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Question
The first order rate constant for the decomposition of CaCO3 at 700 K is 6.36×10−3s−1 and activation energy is 209 kJ mol−1. Its rate constant (in s−1) at 600 Kisx×10−6. The value of is (Nearest integer)
[Given R=8.31 J K−1mol−1;log 6.36×10−3=−2.19, 10−4.79=1.62×10−5
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JEE MAINS 2021
[Given R=8.31 J K−1mol−1;log 6.36×10−3=−2.19, 10−4.79=1.62×10−5
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Solution
k1=6.36×10−3 s−1T1=700 K
Ea=209 kJ mol−1
k2=x×10−6 s−1T2=600 K
lnk2k1=EaR(1T1−1T2)
log(x×10−66.36×10−3)=209×1038.31×2.303(1700−1600)
log (x×10−6)=−4.79x×10−6=10−4.79
x×10−6=1.62×10−5
x=16.2 =16 (Nearest integer)
Ea=209 kJ mol−1
k2=x×10−6 s−1T2=600 K
lnk2k1=EaR(1T1−1T2)
log(x×10−66.36×10−3)=209×1038.31×2.303(1700−1600)
log (x×10−6)=−4.79x×10−6=10−4.79
x×10−6=1.62×10−5
x=16.2 =16 (Nearest integer)
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