |
| 22. |
Change in entropy for one mole water from liquid from to vapour at 373 K is ........... Joule/ Kelvin.(Delta H_("vap") = 2.257 "KJ/gm") |
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Answer» `105.9` |
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| 23. |
Change in entropy during adsorption of gas on the solid substance is.......... . |
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Answer» Increasing |
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| 24. |
Change in entropy for the expansion of 1 mole ideal gas in vacuum is .......... |
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Answer» 1 Joule |
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| 25. |
Change in enthalpy when 11.2 dm^3 of He at NTP is heated in a cylinder to 100^@C is |
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Answer» 623.5J |
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| 26. |
Change in enthalpy during neutralization reaction of one mole HCI with dilute KOH at 298 K is .......... KJ |
| Answer» ANSWER :B | |
| 27. |
Change in enthalpy is |
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Answer» Heat ABSORBED at constant pressure |
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| 28. |
Change in enthalpy and change in internal energy are state functions. The value of DeltaH, DeltaU can be determined by using Kirchoff's equation. Calculate the heat of formation of methane, given that heat of formation of water = -286kJ mol^(-1), heat of combustion of methane = -890kJ mol^(-1) heat of combustion of carbon = -393.5 kJ mol^(-1) |
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Answer» `90.5 kJ mol^(-1)` `DELTA H = Delta H_(f_(CO_(2))) + 2DELTA H_(f_(H_(2)O)) - Delta H_(f_(CH_(4)))` `Delta H_("COMBUSTION"_(CH_(4))) = Delta H_("combustion"_(( C))) + 2 Delta H_(f_(H_(2)O)) - Delta H_(f_(CH_(4)))` `-890 = -393.5 + 2 XX (-286) - Delta H_(f_(CH_(4)))` `Delta H_(f_(CH_(4))) = - 75.5` kJ/mol |
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| 29. |
Change in enthalpy and change in internal energy are state functions. The value of DeltaH, DeltaU can be determined by using Kirchoff's equation. Calculate DeltaH when 10dm^(3) of helium at NTP is heated in a cylinder to 100^(@)C, assuming that the gas behave ideally. |
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Answer» 927.9J |
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| 30. |
Chain isomer of cyclobutene is |
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Answer» `CH_(3)CH_(2)CH=CH_(2)` are chain ISOMERS because there is a differ in the carbon chain. |
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| 31. |
Chain-growth polymerization may proceed by the following mechanism: |
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Answer» free RADICAL POLYMERIZATION |
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| 32. |
Chain growth addition polymerization is an important reaction of: |
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Answer» ALKENES |
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| 33. |
Chadwick got the Nobel Prize for the discovery of _________. |
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Answer» protons |
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| 34. |
CH_(4(g))+2O_(2(g))hArrCO_(2(g))+2H_(2)O_((l)) DeltaH=-170.8" kJ mol"^(-1) which of the following statement is not true ? |
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Answer» At equilibrium, the concentration of `CO_(2(g))` and `H_(2)O_((l))` are not equal |
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| 35. |
CH_(4(g)) + 2O_(2(g)) rarr CO_(2(g)) + 2H_2O_((l)) DeltaH = -200 K.cal, now DeltaE Cal, now DeltaE for the same process at 300 K (in K.Cal) |
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Answer» `-199.8` |
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| 36. |
CH_(4) is not obtained by |
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Answer» WURTZ reaction |
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| 37. |
CH_4is formed when |
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Answer» SODIUM acetate is heated with SODA lime |
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| 38. |
CH_(4) gas is behaving non-ideally. Compressibility factor for gas is 1.5 at 2 atm, 400K. Calculate molar volume for gas: [Given: R=0.08("Litre"-"atm")/(K-"mole")] |
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Answer» 24 litre |
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| 39. |
CH_4 diffuses two times faster than a gas X. The number of molecules present in 32 g of gas X is (N is Avogadro number) |
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Answer» `N` |
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| 40. |
CH_4 + Cl_2 overset(hv)to CH_3 Cl + HCl to obtain high yields of CH_3Cl , the ratio of CH_4 to Cl_2 must be |
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Answer» HIGH |
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| 41. |
CH_4 can be prepared by the reaction of H_2O with |
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Answer» `Mg_2C_3` ` Al_4C_3 + 12H_2O to 3CH_4 + 4AL(OH)_3` |
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| 42. |
CH_3underset(Br)underset(|)CH=underset(Br)underset(|)CH_2 underset( " of alc." KOH)overset("1 equivalent")to The major product is |
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Answer» `CH_3CH = CHBR` |
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| 43. |
CH_3OH_((l)) underset"673 K"overset"50 bar"to CO_((g)) + 2H_(2(g)) Which catalyst is used in the above reaction. |
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Answer» COPPER OXIDE |
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| 44. |
CH_3OH, methanol can be prepared from CO and H_2. CO(g)+2H_2(g)hArr CH_3OH(g) The value of K_p at 500 K is 6.23xx10^(-3) When total pressure (in atm) at equilibrium is required to convert 25% of CO to CH_3OH at 500 K if CO and H_2 comes from CH_4(g)+H_2O(g) to CO(g)+3H_2(g) Given your answer excluding decimal places. |
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Answer» |
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| 45. |
CH_(3)Mgloverset(C_(2)H_(5)OH)toX+C_(2)H_(5)OMgl The product X is........ |
| Answer» Answer :A | |
| 46. |
CH_(3)COOK_((aq))overset("Electrolysis")rarrAunderset(1000^(@)C)overset(Delta)rarrB Bunderset(KMnO_(4))overset("Alkaline")rarrC. Here compound 'C' is |
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Answer» `UNDERSET(COOH)underset("|")(CH_(3)` |
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| 47. |
CH_(3)COOH_((l)) + C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5(l)) + H_(2)O_((l)) In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of little cone. H_(2)SO_(4). On equilibrium being attained |
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Answer» 1 mole of ETHYL acetate is formed |
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| 48. |
CH_(3)COOH(l)+CH_(2)H_(5)OH(l)hArrCH_(3)COOC_(2)O(l) In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of little conc. H_(2)SO_(4). On equilibrium being attained |
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Answer» `1` mole of ETHYL acetate is FORMED |
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| 49. |
CH_(3)COOH_((l)) + C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5(l)) + H_(2)O_((l)) In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of little cone. H_(2)SO_(4). On equilibriumbeing attained |
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Answer» 1 mole of ethyl ACETATE is formed |
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