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:


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:


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 $C$?


1 Answer 1


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.