1
$\begingroup$

enter image description here

I have no idea on how to answer this question. Please help.

$\endgroup$

2 Answers 2

4
$\begingroup$

The concentration must be exponential. Suppose $C(t)$ is the concentration at time $t$ (min). $C(t)$ is of the form $$ C(t) = A2^{-kt} $$ where $A$ and $k$ are constants you need to find. Using known data points, you can find that $A = 0.4$ and $k = \frac 1{120}$.

The slope of $C(t)$ at $t = 0$ is $C'(0)$, so compute $C'(t)$ first: \begin{align} C'(t) & = -kA 2^{-kt} \log 2 \\ C'(0) & = -\frac 1{120} \cdot 0.4 \cdot \log 2 \approx -0.00231. \end{align} Therefore, the tangent line at $0$ would intersect the $x$-axis at \begin{align} -\frac{C(0)}{C'(0)} & \approx \frac{0.4}{0.00231} \approx 173.12 \end{align}

$\endgroup$
5
  • $\begingroup$ How do you know the base is $2$? Does it matter what the base is, or can any base be in some way turned into another base? I was under the impression that the base must be $e$. $\endgroup$
    – Greg
    Sep 18, 2013 at 5:02
  • $\begingroup$ It does not matter what base you choose. You will get the same result. $\endgroup$
    – Tunococ
    Sep 19, 2013 at 1:58
  • $\begingroup$ Incredible! I had never noticed before that $$a^{bx}=e^{\ln a bx}$$ I see it now $\endgroup$
    – Greg
    Sep 19, 2013 at 3:10
  • $\begingroup$ @Greg That was not quite right. It's $a^{bx} = e^{bx \ln a}$. (I have no idea why you need $bx$ instead of just $x$, but this works also.) $\endgroup$
    – Tunococ
    Sep 19, 2013 at 10:19
  • $\begingroup$ You're right. I meant to put $a$ as the only argument of $ln$, it just came out like that in LATEX. $\endgroup$
    – Greg
    Sep 19, 2013 at 13:10
2
$\begingroup$

I'm unsure why this was migrated from chemistry.

The rates of chemical reactions are governed by rate equations, for example (Where $\mathrm{A}$ is the reactant, $k$ is the rate constant, and $n$ is the reaction order which usually must be experimentally determined:

$$rate=-\dfrac{dA}{dt}=k[\mathrm{A}]^n$$

We can solve the differential equation to get a generalized equation for $\mathrm{A}$ as a function of time.

$$-\dfrac{dA}{[\mathrm{A}]^n}=kdt$$ $$-\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{dA}{[\mathrm{A}]^n}=\int_0^tkdt=kt$$

If $n=0$ (zeroth order), then the left integral evaluates to:

$$-\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{dA}{[\mathrm{A}]^0}=[\mathrm{A}]_0-[\mathrm{A}]_t=kt$$

This equation is linear. Since your graph is not linear, your data is not zerothorder.

If $n=1$ (first order):

$$-\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{dA}{[\mathrm{A}]}=\ln{[\mathrm{A}]_t}-\ln{[\mathrm{A}]_0}=\ln{\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}=-kt$$

Solving for $[\mathrm{A}]_t$ gives you:

$$\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=e^{-kt}$$ $${[\mathrm{A}]_t}=[\mathrm{A}]_0e^{-kt}$$

Your graph looks like an exponential decay, so you can find the slope of the tangent line at time $t=0$ by differentiating, except you need to know $k$ to evaluate at $t=0$.

$$\dfrac{d}{dt}[\mathrm{A}]_t=\dfrac{d}{dt}[\mathrm{A}]_0e^{-kt}=-k[\mathrm{A}]_0e^{-kt}$$

However, since the half-life of a first order reaction is given by $$t_{1/2}=\dfrac{\ln{(2)}}{k}$$

you can solve for $k$ since the half-life is marked on the graph.

$\endgroup$
2
  • $\begingroup$ I was the one who migrated the question… I thought, when I first look at it, that it was a "graphical" question, asking to spot a tangent from a series of lines (i.e. that it had nothing to do with chemical kinetics, even though that was the background of the question). I see from the answers that there were other ways to solve it :) $\endgroup$
    – F'x
    Sep 14, 2013 at 12:21
  • $\begingroup$ @F'x: A geometric approach would clearly not be sound here -- given how weird the distances on the $y$-axis are, the graph is certainly not to scale ... $\endgroup$ Sep 14, 2013 at 14:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .