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Chemical Kinetics and Nuclear Chemistry question

2019 · 11 Jan · Shift 1 · Q9
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Chemical Kinetics and Nuclear Chemistry question

2019 · 11 Jan · Shift 1 · Q9

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
If a reaction follows the Arrhenius equation, the plot ln k vs 1(RT){1 \over {\left( {RT} \right)}}(RT)1​ gives straight line with a gradient (−-− y) unit. The energy required to active the reactant is :
  1. A
    y unit
  2. B
    y/R unit
  3. C
    yR unit
  4. D
    −-− y unit
View written solutionFree

Correct answer: A

  1. Write the Arrhenius equation

    The Arrhenius equation is k=Ae−Ea/(RT)k = A e^{-E_a/(RT)}k=Ae−Ea​/(RT)

    Taking natural logarithm on both sides: ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_a}{RT}lnk=lnA−RTEa​​

  2. Compare with the equation of a straight line

    A straight line has the form Y=c+mXY = c + mXY=c+mX

    Here, Y=ln⁡kY = \ln kY=lnk and the variable on the x-axis is given as X=1RTX = \frac{1}{RT}X=RT1​

    So the Arrhenius equation becomes ln⁡k=ln⁡A−Ea(1RT)\ln k = \ln A - E_a\left(\frac{1}{RT}\right)lnk=lnA−Ea​(RT1​)

    Therefore, the slope (gradient) of the plot of ln⁡k\ln klnk vs 1RT\dfrac{1}{RT}RT1​ is m=−Eam = -E_am=−Ea​

  3. Use the given gradient

    The gradient is given as −y-y−y.

    So, −Ea=−y-E_a = -y−Ea​=−y which gives Ea=yE_a = yEa​=y

  4. Conclusion

    The activation energy required is y unit\boxed{y\text{ unit}}y unit​

  5. Check options

    • A: yyy unit ✅
    • B: y/Ry/Ry/R unit ❌
    • C: yRyRyR unit ❌
    • D: −y-y−y unit ❌

Hence, the correct option is A.

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