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I have the rate plot below:

rate plot

I know that $\ce{A}$ and $\ce{B}$ are the reactants, and $\ce{C}$ is the product. The textbook says that the balanced equation is $$\ce{3A + B -> 2C}\\[1em]$$ However, I am not sure how to determine those coefficients based on the graph. Any help is greatly appreciated.

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    $\begingroup$ Which textbook? $\endgroup$ Commented May 1, 2023 at 7:23
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    $\begingroup$ (6.5 - 2 ) : ( 4 - 2.5 ) : ( 3 - 0 ) -> 3 : 1 : 2 $\endgroup$
    – Poutnik
    Commented May 1, 2023 at 9:27
  • $\begingroup$ Cite your references and workings on the problem while posting it on CSE for a better chance at receiving an answer and fewer downvotes. $\endgroup$
    – ananta
    Commented May 1, 2023 at 15:46

1 Answer 1

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By stoichiometry, the number of moles of species $j$ is related to the extent of the reaction via \begin{align} n_j &= n_{j0} + \nu_j \varepsilon \hspace{3 cm} (\text{divide by $V$}) \\ \frac{n_j}{V} &= \frac{n_{j0}}{V_0} + \nu_j \frac{\varepsilon}{V} \\ C_j &= C_{j0} + \nu_j \varepsilon_V \rightarrow \nu_j = \frac{C_j - C_{j0}}{\varepsilon_V} \tag{1} \\ \end{align} where we assumed that the volume of the reacting mixture does not change as the reaction take place, i.e., $V\approx V_0$. This is a good approximation for liquid-phase reactions, where the temperature does not change much, so that the density is almost constant. We also defined for convenience an extent of reaction per unit volume $\varepsilon := \varepsilon_V / V$

Apply Eq. (1) to each component \begin{align} \nu_A = \frac{C_A - C_{A0}}{\varepsilon_V} \tag{2} \\ \nu_B = \frac{C_B - C_{B0}}{\varepsilon_V} \tag{3} \\ \nu_C = \frac{C_C - C_{C0}}{\varepsilon_V} \tag{4} \\ \end{align} Dividing Eq. (2) with Eq. (3), and Eq. (4) with Eq. (3), and using the values shown in your plot, we obtain the ratio of the stoichiometric numbers \begin{align} \require{cancel} \frac{\nu_A}{\nu_B} & = \frac{C_A - C_{A0}}{C_B - C_{B0}} = \frac{(1.2 - 6) \; \cancel{\pu{M}}}{(2.4 - 4) \; \cancel{\pu{M}}} = 3 \rightarrow \nu_A = 3 \nu_B \tag{5} \\ \frac{\nu_C}{\nu_B} &= \frac{C_C - C_{C0}}{C_B - C_{B0}} = \frac{(3.2 - 0) \; \cancel{\pu{M}}}{(2.4 - 4) \; \cancel{\pu{M}}} = -2 \rightarrow \nu_C = -2\nu_B \tag{6} \end{align} And finally we write the chemical equation using Eqs. (5) and (6) \begin{align} \require{cancel} \nu_A A + \nu_B C + \nu_C C = 0 \\ 3\cancel{\nu_B} A + \cancel{\nu_B} B - 2\cancel{\nu_B} C = 0 \\ 3 A + B - 2 C = 0 \rightarrow \boxed{3A + B \rightarrow 2C} \end{align}

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  • $\begingroup$ Good post. But, if I imagine I am a beginner, I am not sure if I would thank you or would be rather running to hills screaming. :-) $\endgroup$
    – Poutnik
    Commented May 1, 2023 at 11:44
  • $\begingroup$ @Poutnik Hello Poutnik. You say that this is too advanced when you are in the first course of general chemistry at the university? $\endgroup$ Commented May 1, 2023 at 14:28
  • $\begingroup$ I think what you ask is more important than where you are. // Initially, I wanted to provide example of Calvin and Susie exchanging fruits. (Calvin & Hobbes) $\endgroup$
    – Poutnik
    Commented May 1, 2023 at 14:32
  • $\begingroup$ @Poutnik Sorry Poutnik, I didn't quite understand... what do you mean? $\endgroup$ Commented May 1, 2023 at 14:52
  • $\begingroup$ 1st part: I answer simply to simple questions and vice versa. Equal question can be asked by skilled high school freshman and ehm, simple college freshmen. // The question is so simple that it can be colourfully illustrated by basic arithmetic task from elementary/basic school, not too far from the Calvin level. $\endgroup$
    – Poutnik
    Commented May 1, 2023 at 15:01

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