Solution.
A→BA\to BA→B
d[B]dt=−d[A]dt=k×[A]\frac{d[B]}{dt} = -\frac{d[A]}{dt} = k \times [A]dtd[B]=−dtd[A]=k×[A]
−d[A][A]=kdt-\frac{d[A]}{[A]} = kdt−[A]d[A]=kdt
Integrate:
−ln([A][A0])=k×t-ln(\frac{[A]}{[A0]}) = k \times t−ln([A0][A])=k×t
ln([A0][A])=k×tln(\frac{[A0]}{[A]}) = k \times tln([A][A0])=k×t
Answer:
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