Question

if a 1st order reaction 75% of reaction complete in 4 hrs then how much time is required to complete 87.5% of reaction

Answer 1

the remaining solution is here

1/4log4 = 1/tlog8

1/4×2log2 = 1/t ×3log2

2/4=3/t

4/2=t/3

4/2×3= t

t=6hours

Answer 2

5.94 hours is required to complete 87.5% of reaction

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 for decomposition = 4 hrs

a = let initial amount of the reactant  = 100

a - x = amount left after decay process =$\frac{25}{100}\times 100=25$  

$4=\frac{2.303}{k}\log\frac{100}{25}$

$k=\frac{2.303}{4}\log\frac{100}{25}=0.35$

b) for completion of 87.5 % of reaction  

$t=\frac{2.303}{0.35}\log\frac{100}{100-87.5}$

$t=\frac{2.303}{0.35}\times \log(8)$

$t_{87.5}=5.94hrs$

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