Question

a first order reaction is 40% complete in 50 minutes. calculate the value of rate constant. in which time will the reaction be 80% complete

Answer 1

Answer:

Explanation:

1) For the first order reaction:

k = (2.303 / t) log (a / a – x)

When x = (40 / 100) a = 0.4 a

t = 50 minutes (given)

Therefore, k = (2.303 / 50) log (a / a – 0.4 a)

k = (2.303 / 50) log (1 / 0.6)

= 0.010216 min-1

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

2) t= ?, when x = 0.8 a

From above, k = 0.010216 min-1

Therefore, t = (2.303 / 0.010216) log (a / a – 0.8 a)

= (2.303 / 0.010216) log (1 / 0.2) = 157.58 min

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

Answer 2

Answer:

157.57

Explanation:

Since we know k=502.303log(a−2a/5a)

⟹k=502.303log35

Now , t=2.3032.303×50log(35)log(5)

⟹t=50×0.2220.699

⟹t=50×3.15

⟹t=157.57 min