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

The correct option is D 2 cal K=Ae^ Ea/RT slope +Ea/R × 2.303=+1/2.303 Ea=REa=2cal ...

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Standard XII

Physics

Dependence of Resistance on Temperature

Graph between...

Question

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:

2.303

\times

2 cal

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\frac{5}{2.303} c a l

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2 cal

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Solution

The correct option is D 2 cal
$K = A e^{- E a / R T}$
slope
$+ \frac{E a}{R \times 2.303} = \frac{+ 1}{2.303}$
Ea=R
Ea=2cal

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Similar questions

Q. 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:

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. 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

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. 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:

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

The correct option is D 2 calK=Ae−Ea/RTslope+EaR×2.303=+12.303Ea=REa=2cal

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:

The correct option is D 2 cal
$K = A e^{- E a / R T}$
slope
$+ \frac{E a}{R \times 2.303} = \frac{+ 1}{2.303}$
Ea=R
Ea=2cal