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Experiments show that if the chemical reaction

$$\mathrm{N_2O_5 \to 2\,NO_2 + \frac{1}{2}\,O_2}$$

takes place at $45\,\rm^\circ C$, the rate of the reaction of dinitrogen pentoxide is proportional to its concentration as follows:

$$-\frac{d[\mathrm{N_2O_5}]}{dt} = 0.0005[\mathrm{N_2O_5}]$$

a) Find an expression for the concentration $[\mathrm{N_2O_5}]$ after $t$ seconds if the initial concentration is $C$.

b) How long will the reaction take to reduce the concentration of $\mathrm{N_2O_5}$ 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:

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

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:

$$Ce^{-0.0005(t)}$$

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

$$Ce^{-0.0005(t)} = -0.9$$

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:

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

What happened to $C$?

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1 Answer 1

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First of all, we denote the concentration of $\mathrm{N_2O_5}$ at time $t$ with $y(t)$. Then your governing equation is

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

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 $\mathrm{N_2O_5}$. Then our solution becomes

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

So the solution of Part a is finished since we have found the concentration of $\mathrm{N_2O_5}$ 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.

$$y({t^*}) = {{90} \over {100}}y(0) = {{90} \over {100}}C = 0.9C$$

where ${t^*}$ is the time we want to find. So we may write

$$C{e^{ - k{t^*}}} = 0.9C$$

Divide the above equation by $C$

$${e^{ - k{t^*}}} = 0.9$$

taking natural logarithm from both sides we have

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

and finally, solving for ${t^*}$ leads to

$${t^*} = - {{\ln (0.9)} \over k}$$

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