Question

3 A first order reaction is 40% complete in 50 mins. Calculate the value of the rate
constant In what time will the reaction be 80% complete?
tant for a first order reaction is 1.54 x 10's Calculatets ha​

Answer 1

Answer:

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.

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Explanation:

Answer 2

Answer: k = 0.01                 t = 160 min

Explanation:

For first order reaction

x = A(1 - e^-kt)

It is 40% complete in 50 min.

A=100%              x =40%

40 = 100 (1 - e^-k(50))

2/5 = 1 - e^-50k

-3/5 = -e^-50k

3/5 = 1/e^50k

e^50k = 5/3

50k = ln(5/3)

k = ln(5/3) / 50  

k = 0.01 --------------- 1

It is 80% complete in t min.

80 = 100(1 -e^-kt)

4/5 = 1-e^-kt

1/5 = e^-kt

1/5 = 1/e^kt

e^kt =5

kt = ln 5

From 1

t = ln 5 / 0.01

t = 160 min

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