Chemical Kinetics

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New answer posted•a year ago
S
a year ago

CN is a strong field ligand, so pairing occurs.

1. [ F e ( C N ) 6 ] 4 −

So; it is diamagnetic

2. [ F e ( C N ) 6 ] 3 −

So; it is paramagnetic.

3. [ T i ( C N ) 6 ] 3 −

Ti3+ = 4s03d1

 It is paramagnetic

4. [ N i ( C N ) 4 ] 2 −

It is diamagnetic

5. [ C o ( C n ) 6 ] 3 −

Hence,  [ F e ( C N ) 6 ] 3 − a n d     [ T i ( C N ) 6 ] 3 −  are paramagnetic.

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New answer posted•a year ago
S
a year ago

M n 2 + → t 2 g 1 1 1 e g 1 1

(Number of unpaired electron = 5)

μ S = 3 5 = 5 . 9 1 ≅ 6

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New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
S
a year ago

First order reaction

K=2.303tlog? a0a0-x

K=2.30390log? a00.25a0

=0.0154

t=60%=2.303Klog? a0a0…. (2)=2.3030.0154× (1-0.602)=59.51mins≈60

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New answer posted•a year ago
S
a year ago

Rate constant, k = 5.5 × 10-14 s-1.

t = 2 . 3 0 3 k l o g [ R ] 0 [ R ]                

= 2 . 3 0 3 k l o g 3 − ( i )  

t 5 0 % = 2 . 3 0 3 k l o g 2 − ( i i )                

From (i) & (ii)

t 6 7 % t 5 0 % = l o g 3 l o g 2                

= 1.58 t50%

So; t67% is 15.8 × 10-1 times half life.

X = 16 (the ne

...Read more

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New answer posted•a year ago
S
a year ago

According to condition

Rate ∝ [X]1 [Y]0

By using experimental data I and II

4×10−32×10−3=L0.1∴L=0.2

And by using experimental data I and II

M×10−32×10−3=0.40.1

∴ M = 8

Ratio of  (ML)=80.2=40

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New answer posted•a year ago