मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Match the graph given in Column I with the order of reaction given in Column II. More than one item in Column I may link to the same item of Column II. Column I Column II

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प्रश्न

Match the graph given in Column I with the order of reaction given in Column II. More than one item in Column I may link to the same item of Column II.

  Column I Column II
(i)  
(ii)  (a) 1st order
(iii) (b) Zero-order
(iv)    
जोड्या लावा/जोड्या जुळवा
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उत्तर

  Column I Column II
(i) (a) 1st order
(ii) (b) Zero-order
(iii)  (b) Zero order
(iv) (a) 1st order

Explanation:

For zero order reaction rate equation may be written as [R] = – kt + [R0]

Which denotes a straight line equation similar to y = mx + c

On transmitting (i) `([R] - [R_0])/t = - k`

`k = ([R_0] - [R])/t`

k = Rate

Rate = `k.[t]^0`

Rate `oo  [t]^0`
For a first order reaction `(dx)/(dt) oo` [concentration]

∴ Graph between rate and concentration may be drawn as


`k = 2.303/t log  ([R]_0)/([R])`

`(kt)/2.303 = log  ([R]_0)/([R])`

`(kt)/2.303 = log[R_0] - log[R]`

\[\ce{log[R] = \underset{(Slope)}{\frac{-k}{2.303}} t +  \underset{(Intercept)}{log[R]_0}}\]

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पाठ 4: Chemical Kinetics - Exercises [पृष्ठ ५७]

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एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 12
पाठ 4 Chemical Kinetics
Exercises | Q IV. 53. | पृष्ठ ५७

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