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Gujarati JEE Chemistry : સંતુલન

Multiple Choice Questions

11.
0.2 વાતાવરણ દબાણે HI(g) ના એક નમૂનાને એક ફલાસ્કમાં ભરવામાં આવે છે, તો સંતુલને HI(g) નું આંશિક દબાણ 0.04  વાતાવરણ છે, તો આપેલી સંતુલિત પ્રક્રિયા માટે KP નું મૂલ્ય કેટલું થાશે ?

bold 2 bold HI subscript bold left parenthesis bold g bold right parenthesis end subscript bold space bold rightwards harpoon over leftwards harpoon bold space bold H subscript bold 2 bold left parenthesis bold g bold right parenthesis end subscript bold space bold plus bold space bold I subscript bold 2 bold left parenthesis bold g bold right parenthesis end subscript
  • 6.0

  • 8.0

  • 4.0

  • 0.4


12.

નિયત તાપમાને નીચેનાં સંતુલનો માટે KC નું મૂલ્ય અનુક્રમે કેટલું થશે ?

bold left parenthesis bold i bold right parenthesis bold space bold 2 bold NOCl subscript bold left parenthesis bold g bold right parenthesis end subscript bold space bold rightwards harpoon over leftwards harpoon bold space bold 2 bold NO subscript bold left parenthesis bold g bold right parenthesis end subscript bold space bold plus bold space bold Cl subscript bold 22 bold left parenthesis bold g bold right parenthesis end subscript bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold K subscript bold P bold space bold equals bold space bold 1 bold. bold 8 bold space bold cross times bold space bold 10 to the power of bold minus bold 2 end exponent bold comma bold space bold T bold space bold equals bold space bold 500 bold space bold K

bold left parenthesis bold ii bold right parenthesis bold space bold CaCO subscript bold 3 bold left parenthesis bold S bold right parenthesis end subscript bold space bold rightwards harpoon over leftwards harpoon bold space bold CaO subscript bold left parenthesis bold s bold right parenthesis end subscript bold space bold plus bold space bold CO subscript bold 2 bold left parenthesis bold s bold right parenthesis end subscript bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold K subscript bold P bold space bold equals bold space bold 167 bold comma bold space bold T bold space bold equals bold space bold 1073 bold space bold K
  • 4.38 space cross times space 10 to the power of negative 5 end exponent comma space 18.9

  • 4.38 space cross times space 10 to the power of negative 5 end exponent comma space 28.9

  • 4.38 space cross times space 10 to the power of negative 4 end exponent comma space 1.89

  • 4.38 space cross times space 10 to the power of negative 4 end exponent comma space 2.89


13.

ચોક્કસ તાપમાને અને 10પાસ્કલ કુલ દ્બાણે આયોનિડની બાષ્પ કદથી 40 % આયોડિન પરમાણુઓ ધરાવે છે, તો આપેલા સંતુલન માટે KP કેટલો થાશે ? bold I subscript bold 2 bold left parenthesis bold g bold right parenthesis end subscript bold space bold rightwards harpoon over leftwards harpoon bold space bold 2 bold I subscript bold left parenthesis bold g bold right parenthesis end subscript

  • 2.67 space cross times space 10 to the power of 6

  • 2.67 space cross times space 10 to the power of 4

  • 2.67 space cross times space 10 to the power of negative 4 end exponent

  • આપેલ પૈકી એક પણ નહી 


14. સક્રિય જથ્થાના નિયમ મુજબ પ્રક્રિયાનો વેગ એ કોના સમપ્રમાણમાં હોય છે ?
  • પ્રક્રિયકોની સાંદ્વતા

  • પ્રક્રિયકનો સ્વભાવ 

  • પાત્રનું કદ 

  • સંતુલન અચલાંક 


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15.

