The time required for $10$ % completion of a first order reaction at $298 K$ is equal to that required for its $25$ % completion at $308 K$ . If the value of A is $4 \times 10^{10} s^{- 1}$ . Calculate k at $318 K$ and $E_{a}$ .

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Solution

For $10 %$ completion of the reaction, we have
$k (298) = \frac{2.303}{t} l o g \frac{100}{90}$

For $25 %$ completion of the reaction, we have
$k (308) = \frac{2.303}{t} l o g \frac{100}{75}$

$\frac{k (308)}{k (298)} = \frac{\frac{2.303}{t} l o g \frac{100}{75}}{\frac{2.303}{t} l o g \frac{100}{90}} = 2.73$

But, $l o g \frac{k (308)}{k (298)} = \frac{E_{a}}{2.303 R [\frac{T^{'} - T}{T T^{'}}]}$

$l o g 2.73 = \frac{E_{a}}{2.303 \times 8.314} \times \frac{308 - 298}{308 \times 298}$

$E_{a} = 76623 J / m o l = 76.623 k J / m o l$

But, $l o g k = l o g A - \frac{E_{a}}{2.303 R T}$

$l o g k (318 K) = l o g 4 \times 10^{10} - \frac{76623}{2.303 \times 8.314 \times 318} = - 1.9823$

$k (318) = 1.042 \times 10^{- 2} / s$

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Similar questions

The time required for $10 %$ completion of a first order reaction at 298 K is equal to that required for its $25 %$ completion at 308K. If the value of A is 4 \times $10^{10} s^{- 1}$ . Calculate K at 318 K and $E_{a}$

Q. The time required for

10

% completion of a first order reaction at

298

K

is equal to that required for its

25

% completion at

308

K

. If the pre-exponential factor for the reaction is

3.56 \times 10^{9} {s e c}^{- 1}

, calculate its rate constant at

318

K

and also the energy of activation.

Q. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the pre-exponential factor for the reaction is

3.56 \times 10^{9} s^{- 1}

, the value of rate constant at 318 K will be:

Q. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the pre-exponential factor for the reaction is

3.56 \times 10^{9} s^{- 1}

, If the rate constant at 318 K is

x \times 10^{- 3} s e c^{- 1}

100 x

Q. The time required for 5% completion of a reaction at 300 K is equal to that required for its 20% completion at 320 K. If the pre -exponential factor for the reaction is

3.56 \times 10^{9}

s e c^{- 1}

. Then which of the following is correct.(given log95 = 1.98, log80 = 1.90, log5 =0.7, R=

\frac{25}{3} J K^{- 1} m o l^{- 1}

)

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The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4×1010s−1. Calculate k at 318K and Ea.

The time required for $10$ % completion of a first order reaction at $298 K$ is equal to that required for its $25$ % completion at $308 K$ . If the value of A is $4 \times 10^{10} s^{- 1}$ . Calculate k at $318 K$ and $E_{a}$ .