# How to write the balanced chemical equation for the system in a rate plot?

I have the rate plot below:

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.

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

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}