Calculate the [H2O2] after 10 minutes if [H2O2]0 is 0.200M.

Question

The decomposition of H2O, to water and oxygen follows first-order kinetics with a rate constant of 0.0400 min^-1.

2H₂O₂(I) -> 2H₂O(I)+ 0₂(g)

Calculate the [H₂O₂] after 10 minutes if [H₂O₂]₀ is 0.200M.

Answer 1

First order has the relationship of...

ln[A] = -kt + ln[A]_o

[A] = ?

k = 0.04 min^-1

t = 10 min

[A]_o = 0.200 M

Solving for [A]...

ln[A] = -(0.04)(10) + ln 0.200

ln[A] = -0.4 -1.61

ln[A] = -2.01

[A] = 0.134 M

(be sure to check the math)

Answer 2

The equation for first-order kinetics is ln[A]_t = -kt + ln[A]₀. In this case, our, t (time) is 10 minutes and our k (rate constant) is 0.0400 min^-1. Our ln[A]₀ (Initial concentration) is 0.200M. First, let's manipulate the equation so the natural logs (ln) are on one side.

ln[A]_t = -kt + ln[A]₀

When we subtract ln[A]₀ from both sides, we get...

ln[A]_t - ln[A]₀ = -kt

This simplifies to...

ln([A]_t / [A]₀ ) = -kt

Now, let's plug in our numbers.

ln([A]_t / [0.200 M] ) = -(0.0400 min^-1) (10 min)

To get rid of the ln, we must use base e.

e^{ln([A]t / [0.200 M] )} = e ^{-(0.0400 min-1) (10 min)}

On the left side, the base e cancels out the ln. On the right side, we can calculate it. So, we are left with...

[A]_t / [0.200 M] = .6703

We then multiply both sides by 0.200M to get [A]_t = 0.134 M.

Hope this helps. If you have any questions, please feel free to comment and I'll get back to you!