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Chapter 1: Problem 5

$$\mathrm{SO}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ At \(600 \mathrm{K}, \mathrm{SO}_{2} \mathrm{Cl}_{2}\) will decompose to form sulfur dioxide and chlorine gas via the above equation. If the reaction is found to be first order overall, which of the following will cause an increase in the half life of \(\mathrm{SO}_{2} \mathrm{Cl}_{2} ?\) (A) Increasing the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (B) Increasing the temperature at which the reaction occurs (C) Decreasing the overall pressure in the container (D) None of these will increase the half life.

Short Answer

Expert verified
(D) None of these will increase the half life.

Step by step solution

01

Understand the dependence of half-life on initial concentration

In a first-order reaction, the half-life is independent of the initial concentration of reactants. This can be derived from the first-order rate law, \( k = [A] / t_{1/2} \). Rearranging gives us \( t_{1/2} = [A] / k \), which shows that the half-life does not depend on [A]. Therefore, changing the initial concentration of the reactant (\(\mathrm{SO}_{2} \mathrm{Cl}_{2}\)) will not affect the half-life. So (A) is not a correct option.
02

Understand the dependence of half-life on temperature

The half-life of a first-order reaction is inversely proportional to the rate constant (\( t_{1/2} = 1/ k \)). As temperature increases, the rate constant increases due to higher average kinetic energy of the molecules, leading to more collisions per unit time. Consequently, the half-life decreases. Therefore, increasing the temperature at which the reaction occurs will actually decrease the half-life, not increase it. So (B) is not a correct option.
03

Understand the dependence of half-life on pressure

Pressure only indirectly affects the half-life through changes in concentration when volume changes. Since we already established that the half-life of a first-order reaction is independent of the initial concentration, changes in pressure should not affect the half-life of the reaction. So (C) is not a correct option.
04

Determine the correct answer

As we have refuted options (A) to (C), it means that none of the options given will increase the half-life of the reaction.

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Most popular questions from this chapter

Use the following information to answer questions 12-15. When heated in a closed container in the presence of a catalyst, potassium chlorate decomposes into potassium chloride and oxygen gas via the following reaction: \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) If 12.25 g of potassium chlorate decomposes, how many grams of oxygen gas will be generated? (A) 1.60 g (B) 3.20 g (C) 4.80 g (D) 18.37 g

Molten \(\mathrm{AlCl}_{3}\) is electrolyzed with a constant current of 5.00 amperes over a period of 600.0 seconds. Which of the following expressions is equal to the maximum mass of Al(s) that plates out? (1 faraday = 96,500 coulombs) (A) \(\frac{(600)(5.00)}{(96,500)(3)(27.0)}\) grams (B) \(\frac{(600)(5.00)(3)(27.0)}{(96,500)}\) grams (C) \(\frac{(600)(5.00)(27.0)}{(96,500)(3)}\) grams (D) \(\frac{(96,500)(3)(27.0)}{(600)(5.00)}\) grams

\(\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2} \mathrm{O}(s)\) Which of the following is true for the above reaction? (A) The value for \(\Delta S\) is positive. (B) The value for \(\Delta G\) is zero. (C) The value for \(\Delta H\) is positive. (D) The reaction is favored at 1.0 \(\mathrm{atm}\) and 298 \(\mathrm{K}\) .

\(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\) Based on the information given in the table below, what is \(\Delta H^{\circ}\) for the above reaction? \(\begin{array}{cc}{\text { Bond }} & {\text { Average bond energy }(\mathrm{kJ} / \mathrm{mol})} \\ {\mathrm{H}-\mathrm{H}} & {500} \\\ {\mathrm{O}=\mathrm{O}} & {500} \\ {\mathrm{O}-\mathrm{H}} & {500}\end{array}\) (A) \(-2,000 \mathrm{kJ}\) (B) \(-500 \mathrm{kJ}\) (C) \(+1,000 \mathrm{kJ}\) (D) \(+2,000 \mathrm{kJ}\)

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