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Second Order Reaction

The temperatu...

Question

The temperature dependence of the rate of a chemical reaction is given by Arrhenius equation, $k = A e^{- E_{a} / R T}$ . Which of the following graphs will be a straight line?

l n A v s 1 / T

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l n A v s E_{a}

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l n k v s 1 / T

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l n k v s - E_{a} / R

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Solution

The correct option is C $l n k v s 1 / T$

Arrhenius equation is given by,

$k = A^{\frac{- E_{a}}{R T}}$

Taking ln on both sides:

The plot of $l n k v s 1 / T$ gives a straight line according to the equation,

$l n k = - \frac{E_{a}}{R T} + l n A$ .

A straight line is in the form of $Y = m X + C$

Here, slope $= \frac{- E_{a}}{R}$

Intercept $= l n A$

Hence, the correct answer is option $C$ .

Suggest Corrections

Similar questions

(k)

k = A e^{- E_{a} / R T}

(E_{a})

k = A e^{- E_{a} / R T}

(k)

(E_{a})

k = A e^{- E / R T}

(E_{a})

k = A e^{- E_{a} / R T}

E_{a}

k = A e^{- E_{a} / R T}

A

{log}_{10} k = {log}_{10} A - \frac{E_{a}}{2.303 R T}

{log}_{10} k = 14 - \frac{1.25 \times 10^{4} K}{T}

k

c a l

{m o l}^{- 1}

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