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Chapter 12: Problem 28

Consider the general reaction $$ \mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC} $$ and the following average rate data over some time period \(\Delta t :\) $$ \begin{aligned}-\frac{\Delta \mathrm{A}}{\Delta t} &=0.0080 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \\\\-& \frac{\Delta \mathrm{B}}{\Delta t}=0.0120 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \\ \frac{\Delta \mathrm{C}}{\Delta t} &=0.0160 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ Determine a set of possible coefficients to balance this general reaction.

Short Answer

Expert verified
The balanced general reaction with the given average rate data is: \(2\mathrm{A} + 3\mathrm{B} \longrightarrow 4\mathrm{C}\)

Step by step solution

01

Write the general reaction with given coefficients

We are given a general reaction with reactants A and B and product C, represented as follows: \[ \mathrm{aA} + \mathrm{bB} \longrightarrow \mathrm{cC} \] Our goal is to find the coefficients a, b, and c that will balance this reaction.
02

Analyze the rate data given

We are given the following average rate data: - \(-\frac{\Delta \mathrm{A}}{\Delta t} = 0.0080 \, \mathrm{mol} \, \mathrm{L}^{-1} \mathrm{s}^{-1}\) - \(-\frac{\Delta \mathrm{B}}{\Delta t} = 0.0120 \, \mathrm{mol} \, \mathrm{L}^{-1} \mathrm{s}^{-1}\) - \(\frac{\Delta \mathrm{C}}{\Delta t} = 0.0160 \, \mathrm{mol} \, \mathrm{L}^{-1} \mathrm{s}^{-1}\)
03

Determine the relationship between the coefficients

To find the relationship between the coefficients, let's divide the reactions rates: By dividing the average rate of A by the average rate of B, we get: \[ \frac{-\frac{\Delta \mathrm{A}}{\Delta t}}{-\frac{\Delta \mathrm{B}}{\Delta t}} = \frac{a}{b} \] Similarly, divide the average rate of C by either the average rate of A or B: \[ \frac{\frac{\Delta \mathrm{C}}{\Delta t}}{-\frac{\Delta \mathrm{A}}{\Delta t}} = \frac{c}{a} \] Now, let's solve for the coefficients.
04

Solve for the coefficients

We can now substitute the rate data into the equations: 1. \(\frac{-\frac{0.0080}{1}}{-\frac{0.0120}{1}} = \frac{a}{b}\) 2. \(\frac{\frac{0.0160}{1}}{-\frac{0.0080}{1}} = \frac{c}{a}\) By solving the equations: 1. \(\frac{0.0080}{0.0120} = \frac{a}{b}\) 2. \(\frac{0.0160}{0.0080} = \frac{c}{a}\) Simplify the fractions: 1. \(\frac{2}{3} = \frac{a}{b}\) 2. \(2 = \frac{c}{a}\)
05

Find the set of coefficients

From the simplified equations, we can find one possible set of coefficients that will balance the reaction: 1. \(a = 2\) and \(b = 3\) 2. \(c = 2 \times a = 2 \times 2 = 4\) So, one possible set of coefficients that will balance the reaction is: \[ 2\mathrm{A} + 3\mathrm{B} \longrightarrow 4\mathrm{C} \] This is one possible balanced general reaction with the given average rate data.

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Most popular questions from this chapter

Chemists commonly use a rule of thumb that an increase of 10 \(\mathrm{K}\) in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to \(35^{\circ} \mathrm{C} ?\)

The reaction $$ 0^{\circ} \mathrm{C}, $$ These relationships hold only if there is a very small amount of \(\mathrm{I}_{3}^{-}\) present. What is the rate law and the value of the rate constant? (Assume that rate $=-\frac{\Delta\left[\mathrm{H}_{2} \mathrm{SeO}_{3}\right]}{\Delta t} )$

Describe at least two experiments you could perform to determine a rate law.

Provide a conceptual rationale for the differences in the halflives of zero-, first-, and second-order reactions.

At \(40^{\circ} \mathrm{C}, \mathrm{H}_{2} \mathrm{O}_{2}(a q)\) will decompose according to the following reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(\mathrm{~g}) $$ The following data were collected for the concentration of $\mathrm{H}_{2} \mathrm{O}_{2}$ at various times. $$ \begin{array}{|cc|} \hline \begin{array}{c} \text { Time } \\ (\mathbf{s}) \end{array} & \begin{array}{c} {\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]} \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 0 & 1.000 \\ \hline 2.16 \times 10^{4} & 0.500 \\ \hline 4.32 \times 10^{4} & 0.250 \\ \hline \end{array} $$ a. Calculate the average rate of decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}\( between 0 and \)2.16 \times 10^{4} \mathrm{~s}$. Use this rate to calculate the average rate of production of \(\mathrm{O}_{2}(g)\) over the same time period. b. What are these rates for the time period \(2.16 \times 10^{4} \mathrm{~s}\) to \(4.32 \times 10^{4} \mathrm{~s} ?\)
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