Gardiolo0j

2022-10-01

Experiments show that if the chemical reaction
${\mathrm{N}}_{2}{\mathrm{O}}_{5}\to 2\phantom{\rule{thinmathspace}{0ex}}\mathrm{N}{\mathrm{O}}_{2}+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{O}}_{2}$
takes place at $45{\phantom{\rule{thinmathspace}{0ex}}}^{\circ }\mathrm{C}$, the rate of the reaction of dinitrogen pentoxide is proportional to its concentration as follows:
$-\frac{d\left[{\mathrm{N}}_{2}{\mathrm{O}}_{5}\right]}{dt}=0.0005\left[{\mathrm{N}}_{2}{\mathrm{O}}_{5}\right]$
a) Find an expression for the concentration [${\mathrm{N}}_{2}{\mathrm{O}}_{5}$] after t seconds if the initial concentration is C.
b) How long will the reaction take to reduce the concentration of ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$ to 90% of its original value?
Part a
I simply used the instantaneous rate of 0.0005 as the constant k (referring to the rate of reaction) and, since C is a constant, I utilized it as the initial value in the formula:
$y\left(t\right)=y\left(o\right){e}^{kt}=C{e}^{kt}$
I replaced k for 0.0005: the book has a negative sign in front of this value, I am curious as to why since the equal sign would indicate otherwise, no? Nonetheless, the equation:
$C{e}^{-0.0005\left(t\right)}$
Part b
I simply utilized the above equation and set it equal to the decimal form of 90% with a negative sign since the question said reduced
$C{e}^{-0.0005\left(t\right)}=-0.9$
However, I really don't understand conceptually why 90% can be considered 0.9 of the concentration. I just need to clear this up in my mind.
From that point I am pretty lost why the C appears on both sides of the equation and why it disappears.
$y\left(t\right)=C{e}^{-0.0005}=0.9C\to C{e}^{-0.0005}=0.9C\to {e}^{-0.0005}=0.9$
What happened to C?

bewagox7

First of all, we denote the concentration of ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$ at time t with y(t). Then your governing equation is
$-\frac{dy\left(t\right)}{dt}=ky\left(t\right)$
The solution to this simple ordinary differential equation (ODE) is
$y\left(t\right)=A{e}^{-kt}$
where A is an arbitrary constant. You can simply check this by substituting into the ODE. Your first mistake was to take $y\left(t\right)=A{e}^{kt}$ as the solution of the ODE, neglecting the minus sign. Now, we have the initial condition
$y\left(0\right)=C$
where C is a known constant which is the initial concentration of ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$. Then our solution becomes
$y\left(t\right)=C{e}^{-kt}$
So the solution of Part a is finished since we have found the concentration of ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$ at time t in terms of its initial concentration. Part b wants to know at what time the concentration is 90% of the initial concentration, i.e.
$y\left({t}^{\ast }\right)=\frac{90}{100}y\left(0\right)=\frac{90}{100}C=0.9C$
where t∗ is the time we want to find. So we may write
$C{e}^{-k{t}^{\ast }}=0.9C$
Divide the above equation by C
${e}^{-k{t}^{\ast }}=0.9$
taking natural logarithm from both sides we have
$-k{t}^{\ast }=\mathrm{ln}\left(0.9\right)$
and finally, solving for t∗ leads to
${t}^{\ast }=-\frac{\mathrm{ln}\left(0.9\right)}{k}$

Do you have a similar question?