Answer to Question #156063 in General Chemistry for Akash

Question

If 15% of a substance decomposes in first 10 minutes in a first order reaction. Derive the rate constant of the 1st order reaction with graphical presentation and calculate how much of it would remain undecomposed after one hour?

Answer 1

We know that the integrated rate equation of 1st order reaction is,

ln(A₀/A)= kt _______(1)

Where, A₀= initial amount of reactant

A = amount of reactant at any time, t

k = rate constant

The graph of equation (1) will be a straight line pussing through the origin with the slope = k

After 15% substance decomposed amount of reactant remain

, A = (100–15)%

= 85%

A₀ = 100%

Graph:

In the above graph,

OC = ln(100/85) and OB = 10 min

So, slope k = OC÷OB

= ln(100/85)÷10

= 0.01625 min^-1

Hence, the rate constant = 0.01625 min^-1

Let, the amount of Reactant = A%, after 1 hrs

t = 1hrs

= 60 min

k = 0.01625 min^-1

Putting these values in equation (1) we get

ln(100/A) =0.01625×60

Or, 100/A= exp(0.01625×60)

Or, A = exp(0.01625×60)÷100

= 0.0265

Hence after 1 hrs 0.0265% of reactant remain.