1127 K તાપમાને અને 1 વાતાવરણ દબાણે CO અને CO2 નું વાયુમય મિશ્રણ એ ઘન કાર્બન સાથે સતુલને વજનથી 90.55 % CO ધરાવે છે, તો આ જ તાપમાને Kનું મૂલ્ય કેટલું હશે ?

bold C subscript bold left parenthesis bold s bold right parenthesis end subscript bold space bold plus bold space bold CO subscript bold 2 bold left parenthesis bold s bold right parenthesis end subscript bold space bold rightwards harpoon over leftwards harpoon bold space bold 2 bold CO subscript bold left parenthesis bold g bold right parenthesis end subscript

  • 0.365

  • 0.153

  • 0.0153

  • 0.283


B.

0.153

CO ના મોલ fraction numerator 9.55 over denominator 28 end fraction space equals space 3.234
CO2 ના મોલ equals space fraction numerator 9.45 over denominator 44 end fraction space equals space 0.215


straight P subscript CO space equals space straight X subscript CO times straight P subscript ક ુ લ space end subscript space space space space space space space space space space space space space space space space space space space space space space space space space straight X subscript CO space equals space fraction numerator 3.234 over denominator 3.234 space plus space 0.215 end fraction

space space space space space space space space equals space 0.938 space cross times space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0. space 938

space space space space space space space space equals space 0.938 space વ ા ત ા વરણ

 straight P subscript CO subscript 2 end subscript space equals space straight X subscript CO subscript 2 end subscript space times space straight P subscript ક ુ લ space space space space space space space space space space space space space space straight X subscript CO subscript 2 end subscript space equals space 1 space minus space 0.938 space

space space space space space space space space space space space equals space space 0.62 space cross times space 1 space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.062 space

space space space space space space space space space space space equals space 0.062 space વ ા ત ા. 

therefore space space straight K subscript straight p space equals space fraction numerator straight P squared co over denominator Pco subscript 2 end fractionમેળવવું 

તથા straight K subscript straight p space equals space straight K subscript straight C left parenthesis RT right parenthesis to the power of increment straight n left parenthesis straight g right parenthesis end exponent પરથી KC મેળવવું.

CO ના મોલ fraction numerator 9.55 over denominator 28 end fraction space equals space 3.234
CO2 ના મોલ equals space fraction numerator 9.45 over denominator 44 end fraction space equals space 0.215


straight P subscript CO space equals space straight X subscript CO times straight P subscript ક ુ લ space end subscript space space space space space space space space space space space space space space space space space space space space space space space space space straight X subscript CO space equals space fraction numerator 3.234 over denominator 3.234 space plus space 0.215 end fraction

space space space space space space space space equals space 0.938 space cross times space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0. space 938

space space space space space space space space equals space 0.938 space વ ા ત ા વરણ

 straight P subscript CO subscript 2 end subscript space equals space straight X subscript CO subscript 2 end subscript space times space straight P subscript ક ુ લ space space space space space space space space space space space space space space straight X subscript CO subscript 2 end subscript space equals space 1 space minus space 0.938 space

space space space space space space space space space space space equals space space 0.62 space cross times space 1 space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.062 space

space space space space space space space space space space space equals space 0.062 space વ ા ત ા. 

therefore space space straight K subscript straight p space equals space fraction numerator straight P squared co over denominator Pco subscript 2 end fractionમેળવવું 

તથા straight K subscript straight p space equals space straight K subscript straight C left parenthesis RT right parenthesis to the power of increment straight n left parenthesis straight g right parenthesis end exponent પરથી KC મેળવવું.


