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I tried finding the slope between two points but I keep getting wrong values, how does one find the initial rate by using such graphs?
So I started by trying to find a slope of two points, $(0,0.4)$ and $(10,0.2)$ after calculations I got $-0.02$, which is the opposite of the right answer in sign. I tried to check myself by trying to find the slope for other points like: $(10,0.2)$ and $(20,0.1)$ which gives me $-0.01$, I'm confused as whether my initial idea is right or not.
It seems like you have the right idea to me,
For the "Initial Rate," you're taking a curve and doing a straight approximation of the curve over a small portion, as you appear to be doing.
Your logic also appears correct to me, a decrease in concentration should correspond to a negative rate.
My guess is that the question, (a poorly worded one?) is actually referring to the rate of product creation, and asking you to assume that there is a 1-1 relationship between the decrease in reactant, and the appearance of product. Thus if your reactant is disappearing at $0.02~M/s$ (as indicated by $rate= -0.02 ~M/s$) , then your product should be appearing at $0.02~M/s$, making A the correct answer.
This is a standard exam question pertaining to the kinetics material that is typically presented in a first year general chemistry course. Usually the coverage is restricted to discussion of zero, first and second order kinetics and this is what gets tested on exams. From the photo of the exam question, the decay is apparently first order, i.e., $$\mathrm C(t) = C(0)e^{-kt}$$ where $\mathrm C(t)$ is the molar concentration of whatever is decaying away, $\mathrm C(0)$ is the initial molar concentration (i.e., at $\mathrm t = 0$) of whatever is decaying away, and $k$ is the decay rate constant in units of $\mathrm s^{-1}$. In the present case, $\mathrm C(0) = 0.4 M.$
But, is the decay really exponential? To determine this, without curve fitting data pairs or linearizing the plot, it suffices to see if the half-life, $\mathrm t_{1/2}$, is constant. From the photo of the graph, the initial concentration declines to $\mathrm 0.2 M$ at about $\mathrm t = 13 s$. Note that there are $\mathrm 10$ dash marks every $\mathrm 10 s$ along the time axis, so this is an aid, albeit imperfect, in estimating times along the time axis. The concentration declines to $\mathrm 0.1 M$ at about the $\mathrm 26 s$ mark and further declines to about $\mathrm 0.05 M$ at about the $\mathrm 39 s$ mark.
So, $\mathrm t_{1/2}$ is constant and approximately equal to $\mathrm 13 s$. Therefore, the decay process is a first order exponential, as shown in the equation at the top. Note that the only other alternatives, in the restricted context of typical course material coverage and exam coverage of said material, are zero and second order decays. But these do not have constant half-lives: for a zero order decay, every subsequent half-life is halved and for a second order decay, every subsequent half-life is doubled. The graph clearly eliminates these two possibilities. It is also obvious that the decay cannot be first order since then the concentration versus time plot would be linear. If the decay was second order, the concentration could not be down to $\mathrm 0.1M$ in less than $\mathrm 30s$, even if the first $\mathrm t_{1/2}$ was clearly underestimated as $\mathrm 10s$.
The relationship between $\mathrm k$ and $\mathrm t_{1/2}$ is $\mathrm k = (ln2)/t_{1/2}$. This is easily verified by substitution into the exponential equation, resulting in $$\mathrm C(t = t_{1/2}) = C(0)/2$$ Hence, with $\mathrm t_{1/2}$ estimated as $\mathrm 13s$, the rate constant estimate is $\mathrm k = 0.053 s^{-1}$.
To find the initial rate of decay, simply differentiate the exponential equation and evaluate it at $\mathrm t = 0$. The derivative is $$\mathrm dC(t)/dt = -kC(0)e^{-kt}$$ so at $\mathrm t = 0$, the slope is simply $\mathrm -kC(0)$ and the magnitude of the slope is $\mathrm kC(0)$. With $\mathrm k = 0.053 s^{-1}$ and $\mathrm C(0) = 0.4 M$, this yields $\mathrm 0.021 M s^{-1}$ as the initial rate magnitude. Rounded to one significant figure, which is all this exam problem deserves, yields $\mathrm = 0.02 M s^{-1}$. This is exam answer A.
One last thing: a small error in estimating the half-life will not matter because the initial decay rate is inversely proportional to it, so it would have to be high by a factor of two in order to get answer B on the exam.
The idea is that you want to find the instantaneous rate of change (rise over run)at the initial value (time = 0). but since this gives .4/0 (can't be done) just use a close number to the initial value to get a good approximation of the slope at the value. So for initial so (.4-.3)/(0-5) =.02 you use the slope because the units are Moles(rise)/Seconds(run). since it doesn't specify the rate of appearance or disappearance then the magnitude is what matters not sign.
Firstly, you're right that the slopes at all points on this curve will be negative as it's a downward sloping curve. The rate of reaction w.r.t. reactants is the negative of the rate of consumption of reactants provided there's 1:1 stoichiometric ratio between the reactants and products. But in this question, nowhere is it mentioned that this curve depicts the concentration vs time curve for reactants. Even if we consider it to be so, there's no mention of a function that describes this graph, from which we could have found out the order and rate constant of the reaction.
Thus we have to proceed in a way @Dan pointed out, you need to use the slope a line that has approximately the same slope as the curve at the initial state of the reaction. But for that we need 2 exact coordinates of very closely spaced points on this graph, just like you tried to do, but as you see, the graph doesn't pass through $(10,0.2)$, actually there's no other point than $(0,0.4)$ whose coordinates we know for sure. But if you look closely, you'd find $(5,0.3)$ is a better match (as far as my eyes tell me). So using that, one can find the slope of line between them to be $\frac{0.3-0.4}{5-0}=-0.02$
Thus the rate of reaction w.r.t. reactants could be-
Rate $=\frac{\mathrm{-d}[\text{reactants}]}{\mathrm{d}t}=-(-0.02)=0.02 \pu{mol L^-1 s^-1}$
Even though my stand is that this is a poor question that lacks a lot of necessary detail. So it's better not to waste time on this one, but yeah there's always a chance of learning something out of everything.
