Question

Hello,
The half life of a substance is 60min. How long will it take to consume the 90% of the reactant. ​

Answer 1

Answer:

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Explanation:

The half-life of first order reactions is 60 minutes. How long will it take to consume 90% of the reactant?

The half life of a first order reaction is given by the equation,

t(1/2) = 0.693/k

Where “k” is the rate constant.

From the equation, we can calculate the rate constant.

60 = 0.693/k

k = 0.693/60 = 0.01155 min^(-1)

Now, use the equation

k= (2.303/t)log(initial conc./final conc.)

After consuming 90% of the reactant, the final concentration will be 10% of the initial concentration.

Assume, initial concentration is x

Final concentration is 10% of x = (10/100)x = 0.1x

Therefore

Initial conc./final conc. = x/0.1x = 10

Substitute this in the above formula

0.01155 = (2.303/t)log(10)

0.01155 = 2.303/t

t = 199.4 minutes.

Therefore it takes 199.4 minutes to consume 90% of the reactant.

Answer 2

Answer:

The Answer is 200 minutes

Step by Step Explanation:

1) For the first order reaction

k=0.693t1/2=0.69360=11.55×10−3min−1k=0.693t1/2=0.69360=11.55×10-3min-1

Applying first order kinetic equation,

t=2.303klog.a(a−x)t=2.303klog.a(a-x)

Given: a=100,x=90a=100,x=90, i.e., (a−x)=(100−90)=10(a-x)=(100-90)=10

Hence, t=2.30311.5×10−3log10=200mint=2.30311.5×10-3log10=200min

2) Alternate method-

t1/2=60mint1/2=60min

t99.9%t1/2=log(100100−90)0.3t99.9%t1/2=log(100100-90)0.3

t90%=t1/2×log100.3=60×10.3=200min