# Answer in General Chemistry for Akash #156063

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?

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.

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