Question

A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. in what time will the reaction be 80% complete?

Answer 1

(i) For the first order reaction 

\(k = \frac{2.303}t \log\frac a{a - x}\)

When \(x = \frac{40}{100} a = 0.4a\)

t = 50 minutes (given)

\(\therefore k = \frac{2.303}{50\,min}\log\frac a{a - 0.4a} \;or \;k\)

\(= \frac{2.303}{50\, min} \log \frac a{0.6}\)

\(= 0.010216\) min^-1

(ii) t = ?, when reaction is 80% complete

i.e., x = 0.8 a

k = 0.010216 min⁻¹ (calculated above)

\(\therefore t = \frac{2.303}k \log\frac a{a - x}\)

\(= \frac{2.303}{0.010216\, min^{-1}} \log \frac a{a - 0.8a}\)

\(= \frac{2.303}{0.010216 \, min^{-1}}\log \frac 1{0.2}\)

\(= 157.58\) min

Answer 2

1. For the first order reaction k

Assume, a = 100 %, x = 40%, t = 50 minutes 

Therefore, a – x = 100 – 40 = 60

k = (2.303/50) log (100/60) 

k = 0.010216 min^-1

Hence the value of the rate constant is 0.010216 min^-1

2. t = ?, when x = 8O% 

Therefore, a – x = 100 – 80 = 20 

From above, k = 0.0102 16 min^-1

t = (2.303 / 0.010216) log (100 / 20) 

t = 157.58 min

The time at which the reaction will be 80% complete is 157.58 min.