Question

2)If rate constant of reaction becomes twice when temperature increases from 300 K to 310 K, then activation energy (in Joules) of the reaction is (Given : log 2 = 0.30)

2.303 x 90 R

2.303 x 31 R

2.303 x 90 × 31

2.303 x 90 x 31 R​

Answer 1

Answer:

For a zero order reaction, rate constant is given by equation 4.7. 0. R. R k t. 0. 1/2.

Answer 2

The Main Answer is: 2.303 × 90 × 31 R

Given: Rate constant gets doubled-- $K_{2} = 2K_{1}$

           Temperature rises from 300K to 310K

           log2 = 0.3

To Find: Activation Energy ($E_{a}$)

Solution:

This question uses the concept of the "Arrhenius Equation"

It states that-

log($\frac{K_{2} }{K_{1} }$) = $\frac{E_{a}}{2.303R}$$(\frac{1}{T_{1} } - \frac{1}{T_{2} } )$

where, $K_{1}$ and $K_{2}$ are original and new rate constants respectively,

$E_{a}$ is the activation energy, $T_{1}$ and $T_{2}$ are the initial and final temperatures respectively, and R is the gas constant.

⇒ log (2) = $\frac{E_{a}}{2.303R}$ $(\frac{1}{300 } - \frac{1}{310 } )$

⇒ 0.3 = $\frac{E_{a}}{2.303R}$ $(\frac{310 - 300}{300 * 310} )$

⇒ $E_{a}$ = 2.303 R × 0.3 × 300 × 310 / 10

⇒ $E_{a}$ = 2.303 R × 90 × 31

Therefore, the activation energy =  2.303 × 90 × 31 R

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