Figure shows a graph in log10K vs- 1-T where- K is rate constant and T is temperature- The straight line BC has slope- tan-1-2-303 and an intercept of 5 on y-axis- Thus Ea- the energy of activation is-

The correct option is C 2 cal #160; E_a #160;, the energy of activation is 2 #160;cal .It can be solved as follows : 2.303 log_10K= E_a/RT ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

Figure shows ...

Question

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:

2.303 \times 2 c a l

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\frac{2}{2.303} c a l

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2 cal

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none of these

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Solution

The correct option is C 2 cal
$E_{a}$ , the energy of activation is $2 c a l$ .
It can be solved as follows :
$2.303 l o g_{10} K = - \frac{E_{a}}{R T} + 2.303 l o g_{10} A^{1}$
Thus, $- \frac{E_{a}}{2.303 R} t a n θ = - \frac{1}{2.303}$
$∴ E_{a} = R = 2 c a l$

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Similar questions

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. 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. 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 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. The activation energy for a reaction at the temperature

T

K was found to be 2.303 RT

J / m o l^{- 1}

. The ratio of the rate constant to the Arrhenius factor is:

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Figure shows a graph in log10Kvs.1T where, K is rate constant and T is temperature. The straight line BC has slope, tanθ=−12.303 and an intercept of 5 on y-axis. Thus Ea, the energy of activation is:

The correct option is C 2 cal Ea , the energy of activation is 2cal.It can be solved as follows :2.303log10K=−EaRT+2.303log10A1Thus, −Ea2.303Rtanθ=−12.303∴Ea=R=2cal

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:

The correct option is C 2 cal
$E_{a}$ , the energy of activation is $2 c a l$ .
It can be solved as follows :
$2.303 l o g_{10} K = - \frac{E_{a}}{R T} + 2.303 l o g_{10} A^{1}$
Thus, $- \frac{E_{a}}{2.303 R} t a n θ = - \frac{1}{2.303}$
$∴ E_{a} = R = 2 c a l$