# Finding rate of reaction from a concentration vs. time graph

A tangent drawn on “conc. of product vs. time” curve of a 1st order reaction makes an angle 30 degrees with the $y$-axis at 20 min. Find the rate after 20 min.

I first tried to find the equation like this: $$k=\frac{2.303}{t}\log\frac{a}{a-x}$$ Let $\frac{k}{2.303}$ be $c$ to make calculations less messy. Here, $a$ is the initial concentration of reactant. $$\frac{a}{a-x}=10^{ct}$$ $$x=a\left(1-10^{-ct}\right)$$ $$\frac{\mathrm dx}{\mathrm dt}=a\left(c\cdot\ln10\cdot10^{-ct}\right)=ak10^{\frac{kt}{2.303}}$$ Now, the tangent makes an angle of 30 degrees with concentration axis (or $y$-axis in this case). So the angle made with time axis (or $x$-axis) is 60 degrees. Hence $\frac{\mathrm dx}{\mathrm dt}=\tan\theta=\sqrt{3}$. But the value of initial concentration is not given for finding rate constant. Should I go for a different approach or am I making any conceptual mistake?

• I think that your approach is ok. If P is product and R$_0$ initial reactant, the $rate$ and slope of the graph is $dP/dt=R_0.k.exp(-kt)$ which is dimensionally ok. I don't see how you can get the $rate$ $constant$, you are not give enough information. The slope has units of concentration/time and so taking the tangent is not strictly correct as the tan should be dimensionless. Jul 10 '16 at 14:26

1. Let the reactant be $S$ and the product be $P$. Let $p$ be the dimenionless product concentration, $\frac{P}{S_0}$. First order kinetics:

$$\frac{P}{S_0} = p = 1- e^{-k t}$$

1. Slope of first-order kinetics:

$$\frac{dp}{dt} = k e ^{-kt}$$

1. Slopes and angles. One confusing thing about this problem is the definition of $\theta$ is different than what you usually see in math class. It's the angle with respect to the $y$ axis, not the $x$ axis. As you note, if the $y$ angle is 30 degrees, then the $x$ angle is 60 degrees, meaning the slope is $\tan~60^{\circ} = \sqrt{3}$.

2. Units problem. However there is a big units problem in this question. The slope is $\sqrt{3}$ what? Gigamolar per nanosecond? Zeptomolar per eon? The angle of the tangent curve will change depending on the units used to plot the data. If you use nanoseconds as the time unit, the angle of the tangent curve will be very flat, but conversely if eons is the time unit, the the angle will be very sharp. The question does not explicitly say what units are used in the plot. And the expression for the derivative of a first-order reaction shows that the slope should have units of concentration per time, but $\sqrt{3}$ is dimensionless. So there is not really enough information to solve the problem.

3. Dicey assumptions about units. Let's assume that the time unit is minutes. Let's further assume that the $y$ axis plots a dimensionless concentration, i.e., $p$. Then the derivative (slope) is $-k e^{-k t}$ and apparently has a value of $\sqrt{3}~\mathbf{{min}^{-1}}$.

The question is asking for the rate, so you are done. But there are a lot of weird assumptions in this problem.