Log rate law problem

r12214001 · Dec 10, 2019

my question is in fig.

Slide showing a solution manual calculation of [N2O5] at 150 s using ln[N2O5] = -kt + ln[N2O5]0, with ln(0.100) written as -2.303 and a red box around 2.303, giving -3.343 and 0.0353 mol/L.

lurflurf · Dec 10, 2019

log0.1=-2.303
because that is what it is when you calculate it
did you calculate it?

you do not need to calculate it thought as the log and antilog cancel
$$\mathrm{antilog}(-1.040+\log(0.010))=\mathrm{antilog}(-1.040)\cdot\mathrm{antilog}(\log(0.100))
=\mathrm{antilog}(-1.040)\cdot(0.100)$$

Borek · Dec 10, 2019

ln means natural logarithm, base e.

r12214001 · Dec 10, 2019

TKS i get it

Chestermiller · Dec 10, 2019

$$\ln{0.1}=\ln{\left(\frac{1}{10}\right)}=\ln{1}-\ln{10}=0-2.303=-2.303$$

Log rate law problem

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