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Chemical Kinetics and Nuclear Chemistry question

2002 · Shift 0 · Q42
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Chemical Kinetics and Nuclear Chemistry question

2002 · Shift 0 · Q42

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The integrated rate equation is Rt = log C0 - log Ct . The straight line graph is obtained by plotting
  1. A
    time vs log Ct
  2. B
    1time{1 \over {time}}time1​ vs Ct
  3. C
    time vs Ct
  4. D
    1time{1 \over {time}}time1​ vs 1Ct{1 \over {{C_t}}}Ct​1​
View written solutionFree

Correct answer: A

  1. Identify the given integrated rate equation

    The question gives: Rt=log⁡C0−log⁡CtRt = \log C_0 - \log C_tRt=logC0​−logCt​

    where RRR is a constant, C0C_0C0​ is the initial concentration, and CtC_tCt​ is the concentration at time ttt.

  2. Rearrange into straight-line form

    We write: log⁡Ct=log⁡C0−Rt\log C_t = \log C_0 - RtlogCt​=logC0​−Rt

    This is of the form: y=c+mxy = c + mxy=c+mx

    with

    • y=log⁡Cty = \log C_ty=logCt​
    • x=tx = tx=t
    • intercept =log⁡C0= \log C_0=logC0​
    • slope =−R= -R=−R
  3. Interpret the graph

    So, a plot of log⁡Ct\log C_tlogCt​ versus ttt gives a straight line.

    Equivalently, plotting time vs log⁡Ct\log C_tlogCt​ also represents a linear relation between the two variables.

  4. Check the options

    • A: time vs log⁡Ct\log C_tlogCt​ → linear relation ✔️
    • B: 1time\dfrac{1}{\text{time}}time1​ vs CtC_tCt​ → not linear ✖️
    • C: time vs CtC_tCt​ → not linear ✖️
    • D: 1time\dfrac{1}{\text{time}}time1​ vs 1Ct\dfrac{1}{C_t}Ct​1​ → not linear ✖️
  5. Final answer

    The correct option is: A\boxed{A}A​

Previous

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