Graph between log k and 1T is a straight line with OX-5-tan- 1 2-303 - Hence Ea will be-

The correct option is D 2 log k =log A E _ a 2.303RT #160;Slope = E _ a 2.303R = 1 2.303 (given) E _ a = 2.303× R× slope=R= ...

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Standard XII

Chemistry

Activation Energy

Graph between...

Question

Graph between $log k$ and $\frac{1}{T}$ is a straight line with $O X = 5, tan θ = (\frac{1}{2.303})$ . Hence $E_{a}$ will be:

2.303 \times 2

cal

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\frac{5}{2.303} c a l

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Solution

The correct option is D $- 2$
$log k = log A - \frac{E_{a}}{2.303 R T}$

Slope $= \frac{- E_{a}}{2.303 R} = \frac{1}{2.303} (g i v e n)$

$E_{a} = - 2.303 \times R \times s l o p e = R = - 2 c a l$

Hence, the correct option is $C$ .

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Similar questions

Q. Graph between logk and

\frac{1}{T}

is a straight line reaction as with

O X = 5, tan Θ = (\frac{1}{2.303})

. Hence

E_{a}

will be:

Q. Graph between log K and

\frac{1}{T}

[Where K is rate constant

(S^{- 1})

and T is temperature (K)] is a straight line with

O X = 5, θ = t a n^{- 1} [- \frac{1}{2.303}]

Hence

E_{a}

and log A respectively will be:

Q. Figure shows a graph in

l o g_{10} K v s . \frac{1}{T}

where, K is rate constant and T is temperature. The straight line BC has slope,

t a n θ = - \frac{1}{2.303}

and an intercept of 5 on y-axis. Thus

E_{a}

, the energy of activation is:

Q. Rate constant

k

of a reaction varies with temperature according to the equation

log k = c o n s t a n t - \frac{E_{a}}{2.303 R} \times \frac{1}{T}

where

E_{a}

is the energy of activation for the reaction. When a graph is plotted for

log k

\frac{1}{T}

a straight line with a slope

- 6670 k

is obtained. The activation energy for this reaction will be:

(

R = 8.314 J K^{- 1} {m o l}^{- 1})

Q. Milk turns sour at

40^{\circ}

C three times as faster as

0^{\circ}

C. Hence,

E_{a}

in the process of turning of milk sour is

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Graph between logk and 1T is a straight line with OX=5,tanθ=(12.303). Hence Ea will be:

The correct option is D −2logk=logA−Ea2.303RT Slope =−Ea2.303R=12.303(given)Ea=−2.303×R×slope=R=−2 calHence, the correct option is C.

Graph between $log k$ and $\frac{1}{T}$ is a straight line with $O X = 5, tan θ = (\frac{1}{2.303})$ . Hence $E_{a}$ will be:

The correct option is D $- 2$
$log k = log A - \frac{E_{a}}{2.303 R T}$

Slope $= \frac{- E_{a}}{2.303 R} = \frac{1}{2.303} (g i v e n)$

$E_{a} = - 2.303 \times R \times s l o p e = R = - 2 c a l$

Hence, the correct option is $C$ .