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Chapter 8: Problem 32

Hydroquinone, \(\mathrm{HOC}_{4} \mathrm{H}_{4} \mathrm{OH}\), can be formed by the reaction with acetylene below: \(3 \mathrm{HC} \equiv \mathrm{CH}+3 \mathrm{CO}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow\) \(2 \mathrm{HOC}_{4} \mathrm{H}_{4} \mathrm{OH}+\mathrm{CO}_{2}\) How is the rate of disappearance of acetylene \((\mathrm{HC} \equiv \mathrm{CH})\) related to the appearance of hydroquinone (Hq)? \(-\frac{\Delta[\mathrm{HC} \equiv \mathrm{CH}]}{\Delta \mathrm{t}}=?\) a. \(+\frac{2 \Delta[\mathrm{Hq}]}{3 \Delta t}\) b. \(+\frac{3 \Delta[\mathrm{Hq}]}{2 \Delta t}\) c. \(+\frac{\Delta[\mathrm{Hq}]}{\Delta \mathrm{t}}\) d. \(+\frac{1 \Delta[\mathrm{Hq}]}{2 \Delta t}\)

Short Answer

Expert verified
The rate is \(+\frac{3 \Delta[\mathrm{Hq}]}{2 \Delta t}\), so option b is correct.

Step by step solution

01

Understand the Reaction Stoichiometry

The given chemical equation is \(3\ \mathrm{HC} \equiv \mathrm{CH}+3\ \mathrm{CO}+3\ \mathrm{H}_{2}\mathrm{O} \rightarrow 2\ \mathrm{HOC}_{4} \mathrm{H}_{4} \mathrm{OH}+\mathrm{CO}_{2}\). It indicates that 3 moles of acetylene \((\mathrm{HC} \equiv \mathrm{CH})\) react to form 2 moles of hydroquinone \((\mathrm{HOC}_{4} \mathrm{H}_{4} \mathrm{OH})\).
02

Define the Rate of Disappearance and Appearance

The rate of disappearance of a reactant is expressed as \(-\frac{\Delta[\text{Reactant}]}{\Delta t}\), while the rate of appearance of a product is \(+\frac{\Delta[\text{Product}]}{\Delta t}\). Here, we want to relate the rate of disappearance of acetylene to the rate of appearance of hydroquinone.
03

Relate the Rates Using Stoichiometry

Based on stoichiometry, we know that every 3 moles of acetylene disappear, 2 moles of hydroquinone form. Therefore, the rate of disappearance of acetylene is related to the rate of appearance of hydroquinone by the factor of their coefficients from the balanced equation. It is calculated as follows: \(-\frac{\Delta[\mathrm{HC} \equiv \mathrm{CH}]}{\Delta t} = \frac{3}{2} \times \frac{\Delta[\mathrm{Hq}]}{\Delta t}\).
04

Select the Correct Option

The expression \(+\frac{3 \Delta[\mathrm{Hq}]}{2 \Delta t}\) matches option \(b\). Hence, the rate of disappearance of acetylene is \(-\frac{\Delta[\mathrm{HC} \equiv \mathrm{CH}]}{\Delta t} = \frac{3 \Delta[\mathrm{Hq}]}{2 \Delta t}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is the heart of balancing chemical reactions and calculating the relationships between reactants and products. It represents the quantitative relationships and helps us understand how much of each substance is involved in a reaction. In the given reaction, stoichiometry is used to relate the amounts of acetylene and hydroquinone. We begin by balancing the chemical equation to ensure that the number of atoms for each element is equal on both reactant and product sides. In this case, the balanced equation shows that 3 moles of acetylene
  • React to produce 2 moles of hydroquinone.
  • Requires 3 moles each of carbon monoxide and water.
Only with this balance can we correctly calculate how much product forms from a given amount of reactant.
Rate of Disappearance
The rate of disappearance refers to how fast a reactant is consumed in a reaction. This rate is represented as the negative change in concentration over time for a reactant, denoted mathematically as 't. In practical terms, it answers the question, "How quickly is the acetylene used up as the reaction proceeds?" In the example provided, three moles of acetylene are consumed to form products, so understanding this rate helps in predicting how long a reaction will take to complete or reach equilibrium.
Rate of Appearance
The rate of appearance, on the other hand, describes how quickly a product forms in a chemical reaction. It is expressed as the positive change in concentration of a product over time, written as '+',), icationlly+ rac{ WC][qt. For hydroquinone in the given reaction, the rate at which it appears is directly proportional to the amount consumed of acetylene due to their stoichiometric relationship. As acetylene disappears, hydroquinone appears based on the stoichiometry of the equation, guiding chemists in scaling up reactions for industrial applications.
Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction, meaning no net change is observed in the concentrations of reactants and products over time. In reactions like the one presented, equilibrium might not be reached immediately, especially if a reaction is far from completion. Understanding when equilibrium is reached helps scientists control reactions better, optimize conditions, or drive the reaction towards desired products. For example, by manipulating conditions such as pressure or concentration, one might favor the formation of more hydroquinone if needed.

