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 $x \times 10^{- 3} s e c^{- 1}$ . $100 x$ is

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Solution

Given,
For $10 %$ completion of the reaction, at $T_{1} = 298 K$

$t = \frac{2.303}{k_{298}} l o g_{10} \frac{100}{90}$

For $25 %$ completion of the reaction, at $T_{2} = 308 K$

$t = \frac{2.303}{k_{308}} l o g_{10} \frac{100}{75}$

$\frac{k_{308}}{k_{298}} = 2.73$

Also, $2.303 l o g_{10}$ $\frac{k_{298}}{k_{308}}$ = $\frac{E_{a}}{R} \times \frac{(T_{2} - T_{1})}{T_{1} \times T_{2}}$

$△ E = 76.6227 k J m o l^{- 1}$

We know from Arhenius equation,

$K = A e^{- E_{a} / R T}$

Take $l n$ both side we get,

$l n K = l n A - E_{A} / R T$
$l n K = l n (3.56 \times 10^{9}) - (76.6 \times 10^{3}) / (8.314 \times 318) = - 7$

$l o g K_{318} = - 7 / 2.303 = - 3.0395$
$K_{318} = 1.01 \times 10^{- 3} s e c^{- 1}$

$x = 1.01 \Rightarrow 100 x = 101$

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Similar questions

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:

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is equal to that required for its

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

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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}

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\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 308K. If the value of A is 4 \times $10^{10} s^{- 1}$ . Calculate K at 318 K and $E_{a}$

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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 pre-exponential factor for the reaction is 3.56×109s−1, If the rate constant at 318 K is x×10−3sec−1 . 100x is

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 $x \times 10^{- 3} s e c^{- 1}$ . $100 x$ is