Express the rate of this reaction in terms of the change in concentration of each of the reactants and products: $$ 2 \mathrm{D}(g)+3 \mathrm{E}(g)+\mathrm{F}(g) \longrightarrow 2 \mathrm{G}(g)+\mathrm{H}(g) $$ When [D] is decreasing at \(0.1 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), how fast is [H] increasing?

Short Answer

Expert verified
0.05 mol/L·s

Step by step solution

01

- Write the reaction equation

The reaction equation given is:\[ 2D(g) + 3E(g) + F(g) \rightarrow 2G(g) + H(g) \]
02

- Express the rate of the reaction

The rate of the reaction can be expressed in terms of the change in concentration of the reactants and products using the stoichiometric coefficients:\[ \text{Rate} = -\frac{1}{2} \frac{d[D]}{dt} = -\frac{1}{3} \frac{d[E]}{dt} = -\frac{d[F]}{dt} = \frac{1}{2} \frac{d[G]}{dt} = \frac{d[H]}{dt} \]
03

- Identify the given rates

We know that the concentration of [D] is decreasing at a rate of 0.1 mol/L\cdot s. This can be written as:\[ -\frac{d[D]}{dt} = 0.1 \text{ mol/L} \cdot \text{s} \]
04

- Relate the rate of disappearance of [D] to the rate of formation of [H]

Using the stoichiometric relationship from Step 2:\[ \frac{d[H]}{dt} = -\frac{1}{2} \frac{d[D]}{dt} \]
05

- Substitute the given rate into the equation

Substitute \( \frac{d[D]}{dt} = -0.1 \text{ mol/L} \cdot \text{s} \) into the equation from Step 4:\[ \frac{d[H]}{dt} = \frac{1}{2} \times 0.1 \text{ mol/L} \cdot \text{s} \]
06

- Calculate the rate of increase of [H]

Perform the calculation:\[ \frac{d[H]}{dt} = 0.05 \text{ mol/L} \cdot \text{s} \]

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Key Concepts

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

reaction kinetics
Understanding reaction kinetics is crucial in determining the speed at which a chemical reaction occurs. Reaction kinetics studies the factors that affect the rate of a chemical reaction and provides insight into the reaction mechanism.

The rate of a reaction can be expressed as the change in concentration of reactants or products per unit time. It is commonly measured in terms of molarity per second (mol/L·s).

Factors influencing reaction rates include temperature, concentration, surface area, and the presence of catalysts. Higher temperatures and concentrations typically increase reaction rates, as they lead to more frequent and energetic collisions between reactant molecules.

For the given reaction (2D(g) + 3E(g) + F(g) → 2G(g) + H(g)), the rate of the reaction can be analyzed by looking at how the concentrations of 'D', 'E', and 'F' decrease, and how the concentrations of 'G' and 'H' increase over time.
stoichiometric coefficients
Stoichiometric coefficients are the numbers placed in front of the molecules in a chemical equation. These coefficients indicate the relative amounts of each substance involved in the reaction and are essential for balancing chemical equations.

In the reaction 2D(g) + 3E(g) + F(g) → 2G(g) + H(g), the coefficients 2, 3, 1, 2, and 1 represent the mole ratios of D, E, F, G, and H, respectively. These ratios are vital in expressing the rate of reaction.

When expressing the reaction rate, we use the stoichiometric coefficients to relate the rate of change in concentration of one species to another. For example, if D is consumed at a rate of 0.1 mol/L·s, we can derive the rate at which H is formed using its stoichiometric ratio relative to D. Since H is formed at one-half the rate that D is consumed, the rate of formation of H would be 0.05 mol/L·s.
concentration change
Concentration change is a key aspect of reaction kinetics. It refers to how the molarity of reactants and products changes over time.

During the reaction, as reactants are converted to products, the concentration of reactants decreases and the concentration of products increases. This change can be measured to determine the reaction rate.

For the reaction 2D(g) + 3E(g) + F(g) → 2G(g) + H(g), if we know that the concentration of D decreases by 0.1 mol/L·s, we can use the stoichiometric coefficients to find how the concentrations of E, F, G, and H change. By using the relationships from the balanced chemical equation, we can affirm that:
  • -d[D]/dt = 0.1 mol/L·s
  • -d[E]/dt = (3/2) * 0.1 mol/L·s = 0.15 mol/L·s
  • -d[F]/dt is at the same rate as it appears with a coefficient of 1
  • d[G]/dt = (1/2) * 0.1 mol/L·s = 0.05 mol/L·s
  • d[H]/dt = 0.05 mol/L·s
This allows us to see the interrelationship of concentration changes among reactants and products.

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Most popular questions from this chapter

Give the individual reaction orders for all substances and the overall reaction order from this rate law: $$ \text { Rate }=k\left[\mathrm{NO}_{2}\right]^{2}\left[\mathrm{Cl}_{2}\right] $$

In a study of ammonia production, an industrial chemist discovers that the compound decomposes to its elements \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in a first-order process. She collects the following data: $$ \begin{array}{llll} \text { Time (s) } & 0 & 1.000 & 2.000 \\ {\left[\mathrm{NH}_{3}\right](\mathrm{mol} / \mathrm{L})} & 4.000 & 3.986 & 3.974 \end{array} $$ (a) Use graphical methods to determine the rate constant. (b) What is the half-life for ammonia decomposition?

The mathematics of the first-order rate law can be applied to any situation in which a quantity decreases by a constant fraction per unit of time (or unit of any other variable). (a) As light moves through a solution, its intensity decreases per unit distance traveled in the solution. Show that \(\ln \left(\frac{\text { intensity of light leaving the solution }}{\text { intensity of light entering the solution }}\right)\) \(=-\) fraction of light removed per unit of length \(\times\) distance traveled in solution (b) The value of your savings declines under conditions of constant inflation. Show that \(\ln \left(\frac{\text { value remaining }}{\text { initial value }}\right)\) \(=-\) fraction lost per unit of time \(\times\) savings time interval

Express the rate of this reaction in terms of the change in concentration of each of the reactants and products: $$ \mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g}) $$ When [B] is decreasing at \(0.5 \mathrm{~mol} / \mathrm{L}\) -s, how fast is \([\mathrm{A}]\) decreasing?

Consider the following general reaction and data: \(2 \mathrm{~A}+2 \mathrm{~B}+\mathrm{C} \longrightarrow \mathrm{D}+3 \mathrm{E}\) $$ \begin{array}{ccccc} \text { Expt } & \begin{array}{c} \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \end{array} & \begin{array}{c} \text { Initial [A] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [B] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [C] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 1 & 6.0 \times 10^{-6} & 0.024 & 0.085 & 0.032 \\ 2 & 9.6 \times 10^{-5} & 0.096 & 0.085 & 0.032 \\ 3 & 1.5 \times 10^{-5} & 0.024 & 0.034 & 0.080 \\ 4 & 1.5 \times 10^{-6} & 0.012 & 0.170 & 0.032 \end{array} $$ (a) What is the reaction order with respect to each reactant? (b) Calculate the rate constant. (c) Write the rate law for this reaction. (d) Express the rate in terms of changes in concentration with time for each of the components.

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