If activation energy of reaction is given as \( (2500+3 T) \mathrm{R} \) then caluclate the valu...

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If activation energy of reaction is given as \( (2500+3 T) \mathrm{R} \) then caluclate the value of \( \ln \mathrm{k} \) (rate constant) at \( 1000 \mathrm{~K} \). Give your answer by multiplying with 10 .
[Given: \( \ln 1000=7, \ln \mathrm{k}=24.775 \) at \( \mathrm{T}=2500 \mathrm{~K} \) and \( \ln 2500 \) \( =7.824] \)
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