Question

Suppose a hypothetical reaction A ---------> D + E

has a t_1/2 of 8.50 min when the initial concentration of A is 0.150 M.

How long will it take for the concentration to drop to 0.035 M if the reaction is first order with respect to A.

Please write down the steps

Thank you

Answer 1

Half life time t_{​​​​​​1/2} = 8.50 min .......(given)

Initial concentration (C_{​​​​​​0)} = 0.150 M .....(given)

Reduced concentration (C_{​​​​​​t}) = 0.035 M ....(given)

Time =?

We know, half life time for a first order reaction is calculated by formula -

t_{​​​​​​1/2} = 0.693/K

Where, K = rate constant ; t_{​​​​​​1/2} half life time

Or K = 0.693/t_{​​​​​​1/2}

So, k = 0.693/8.50 min

= 0.0815 min^-1

First order reaction is calculated by formula -

t = 1/K × ln (C_{​​​​​​0} / C_{​​​​​​t})

t = 1/ 0.0815 × ln (0.150 M/0.035 M)

= 12.269 min × 1.455

= 17.85 min.

Hence time it will take for the concentration reduced to 0.035 M is = 17.85 min.

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