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Standard XII

Chemistry

First Order Reaction

For a first o...

Question

For a first order reaction $A \to P$ , the temperature $(T)$ and dependent rate constant $(k)$ , was found to follow the equation $log k = - (2000) \frac{1}{T} + 6.0$ . What is activation energy?

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Solution

Equate with this formula, $l o g K = l o g A - \frac{E_{a}}{R T}$ , Now
$- \frac{E_{a}}{R T} = - \frac{2000}{T}$ or $E_{a} = 2000 \times 8.314 = 16.6 K J$

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For a first order reaction A→P, the temperature (T) and dependent rate constant (k), was found to follow the equation logk=−(2000)1T+6.0. What is activation energy?

Equate with this formula, logK=logA−EaRT , Now−EaRT=−2000T or Ea=2000×8.314=16.6KJ

For a first order reaction $A \to P$ , the temperature $(T)$ and dependent rate constant $(k)$ , was found to follow the equation $log k = - (2000) \frac{1}{T} + 6.0$ . What is activation energy?

Equate with this formula, $l o g K = l o g A - \frac{E_{a}}{R T}$ , Now
$- \frac{E_{a}}{R T} = - \frac{2000}{T}$ or $E_{a} = 2000 \times 8.314 = 16.6 K J$