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Most popular questions from this chapter

In Arrhenius equation, \(\mathrm{k}=\mathrm{A} \exp (-\mathrm{Ea} / \mathrm{RT})\). A may be regarded as the rate constant at a. Very high temperature b. Very low temperature c. High activation energy d. Zero activation energy

The equation tris(1,10-phenanthroline) iron(II) in acid solution takes place according to the equation: \(\mathrm{Fe}(\text { phen })_{3}^{2+}+3 \mathrm{H}_{3} \mathrm{O}^{+}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow\) $$ \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}+3 \text { (Phen) } \mathrm{H}^{+} $$ If the activation energy (Ea) is \(126 \mathrm{~kJ} / \mathrm{mol}\) and the rate constant at \(30^{\circ} \mathrm{C}\) is \(9.8 \times 10^{-3} \mathrm{~min}^{-1}\), what is the frequency factor (A)? a. \(9.5 \times 10^{18} \mathrm{~min}^{-1}\) b. \(2.5 \times 10^{19} \mathrm{~min}^{-1}\) c. \(55 \times 10^{19} \mathrm{~min}^{-1}\) d. \(5.0 \times 10^{19} \mathrm{~min}^{-1}\)

Which of the following is incorrect about order of reaction? a. it is calculated experimentally b. it is sum of powers of concentration in rate law expression c. the order of reaction cannot be fractional d. there is not necessarily a connection between order and stoichiometry of a reaction.

From the following data for the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) \(\begin{array}{llll}{[\mathrm{A}]} & {[\mathrm{B}]} & \text { initial rate } & \left.(\mathrm{mol}]^{-1} \mathrm{~s}^{-1}\right) \\ \mathrm{mol} 1^{-1} & \mathrm{~mol} & \mathrm{l}^{-1} 300 \mathrm{~K} & 320 \mathrm{~K} \\ 2.5 \times 10^{-4} & 3.0 \times 10^{-5} & 5.0 \times 10^{-4} & 2.0 \times 10^{-3} \\\ 5.0 \times 10^{-4} & 6.0 \times 10^{-5} & 4.0 \times 10^{-3} & \- \\ 1.0 \times 10^{-3} & 6.0 \times 10^{-5} & 1.6 \times 10^{-2} & -\end{array}\) Calculate the rate of the equation. a. \(\mathrm{r}=\mathrm{k}[\mathrm{B}]^{1}\) b. \(\mathrm{r}=\mathrm{k}[\mathrm{A}]^{2}\) c. \(r=k[A]^{2}[B]^{1}\) d. \(\mathrm{r}=\mathrm{k}[\mathrm{A}][\mathrm{B}]\)

In the formation of sulphur trioxide by the contact process,\(2 \mathrm{SO}_{2}+\mathrm{O}_{2}=2 \mathrm{SO}_{3}\), the rate of reaction can be measured as \(-\mathrm{d}\left(\mathrm{SO}_{2}\right)=6.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \frac{\mathrm{s}^{-1}}{\mathrm{dt}}\). Here the incorrect statements are a. The rate of reaction expressed in terms of \(\mathrm{O}_{2}\) will be \(4.0 \times 10^{-4}\) mole \(\mathrm{L}^{-1} \mathrm{~s}^{-1}\) b. The rate of reaction expressed in terms of \(\mathrm{O}_{2}\) will be \(6.0 \times 10^{-6}\) mole \(\mathrm{L}^{-1} \mathrm{~s}^{-1}\) c. The rate of reaction expressed in terms of \(\mathrm{SO}_{3}\) will be \(6.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). d. The rate of reaction expressed in terms of \(\mathrm{O}_{2}\) will be \(3.0 \times 10^{-4}\) mole \(\mathrm{L}^{-1} \mathrm{~s}^{-1}\)
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