At 400 K- the rate constant of a chemical reaction is 3-10-4s-1 and at 300 K- the rate constant is 6-10-5s-1- Find the value of Ea i-e- activation energy required for the reaction- Take-log10 0-2 - 0-7

Expression for rate constant at two different temperature is given by, nbsp; nbsp;log_10k_2k_1=E_a2.303× R[T_2 T_1T_1T_2] Given: nbsp; k_1=3×10^ 4s^ 1 ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

At 400 K, the...

Question

At $400 K$ , the rate constant of a chemical reaction is $3 \times 10^{- 4} s^{- 1}$ and at $300 K$ , the rate constant is $6 \times 10^{- 5} s^{- 1}$ . Find the value of $E_{a}$ i.e. activation energy required for the reaction.
Take:
${log}_{10} (0.2) = 0.7$

66.31 J m o l^{- 1}

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16.08 k J m o l^{- 1}

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66.31 k J m o l^{- 1}

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160.8 J m o l^{- 1}

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Solution

The correct option is D $160.8 J m o l^{- 1}$
Expression for rate constant at two different temperature is given by,
$l o g_{10} \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
Given:
$k_{1} = 3 \times 10^{- 4} s^{- 1} k_{2} = 6 \times 10^{- 5} s^{- 1};$
$R = 8.314 J m o l^{- 1} K^{- 1};$
$T_{1} = 400 K$
$T_{2} = 300 K$
Substituting the values in above equation
$l o g_{10} \frac{6 \times 10^{- 5}}{3 \times 10^{- 4}} = \frac{E_{a}}{2.303 \times 8.314} [\frac{300 - 400}{300 \times 400}]$

$E_{a} = - l o g (0.2) \times 2.303 \times 8.314 \times 12$

$E_{a} = 160.836 J m o l^{- 1}$

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At 400 K, the rate constant of a chemical reaction is 3×10−4s−1 and at 300 K, the rate constant is 6×10−5s−1. Find the value of Ea i.e. activation energy required for the reaction. Take: log10(0.2)=0.7

At $400 K$ , the rate constant of a chemical reaction is $3 \times 10^{- 4} s^{- 1}$ and at $300 K$ , the rate constant is $6 \times 10^{- 5} s^{- 1}$ . Find the value of $E_{a}$ i.e. activation energy required for the reaction.
Take:
${log}_{10} (0.2) = 0.7$