Question

rate constant for first order reaction is 5.78*10^-5 s-1 .what percentage of initial reactant will react in 10 hours?​

Answer 1

Answer:

12.5%

25%

87.5%

75%

Answer :

C

Explanation:

Solution :

5.78×10−5=2.30310×3600×log(A0At)

A0=8A⇒=8A⇒At=A08=12.5%(A0)

Answer 2

Percentage of initial reactant that will react in 10 hours is 12.5 %

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  =$5.78\times 10^{-5}s^{-1}$

t = age of sample = 10 hours = $10\times 3600s=36000s$   (1hour = 3600 s)

a = let initial amount of the reactant  = 100

a - x = amount left after decay process  = ?

$36000=\frac{2.303}{5.78\times 10^{-5}s^{-1}}\log\frac{100}{(a-x)}$

$\log\frac{100}{(a-x)}=0.903$

$\frac{100}{(a-x)}=8.00$

$(a-x)=12.5$

Percentage of initial reactant that will react in 10 hours is $\frac{12.5}{100}\times 100\%=12.5\%$

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