lnC=lnC0-kt(For first order reactions)
a)From the plots, we get k =0.04 (1/min.)
X=1-exp(-k*t)
=1-exp(-0.0401*30)
=0.699
b)k1=0.04(1/min.) k2=0.024(1/min.) [From slopes of the plots]
c)From plot of
lnK=lnA-(Ea/RT)
From slope and intercept values.we get Ea=(85301 J/mol),A=0.024(1/min.)
d)By using of arrhenius equation,we get
ln(k2/k1)=E/R((1/T1)-(1/T2))
k1=0.0401 (1/min.) at T1=303.15K
k2 be at T2=313.15K
By solving from the above equation,we get
k2(400C)=0.118(1/min.)
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