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16.
bold A subscript bold left parenthesis bold g bold right parenthesis end subscript bold plus bold space bold 3 bold B subscript bold left parenthesis bold g bold right parenthesis end subscript bold space bold rightwards harpoon over leftwards harpoon bold space bold 4 bold C subscript bold left parenthesis bold g bold right parenthesis end subscriptપ્રક્રિયામાં A ની શરૂઆતની સાંદ્વતા B ને સમાન હોય તથા સંતુલને A ની સાંદ્વતા  C ને સમાન હોય, તો KC નું મૂલ્ય કેટલું થશે ?
  • 1/8

  • 8

  • 0.8

  • 0.08


17.
1000 K તાપમાને પ્રક્રિયા નીચે મુજબ છે  : 
bold CO subscript bold 2 bold left parenthesis bold g bold right parenthesis end subscript bold space bold plus bold space bold C subscript bold left parenthesis bold s bold right parenthesis end subscript bold space bold rightwards harpoon over leftwards harpoon bold space bold 2 bold CO subscript bold left parenthesis bold g bold right parenthesis end subscript bold comma bold space bold K subscript bold P bold space bold equals bold space bold 3 bold. bold 0 બાર છે. 
જો શરૂઆતમાં bold P subscript bold CO subscript bold 2 end subscript bold space bold equals bold space bold 0 bold. bold 48 બાર અને bold P subscript bold CO bold space bold equals bold space bold 0 બાર અને ગ્રેફાઇટ હાજર હોય, તો સંતુલને CO અને CO2 ના આંશિક દબાણ અનુક્રમે કેટલા થાય ?
  • 0.15 અને 0.66  બાર 

  • 6.6 અને 1.5 બાર  

  • 0.66 અને 0.15  બાર

  • 0.066 અને 0.015 બાર 


18.
0.78M શરૂઆતની સાંદ્વતા ધરાવતા ICl  ના વિઘટનથી ઉત્પન્ન થતા Iઅને ICl  ની સંતુલનને સા6દ્વતા અનુક્રમે કેટલી થશે ?  bold 2 bold ICl subscript bold left parenthesis bold g bold right parenthesis end subscript bold space bold rightwards harpoon over leftwards harpoon bold space bold I subscript bold 2 bold left parenthesis bold g bold right parenthesis end subscript bold space bold plus bold space bold Cl subscript bold 2 bold left parenthesis bold g bold right parenthesis end subscript bold comma bold space bold K subscript bold C bold space bold equals bold space bold 0 bold. bold 14
  • 0.80 અને 0.17

  • 0.339 અને 0.104

  • 0.446 અને 0.167

  • 0.167 અને 0.446


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19. 450 K તાપમાને નીચેની પ્રક્રિયાનો bold K subscript bold P bold space bold equals bold space bold 2 bold space bold cross times bold space bold 10 to the power of bold 10 બાર છે, તો આ જ તાપમાને KC નું મૂલ્ય કેટલું થશે ?
bold 2 bold SO subscript bold 2 bold left parenthesis bold g bold right parenthesis end subscript bold space bold plus bold space bold O subscript bold 2 bold left parenthesis bold g bold right parenthesis end subscript bold space bold rightwards harpoon over leftwards harpoon bold space bold 2 bold SO subscript bold 3 bold left parenthesis bold g bold right parenthesis end subscript
  • 5.41 space cross times 10 to the power of 8
  • 5.24 space cross times space 10 to the power of 10 space
  • 7.38 space cross times space 10 to the power of 11 space
  • આપેલ પૈકી એક પણ નહી 


20.
400 K તાપમાને PCl5(g) નું દબાણ 1  વાતાવરણ છે. જો તેનું નીચે મુજબ વિઘટન થતું હોય તથા તેનો વિયોજન-અંશ 0.4 હોય, તો સંતુલને મિશ્રણની ઘનતા કેટલી થશે ?

પ્રક્રિયા : bold PCl subscript bold 5 bold left parenthesis bold g bold right parenthesis end subscript bold space bold rightwards harpoon over leftwards harpoon bold space bold PCl subscript bold 3 bold left parenthesis bold g bold right parenthesis end subscript bold space bold plus bold space bold Cl subscript bold 2 bold left parenthesis bold g bold right parenthesis end subscript
  • 3.45 ગ્રામ/લિટર
  • 4.54 ગ્રામ/લિટર
  • 45.4 ગ્રામ/લિટર
  • 5.54 ગ્રામ/લિટર

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