Question

calculate the two-third life of a first order reaction having k= 6.909× 10 raised to the power -14/sec
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Answer 1

Answer:

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Answer 2

Answer:  $0.16\times 10^{14}sec$

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  

t = time taken for decomposition

a = let initial amount of the reactant  = 100

x = amount decomposed = $\frac{2}{3}\times 100=66.7$

(a - x) = amount left after decay process =( 100-66.7)) = 33.3

for completion of $\frac{2}{3}$ life:

$t=\frac{2.303}{6.909\times 10^{-14}}\log\frac{100}{33.3}$

$t=0.16\times 10^{14}sec$

Thus the two-third life of a first order reaction is $0.16\times 10^{14}sec$