[Boxxxed]

I haven't done calculus in years so I'm confused with the following.

2H2O2 ----> 2H2O + O2

if

In[H2O2]t/[H2O2]0 = -kt = -0.106

Then

[H2O2]t/[H2O2]0 = 0.90 --->How do you arrive at this? How is the integral eliminated?

[rabolisk]

That's not an integral. It's a natural log. You just do the inverse of natural log (e^-kt)
If your question is how does one derive the integrated rate law, then...

This is a separable differential equation.

You have the reaction 2H₂O₂  2H₂O + O₂ which you know to be first-order with respect to H₂O₂.

This means that the rate of disappearance of H₂O₂ (also known as rate of reaction) is proportional to the concentration of H₂O₂. Mathematically, this can be represented as

$$ \frac{-d[A]}{dt} = k[A] /$$ where A is H₂O₂ and k is the rate constant.

Now I'm going to rearrange, then integrate both sides.

$$ \frac{-d[A]}{dt} = k[A] /$$
$$ \frac{d[A]}{[A]} = -kdt /$$
$$ \int \frac{d[A]}{[A]} = \int -kdt /$$
$$ \ln[A] = -kt + C /$$ where C is a constant (due to integration)
Left hand side always equals right hand side, and when t = 0, left hand side is
$$ \ln[A]_{0} = C /$$
So the overall equal is
$$ \ln[A]_{t} = -kt + \ln[A}_{0} /$$
$$ \ln\frac{[A]_{t}}{[A]_{0}} = -kt /$$