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Grade 12Physical Chemistry

A frist order reaction completes 60% in 20 minutes. The time required for the completion of 90% of the reactions is approx....

Profile image of Neelam Mandani
9 Years agoGrade 12
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2 Answers

Profile image of Vikas TU
9 Years ago
From the first order reaction,
we get,
k = 2.303log(C0/0.4C)/20 => 0.916/20 => 0.04582
Now for 90%,
0.04582 = 2.303log(C/0.1C)/t
t = 50.25 minutes
Profile image of srinu
7 Years ago
given:
a=100 ,x=60, time taken for 60percent =20min
time taken for 90 percent=? 
we know that for first order ,
k=2.303/t(log([a]/[a-x])
k=2.303/20(log[100/40])
k=2.303/20((log5)-(log2)
we know that k is same throughout the reaction 
t=2.303/k(log[100/100-99]
2.303/20(log5-log2)=2.303/t(log[100/10])
log5-log2)/20=log10/t
t=20/(0.6991-0.3010)
t=50.26