Rate constant of the reactions at 37oC and 77oC are 2-105 s-1 and 8-105 s-1 respectively- Find the value of activation energy- Take-log 4 - 0-602

Expression for rate constant at two different temperature is given by nbsp; nbsp; nbsp;logk_2k_1=E_a2.303 R[1T_1 1T_2] nbsp; nbsp; nbsp;log8×10 ...

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Byju's Answer

Standard XII

Chemistry

Rate of Reaction

Rate constant...

Question

Rate constant of the reactions at $37^{o} C$ and $77^{o} C$ are $2 \times 10^{5} s^{- 1}$ and $8 \times 10^{5} s^{- 1}$ respectively. Find the value of activation energy.
Take,
$l o g 4 = 0.602$

12.1 k J m o l^{- 1}

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31.3 k J m o l^{- 1}

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4.6 k J m o l^{- 1}

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84.7 k J m o l^{- 1}

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Solution

The correct option is B $31.3 k J m o l^{- 1}$
Expression for rate constant at two different temperature is given by
$l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$l o g \frac{8 \times 10^{5}}{2 \times 10^{5}} = \frac{E_{a}}{2.303 \times 8.314} (\frac{1}{310} - \frac{1}{350})$

$l o g 4 = \frac{E_{a}}{2.303 \times 8.314} (\frac{1}{310} - \frac{1}{350})$

$0.602 = \frac{E_{a}}{2.303 \times 8.314} (\frac{40}{310 \times 350})$

$E_{a} = 31.3 k J m o l^{- 1}$

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Rate constant of the reactions at 37oC and 77oC are 2×105s−1 and 8×105s−1 respectively. Find the value of activation energy. Take, log 4=0.602

The correct option is B 31.3 kJ mol−1 Expression for rate constant at two different temperature is given by logk2k1=Ea2.303R[1T1−1T2] log8×1052×105=Ea2.303×8.314(1310−1350) log 4=Ea2.303×8.314(1310−1350) 0.602=Ea2.303×8.314(40310×350) Ea=31.3 kJ mol−1

Rate constant of the reactions at $37^{o} C$ and $77^{o} C$ are $2 \times 10^{5} s^{- 1}$ and $8 \times 10^{5} s^{- 1}$ respectively. Find the value of activation energy.
Take,
$l o g 4 = 0.602$