Book Store

Download books and chapters from book store.
Currently only available for.
CBSE

Gujarati JEE Chemistry : રસાયણિક ગતિકી

Multiple Choice Questions

91.

ΛH àª¬àª°àª¾àª¬àª°
àª¶à«‚àª¨à«àª¯
ΛH àª¥à«€ àª“àª›à«àª‚
ΛH àª¥à«€ àªµàª§à«

92.

64 àª—àª£à«‹
10 àª—àª£à«‹
32 àª—àª£à«‹
24 àª—àª£à«‹

93.

0.1
0.002
0.01
0.001

94.

I. $bold A bold space bold rightwards arrow with bold space bold space bold space bold space bold space bold space bold space bold space bold space bold K subscript bold 1 bold space bold space bold space bold space bold space bold space bold space bold space on top bold space bold B$ àª¸àª•à«àª°àª¿àª¯àª•àª°àª£ àªŠàª°à«àªœàª¾ E₁
II. $bold A bold space bold rightwards arrow with bold space bold space bold space bold space bold space bold space bold space bold space bold space bold K subscript bold 2 bold space bold space bold space bold space bold space bold space bold space bold space on top bold space bold C$ àª¸àª•à«àª°àª¿àª¯àª•àª°àª£ àªŠàª°à«àªœàª¾ E₂ àª›à«‡, àªªàª°àª‚àª¤à« E₂ = 2E₁ àªœà«‹ àª¹à«‹àª¯, àª¤à«‹ K₁ àª…àª¨à«‡ K₂ àªµàªšà«àªšà«‡ àª•àª¯à«‹ àª¸àª‚àª¬àª‚àª§ àª›à«‡ ?

$straight K subscript 1 space equals space 2 straight K subscript 2 space times space straight e to the power of begin inline style bevelled straight E subscript 2 over RT end style end exponent$
$straight K subscript 1 space equals space straight K subscript 2 space times space straight A space times space straight e to the power of begin inline style bevelled straight E subscript 1 over RT end style end exponent$
$straight K subscript 2 space equals space straight K subscript 1 space times space straight e to the power of begin inline style bevelled straight E subscript 2 over straight T end style end exponent$
àª†àªªà«‡àª² àªªà«ˆàª•à«€ àªàª• àªªàª£ àª¨àª¹à«€

95.

$bold C subscript bold 2 bold H subscript bold 5 bold I bold space bold plus bold space bold OH to the power of bold minus bold space bold rightwards arrow bold space bold C subscript bold 2 bold H subscript bold 5 bold OH bold space bold plus bold space bold I to the power of bold minus$ àªªà«àª°àª•à«àª°àª¿àª¯àª¾ àª®àª¾àªŸà«‡ 30 $degree$ àª…àª¨à«‡ 60 $degree$ àª¸à«‡. àª¤àª¾àªªàª®àª¾àª¨à«‡ àªµà«‡àª—-àª…àªšàª³àª¾àª‚àª• àª…àª¨à«àª•à«àª°àª®à«‡ 0.325 àª…àª¨à«‡ 6.735 àª²àª¿àªŸàª° àª®à«‹àª²^-1 àª¸à«‡àª•àª¨à«àª¡^-1 àª¹à«‹àª¯ àª¤à«‹ àªªà«àª°àª•à«àª°àª¿àª¯àª¾àª¨à«€ àª¸àª•à«àª°àª¿àª¯àª•àª°àª£ àªŠàª°à«àªœàª¾àª¨à«àª‚ àª®à«‚àª²à«àª¯ àª•à«‡àªŸàª²à«àª‚ àª¹àª¶à«‡ ?

84773 àª•à«…àª²àª°à«€
20260 àª•àª¿. àª•à«…àª²àª°à«€
361.44 àª•àª¿. àª•à«…àª²àª°à«€
20260 àª•à«…àª²àª°à«€

96. એક ઉષ્માક્ષેપક પ્રક્રિયાની પુરોગામી પ્રક્રિયા અને પ્રતિગામી પ્રક્રિયાની સક્રિયકરણ ઊર્જા અનુક્રમે E_f અને E_r હોય, તો

E_f = E_r
E_f < E_r
$straight E subscript straight f space greater than space straight E subscript straight f$
$straight E subscript straight f space greater than greater than greater than space straight E subscript straight r$

97.

1.289 àª—àª£à«‹
12.89 àª—àª£à«‹
0.129 àª—àª£à«‹
25 %

98.

10 àª•àª¿.àª•à«…àª²àª°à«€ àª®à«‹àª²^-1
20àª•àª¿.àª•à«…àª²àª°à«€ àª®à«‹àª²^-1
15 àª•àª¿.àª•à«…àª²àª°à«€ àª®à«‹àª²^-1
àª¶à«àª¨à«àª¯

99.

àª¬à«‡ àªœà«àª¦à«€-àªœà«àª¦à«€ àª°àª¾àª¸àª¾àª¯àª£àª¿àª• àªªà«àª°àª•à«àª°àª¿àª¯àª¾àª“àª¨àª¾ àªµà«‡àª—-àª…àªšàª³àª¾àª‚àª• àª…àª¨à«àª•à«àª°àª®à«‡ K₁ = 10¹⁶ $bold times bold space bold e to the power of begin inline style bevelled fraction numerator bold minus bold 2000 over denominator bold T end fraction end style end exponent$ àª…àª¨à«‡ $bold K subscript bold 2 bold space bold equals bold space bold 10 to the power of bold 15 bold space bold times bold space bold e to the power of begin inline style bevelled fraction numerator bold minus bold 1000 over denominator bold T end fraction end style end exponent$ àª›à«‡. àª•àª¯àª¾ àª•à«…àª²à«àªµàª¿àª¨ àª¤àª¾àªªàª®àª¾àª¨à«‡ K₁ = K₂ àª¥àª¾àª¯ ?

2000 K
1000 K
$fraction numerator 2000 over denominator 2.303 end fraction space straight K$
$fraction numerator 1000 over denominator 2.303 end fraction space straight K$

100.

$fraction numerator straight K subscript 1 apostrophe over denominator straight K subscript 1 end fraction space space equals space fraction numerator straight K subscript 2 apostrophe over denominator straight K subscript 2 end fraction space equals space 0$
$fraction numerator straight K subscript 1 apostrophe over denominator straight K subscript 1 end fraction space equals space fraction numerator straight K subscript 2 apostrophe over denominator straight K subscript 2 end fraction$
$fraction numerator straight K subscript 1 apostrophe over denominator straight K subscript 1 end fraction space greater than space fraction numerator straight K subscript 2 apostrophe over denominator straight K subscript 2 end fraction$
àª†àªªà«‡àª² àªªà«ˆàª•à«€ àªàª• àªªàª£ àª¨àª¹à«€

Switch

Book Store

Subject

Gujarati JEE Chemistry : રસાયણિક ગતિકી

Multiple Choice Questions

Switch