The gas phase decomposition of COCl2, …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Related Questions
Posted by Shikhar Manav 1 year ago
- 0 answers
Posted by Mahi Sharma 1 year ago
- 0 answers
Posted by Priya Dharshini B 11 months, 2 weeks ago
- 4 answers
Posted by Kashish Baisla 3 months, 4 weeks ago
- 1 answers
Posted by Roshni Gupta 1 year ago
- 0 answers
Posted by Bhavishaya 2009 1 year ago
- 0 answers
Posted by Karan Kumar Mohanta 11 months, 1 week ago
- 1 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Dr. Kamlapati Bhatt 7 years, 10 months ago
The order of reaction representing rate law as ,
Rate = k [ COCl2 ]3/2 is 1.5
∴The unit for the rate constant is M -0.5 s-1
(where M represents mole per litre or Molarity of the reactant )
* It should be noted that even though it is a gas phase reaction , where the change in pressure is conveniently measured during the course of a reaction , but still the rate constant is expressed in terms of change in concentration of the reactant ( as per rate law ) in terms of moles per Litre (ie .M or molarity ) / unit of time . The reason being that pressure is temperature as well as unit dependent , where as concentration or molarity is always in moles per litre .
** The concentration of the reactant is easily calculated using the relation PV = nRT , or P = ( n/V ) RT
ie . P = C RT
Explanation :
THe unit for rate constant of a reaction is derived through dimensional analysis .
In this case since ,
k = Rate / [ A ] n
(where , n represents the order of reaction
k =Rate / [ COCl2 ]3/2
or,= Rate / [ COCl2 ]1.5
= ( concentration / time ) x 1 / ( concentration )1.5
= ( mol L-1 ) s-1 x 1 / ( mol L-1 )1.5
= [ ( mol L-1 / { mol L-1 ) x ( 1 / mol L-1 )0.5 }s-1
={ ( 1 / M )0.5 } s -1
or , = M -0.5 s-1
[ where , mol L -1 , represents molarity (M) of the reactant ]
0Thank You