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Chapter 11: Problem 5

Consider the combustion of ethane: $$\mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ If the ethane is burning at the rate of \(0.20 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), at what rates are \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) being produced?

Short Answer

Expert verified
Answer: Carbon dioxide is being produced at a rate of 0.80 mol/L·s, and water is being produced at a rate of 1.20 mol/L·s.

Step by step solution

01

Identify the stoichiometric coefficients of the reactants and products

The coefficients of the reactants and products in the balanced equation are: - Ethane (\(\mathrm{C}_{2}\mathrm{H}_{6}\)): 1 - Oxygen (\(\mathrm{O}_{2}\)): 7 - Carbon dioxide (\(\mathrm{CO}_{2}\)): 4 - Water (\(\mathrm{H}_{2}\mathrm{O}\)): 6
02

Calculate the rate of CO2 production

We can use the stoichiometric coefficients in the balanced equation to find the rate of CO2 production relative to the rate at which ethane is burning. To find the rate of CO2 production, we can use the given rate of ethane consumption and the ratio of the stoichiometric coefficients: $$\text{Rate of} \; \mathrm{CO}_{2} \; \text{production} =\frac{\text { Stoichiometric coefficient of} \; \mathrm{CO}_{2}}{\text { Stoichiometric coefficient of} \; \mathrm{C}_{2}\mathrm{H}_{6}} \times \text {Rate of ethane burning}$$ $$\text{Rate of} \; \mathrm{CO}_{2} \; \text{production} =\frac{4}{1} \times 0.20 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} = 0.80 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}$$ So, carbon dioxide is being produced at a rate of \(0.80 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\).
03

Calculate the rate of H2O production

We can use a similar approach to find the rate of H2O production using the stoichiometric coefficients: $$\text{Rate of} \; \mathrm{H}_{2}\mathrm{O} \; \text{production} =\frac{\text { Stoichiometric coefficient of} \; \mathrm{H}_{2}\mathrm{O}}{\text { Stoichiometric coefficient of} \; \mathrm{C}_{2}\mathrm{H}_{6}} \times \text {Rate of ethane burning}$$ $$\text{Rate of} \; \mathrm{H}_{2}\mathrm{O} \; \text{production} =\frac{6}{1} \times 0.20 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} = 1.20 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}$$ Therefore, water is being produced at a rate of \(1.20 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). To summarize, CO2 is being produced at a rate of \(0.80 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), and H2O is being produced at a rate of \(1.20 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) during the combustion of ethane at the given rate of \(0.20 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\).

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Key Concepts

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

Stoichiometric Coefficients
When we talk about chemical reactions, the term stoichiometric coefficients refers to the numbers in front of the chemical formulas that balance the equation. These numbers convey the ratio in which reactants combine and products form. In the combustion of ethane, the coefficients indicate that one molecule of ethane reacts with seven molecules of oxygen to produce four molecules of carbon dioxide and six molecules of water.

For better understanding, think of a simple analogy: baking a cake. The recipe—like a chemical equation—tells you that to bake one cake, you might need two eggs. If you plan to double the recipe, you'd need four eggs. Similarly, if twice as much ethane were to burn, it would produce twice as much CO2 and water, following the ratios indicated by the stoichiometric coefficients.
Rate of Reaction
The rate of reaction is a measure of how fast a reactant is consumed or a product is formed. It's comparable to the speedometer in a car—a higher rate means the reactants are being used up or the products are being created more rapidly.

In the given exercise, the rate of ethane combustion is provided as 0.20 mol/L·s. This rate, coupled with the stoichiometric coefficients, allows us to calculate the rates at which CO2 and H2O are produced. It's vital to understand that the rate of a reaction can be affected by various factors, including temperature, pressure, and the presence of catalysts, but these aspects are outside the scope of the given problem.
Chemical Reaction Balancing
Balancing a chemical equation, such as the combustion of ethane, ensures that the law of conservation of mass is upheld. It means that the same number of each type of atom appears on both the reactant and product sides of the equation.

A balanced equation is crucial for correctly understanding and calculating reaction stoichiometry—as it's done in the exercise. It's the road map that tells us how many moles of each product form when a certain amount of reactant is used. Without a balanced equation, determining the rate of product formation, like in our cake analogy, would be like trying to bake without knowing the quantity of the ingredients needed.

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Most popular questions from this chapter

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) hydrolyzes into glucose and fructose. The hydrolysis is a first-order reaction. The half-life for the hydrolysis of sucrose is \(64.2\) min at \(25^{\circ} \mathrm{C}\). How many grams of sucrose in \(1.25 \mathrm{~L}\) of a \(0.389 \mathrm{M}\) solution are hydrolyzed in \(1.73\) hours?

The first-order rate constant for the decomposition of a certain hormone in water at \(25^{\circ} \mathrm{C}\) is \(3.42 \times 10^{-4}\) day \(^{-1}\). (a) If a \(0.0200 \mathrm{M}\) solution of the hormone is stored at \(25^{\circ} \mathrm{C}\) for two months, what will its concentration be at the end of that period? (b) How long will it take for the concentration of the solution to drop from \(0.0200 M\) to \(0.00350 \mathrm{M} ?\) (c) What is the half-life of the hormone?

For a reaction involving the decomposition of \(\mathrm{Y}\), the following data are obtained: $$\begin{array}{lllll}\hline \text { Rate }(\mathrm{mol} / \mathrm{L} \cdot \min ) & 0.288 & 0.245 & 0.202 & 0.158 \\ {[\mathrm{Y}]} & 0.200 & 0.170 & 0.140 & 0.110 \\ \hline\end{array}$$ (a) Determine the order of the reaction. (b) Write the rate expression for the decomposition of Y. (c) Calculate \(k\) for the experiment above.

A catalyst lowers the activation energy of a reaction from \(215 \mathrm{~kJ}\) to \(206 \mathrm{~kJ} .\) By what factor would you expect the reaction-rate constant to increase at \(25^{\circ} \mathrm{C} ?\) Assume that the frequency factors \((\mathrm{A})\) are the same for both reactions. (Hint: Use the formula \(\ln k=\ln \mathrm{A}-E_{2} / R T .\) )

For the zero-order decomposition of HI on a gold surface $$\mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{I}_{2}(g)$$ it takes \(16.0 \mathrm{~s}\) for the pressure of HI to drop from \(1.00\) atm to \(0.200 \mathrm{~atm}\). (a) What is the rate constant for the reaction? (b) How long will it take for the pressure to drop from \(0.150 \mathrm{~atm}\) to \(0.0432\) atm? (c) What is the half-life of \(\mathrm{HI}\) at a pressure of \(0.500 \mathrm{~atm}\) ?
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