Answer :
Answer : The value of rate constant is, [tex]0.0949\text{ min}^{-1}[/tex]
Explanation :
First we have to calculate the rate constant, we use the formula :
Expression for rate law for first order kinetics is given by:
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = ?
t = time passed by the sample = 20 min
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 100 - 85 = 15 g
Now put all the given values in above equation, we get
[tex]k=\frac{2.303}{20}\log\frac{100}{15}[/tex]
[tex]k=0.0949\text{ min}^{-1}[/tex]
Therefore, the value of rate constant is, [tex]0.0949\text{ min}^{-1}[/tex]
For the first order a user can do this.
According to the question,
- Time = 20 min
- Decomposition percent = 85%
Now,
The First order rate law will be:
→ [tex]-r = k \ C[/tex]
By differentiating it, we get
[tex]-\frac{dC}{dt} = k \ C[/tex]
[tex]- ln (\frac{C}{Co} ) = kt[/tex]
By substituting the values, we get
[tex]-ln (\frac{0.85 Co}{Co} ) = k\times 20[/tex]
[tex]-ln (0.85) = k\times 20[/tex]
[tex]k = 8.126\times 10^{-3} / min[/tex]
Thus the above answer is right.
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