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:

$$-\frac{d[{\mathrm{N}}_{2}{\mathrm{O}}_{5}]}{dt}=0.0005[{\mathrm{N}}_{2}{\mathrm{O}}_{5}]$$

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:

$$y(t)=y(o){e}^{kt}=C{e}^{kt}$$

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:

$$C{e}^{-0.0005(t)}$$

Please comment on my reasoning.

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

$$C{e}^{-0.0005(t)}=-0.9$$

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.

Answer from the book:

$$y(t)=C{e}^{-0.0005}=0.9C\to C{e}^{-0.0005}=0.9C\to {e}^{-0.0005}=0.9$$

What happened to C?

$${\mathrm{N}}_{2}{\mathrm{O}}_{5}\to 2\phantom{\rule{thinmathspace}{0ex}}\mathrm{N}{\mathrm{O}}_{2}+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{O}}_{2}$$

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:

$$-\frac{d[{\mathrm{N}}_{2}{\mathrm{O}}_{5}]}{dt}=0.0005[{\mathrm{N}}_{2}{\mathrm{O}}_{5}]$$

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:

$$y(t)=y(o){e}^{kt}=C{e}^{kt}$$

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:

$$C{e}^{-0.0005(t)}$$

Please comment on my reasoning.

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

$$C{e}^{-0.0005(t)}=-0.9$$

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.

Answer from the book:

$$y(t)=C{e}^{-0.0005}=0.9C\to C{e}^{-0.0005}=0.9C\to {e}^{-0.0005}=0.9$$

What happened to C?

bewagox7

Beginner2022-10-02Added 10 answers

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

$$-\frac{dy(t)}{dt}=ky(t)$$

The solution to this simple ordinary differential equation (ODE) is

$$y(t)=A{e}^{-kt}$$

where A is an arbitrary constant. You can simply check this by substituting into the ODE. Your first mistake was to take $y(t)=A{e}^{kt}$ as the solution of the ODE, neglecting the minus sign. Now, we have the initial condition

$$y(0)=C$$

where C is a known constant which is the initial concentration of ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$. Then our solution becomes

$$y(t)=C{e}^{-kt}$$

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.

$$y({t}^{\ast})=\frac{90}{100}y(0)=\frac{90}{100}C=0.9C$$

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

$$-k{t}^{\ast}=\mathrm{ln}(0.9)$$

and finally, solving for t∗ leads to

$${t}^{\ast}=-\frac{\mathrm{ln}(0.9)}{k}$$

$$-\frac{dy(t)}{dt}=ky(t)$$

The solution to this simple ordinary differential equation (ODE) is

$$y(t)=A{e}^{-kt}$$

where A is an arbitrary constant. You can simply check this by substituting into the ODE. Your first mistake was to take $y(t)=A{e}^{kt}$ as the solution of the ODE, neglecting the minus sign. Now, we have the initial condition

$$y(0)=C$$

where C is a known constant which is the initial concentration of ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$. Then our solution becomes

$$y(t)=C{e}^{-kt}$$

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.

$$y({t}^{\ast})=\frac{90}{100}y(0)=\frac{90}{100}C=0.9C$$

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

$$-k{t}^{\ast}=\mathrm{ln}(0.9)$$

and finally, solving for t∗ leads to

$${t}^{\ast}=-\frac{\mathrm{ln}(0.9)}{k}$$

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