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For the decomposition of HI, the activation energy is \(182 \mathrm{~kJ} / \mathrm{mol}\). The rate constant at \(850^{\circ} \mathrm{C}\) is \(0.0174 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{h}\). (a) What is the rate constant at \(700^{\circ} \mathrm{C} ?\) (b) At what temperature will the rate constant be a fourth of what it is at \(850^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Question: Calculate the rate constant at \(700^{\circ} \mathrm{C}\) and find the temperature for the rate constant to be a fourth of what it is at \(850^{\circ} \mathrm{C}\) given an activation energy of 182 kJ/mol and a rate constant of 0.0174 L/mol*h at \(850^{\circ} \mathrm{C}\). Answer: The rate constant at \(700^{\circ} \mathrm{C}\) is approximately \(5.32 * 10^{-4} \mathrm{L/mol} \cdot \mathrm{h}\). The rate constant will be a fourth of what it is at \(850^{\circ} \mathrm{C}\) at a temperature of approximately \(750.92^{\circ} \mathrm{C}\).

Step by step solution

01

Write the Arrhenius equation and convert temperatures to Kelvin

The Arrhenius equation is: k = A * e^(-Ea / (R * T)) Where k is the rate constant, A is the proportionality constant, Ea is the activation energy (182 kJ/mol), R is the gas constant (8.314 J/mol*K), and T is the temperature in Kelvin. Convert the given temperatures to Kelvin: \(850^{\circ} \mathrm{C} = 850 + 273.15 = 1123.15 \mathrm{K}\) \(700^{\circ} \mathrm{C} = 700 + 273.15 = 973.15 \mathrm{K}\)
02

Find the proportionality constant, A

Using the given rate constant at \(850^{\circ} \mathrm{C}\), k = 0.0174 L/mol*h, we can find the constant A: 0.0174 = A * e^(-(182000) / (8.314 * 1123.15)) Solve for A: A ≈ 3.61 * 10^13 L/mol*h
03

Calculate the rate constant at \(700^{\circ} \mathrm{C}\)

Now we can find the rate constant at \(700^{\circ} \mathrm{C}\) using the value of A: k = (3.61 * 10^13) * e^(-(182000) / (8.314 * 973.15)) k ≈ 5.32 * 10^-4 L/mol*h Answer for (a): The rate constant at \(700^{\circ} \mathrm{C}\) is approximately \(5.32 * 10^{-4} \mathrm{L/mol} \cdot \mathrm{h}\). (b) Finding the temperature for the rate constant to be a fourth of what it is at \(850^{\circ} \mathrm{C}\).
04

Set up the equation with the new rate constant

The new rate constant is a fourth of the given rate constant at \(850^{\circ} \mathrm{C}\): k_new = 0.0174 / 4 = 0.00435 L/mol*h.
05

Solve for T

Use the Arrhenius equation and solve for the new temperature T_new: 0.00435 = (3.61 * 10^13) * e^(-(182000) / (8.314 * T_new)) Solve for T_new: T_new ≈ 1024.07 K
06

Convert the temperature back to Celsius

Convert the temperature from Kelvin to Celsius: \(T_new^{\circ} \mathrm{C} = 1024.07 - 273.15 = 750.92^{\circ} \mathrm{C}\) Answer for (b): The rate constant will be a fourth of what it is at \(850^{\circ} \mathrm{C}\) at a temperature of approximately \(750.92^{\circ} \mathrm{C}\).

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Key Concepts

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

Activation Energy
When we consider the decomposition of hydrogen iodide (HI) and the need to determine how fast the reaction proceeds at different temperatures, the concept of activation energy becomes crucial. Activation energy, often symbolized as \(E_a\), refers to the minimum amount of energy required to initiate a chemical reaction. It's like the energy needed to push a ball up a hill before it can roll down; for a chemical reaction, molecules must be energized to a certain point before they can react.

Looking at the provided exercise, we're given that the activation energy for the decomposition of HI is \(182 \mathrm{~kJ} / \mathrm{mol}\). This value tells us how much energy needs to be overcome for the reaction to occur. It's a key parameter in the Arrhenius equation, which links the rate constant of a reaction with the temperature. The higher the activation energy, the lower the rate of reaction at a given temperature, as fewer molecules will have sufficient energy to overcome this barrier.
Rate Constant
The rate constant, symbolized as \(k\), is fundamental in the field of chemical kinetics since it quantifies the speed of a chemical reaction. It is influenced by temperature, as captured by the Arrhenius equation: \(k = A * e^(-E_a / (R * T))\), where \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

In our exercise, we learned that the rate constant of HI decomposition at \(850^{\textdegree}C\) was given to be \(0.0174 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{h}\). Using the Arrhenius equation, we can compute the rate constant at other temperatures, aiming to understand how the reaction speed changes. When asked to calculate the rate constant at \(700^{\textdegree}C\), this equation becomes vital. It essentially dictates how likely it is for reactant particles to collide and form products under different temperature conditions.
Chemical Kinetics
Chemical kinetics is the study of rates of chemical processes and the factors affecting them. It explains how different conditions, like temperature and the presence of a catalyst, change the speed of reactions. The heart of chemical kinetics is the rate of reaction: primarily, how fast reactants are consumed or products formed.

In the example at hand, the Arrhenius equation showcases chemical kinetics in practice. It is used to predict the rate constant \(k\) at \(700^{\textdegree}C\) and to determine at what temperature the rate constant would be a quarter of its value at \(850^{\textdegree}C\). Through this exercise, students see how the kinetics concepts apply to real chemical reactions, such as the temperature-dependence of reaction rates, emphasizing the practical importance of understanding kinetics in predicting and controlling chemical processes.

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Most popular questions from this chapter

The decomposition of dimethyl ether \(\left(\mathrm{CH}_{3} \mathrm{OCH}_{3}\right)\) to methane, carbon monoxide, and hydrogen gases is found to be first-order. At \(500^{\circ} \mathrm{C}\), a \(150.0\) -mg 35 sample of dimethyl ether is reduced to \(43.2 \mathrm{mg}\) after three quarters of an hour. Calculate (a) the rate constant. (b) the half-life at \(500^{\circ} \mathrm{C}\). (c) how long it will take to decompose \(95 \%\) of the dimethyl ether.

For the reaction $$5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}$$ it was found that at a particular instant bromine was being formed at the rate of \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). At that instant, at what rate is (a) water being formed? (b) bromide ion being oxidized? (c) \(\mathrm{H}^{+}\) being consumed?

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 \(0.500 \mathrm{~atm}\) ?

The decomposition of nitrogen dioxide is a second-order reaction. At \(550 \mathrm{~K}\), a \(0.250 M\) sample decomposes at the rate of \(1.17 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{min} .\) (a) Write the rate expression. (b) What is the rate constant at \(550 \mathrm{~K}\) ? (c) What is the rate of decomposition when \(\left[\mathrm{NO}_{2}\right]=0.800 \mathrm{M?}\)

Express the rate of the reaction $$2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ in terms of (a) \(\Delta\left[\mathrm{C}_{2} \mathrm{H}_{6}\right]\) (b) \(\Delta\left[\mathrm{CO}_{2}\right]\)

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