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Varvara68 [4.7K]
2 years ago
11

What is the half-life of a first-order reaction if it takes 4.4 x 102 seconds for the concentration to decrease from 0.50 M to 0

.20 M? 2.5 x 102 s 3.3 x 102s 1.6s 21 s 27 s
Chemistry
1 answer:
elena55 [62]2 years ago
4 0

Answer: The half-life of a first-order reaction is, 3.3\times 10^2s

Explanation:

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,  

k = rate constant = ?

t = time taken = 440 s

[A_o] = initial amount of the reactant = 0.50 M

[A] = left amount =  0.20 M

Putting values in above equation, we get:

k=\frac{2.303}{440s}\log\frac{0.50}{0.20}

k=2.083\times 10^{-3}s^{-1}

The equation used to calculate half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

Putting values in this equation, we get:

t_{1/2}=\frac{0.693}{2.083\times 10^{-3}s^{-1}}=332.69s=3.3\times 10^2s

Therefore, the half-life of a first-order reaction is, 3.3\times 10^2s

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Answer: The new concentration of a solution of CaSO_{3} is 0.2 M 10.0 mL of a 2.0 M CaSO_{3} solution is diluted to 100 mL.

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