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AveGali [126]
3 years ago
9

For the second-order reaction below, the rate constant of the reaction is 9.4 x 10^-3 M^-1 s^-1. How long (in sec) is required t

o decrease the concentration of A from 2.16 M to 0.40M?
2A --> B
rate = K[A]^2
Chemistry
1 answer:
Archy [21]3 years ago
7 0

Answer : The time taken by the reaction is 2.2\times 10^2s

Explanation :

The expression used for second order kinetics is:

kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}

where,

k = rate constant = 9.4\times 10^{-3}M^{-1}s^{-1}

t = time = ?

[A_t] = final concentration = 0.40 M

[A_o] = initial concentration = 2.16 M

Now put all the given values in the above expression, we get:

(9.4\times 10^{-3})\times t=\frac{1}{0.40}-\frac{1}{2.16}

t=216.706s\aprrox 2.2\times 10^2s

Therefore, the time taken by the reaction is 2.2\times 10^2s

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