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Standard XII

Chemistry

Half life

The time requ...

Question

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 × 10¹⁰s⁻¹. Calculate k at 318 K and E_a.

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Solution

For a first order reaction,

At 298 K,

At 308 K,

According to the question,

From Arrhenius equation, we obtain

To calculate k at 318 K,

It is given that,

Again, from Arrhenius equation, we obtain

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

For a first order reaction, At 298 K, At 308 K, According to the question, From Arrhenius equation, we obtain To calculate k at 318 K, It is given that, Again, from Arrhenius equation, we obtain

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 × 10¹⁰s⁻¹. Calculate k at 318 K and E_a.

For a first order reaction,

At 298 K,

At 308 K,

According to the question,

From Arrhenius equation, we obtain

To calculate k at 318 K,

It is given that,

Again, from Arrhenius equation, we obtain