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Chapter 1: Problem 7

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) Which of the following would cause the reverse reaction to speed up? (A) Adding more \(\mathrm{SO}_{3}\) (B) Raising the pressure (C) Lowering the temperature (D) Removing some \(\mathrm{SO}_{2}\)

Short Answer

Expert verified
Option (C) Lowering the temperature would cause the reverse reaction to speed up.

Step by step solution

01

Analyzing Options

First, consider each of the options individually. (A) If more \(\mathrm{SO}_{3}\) is added, it would favor the forward reaction and not the reverse. (B) If the pressure is raised, according to Le Chatelier’s Principle, the reaction will shift towards the side with fewer moles of gas. In this case, that's the side of \(\mathrm{SO}_{3}\), which again favors the forward reaction. (D) If some \(\mathrm{SO}_{2}\) is removed, the reverse reaction will be favored to compensate for its loss.
02

Evaluating Option C

Now, let's consider option (C). If the temperature is lowered, an exothermic reaction (\(\Delta H=-198 \) kJ/mol shows that the reaction is exothermic as heat is released) will be favored according to the Le Chatelier’s Principle. In this case, that is the reverse reaction.
03

Conclusion

Based on the above analysis, lowering the temperature will cause the reverse reaction to speed up.

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Most popular questions from this chapter

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) What is the enthalpy change for reaction 1\(?\) (A) \(-148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (B) \(-56 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (C) \(-18 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (D) \(+148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\)

Which of the following could be added to an aqueous solution of weak acid HF to increase the percent dissociation? (A) \(\operatorname{NaF}(s)\) (B) \(\mathrm{H}_{2} \mathrm{O}(l)\) (C) \(\mathrm{NaOH}(\mathrm{s})\) (D) \(\mathrm{NH}_{3}(a q)\)

Which of the following substances has an asymmetrical molecular structure? (A) \(\mathrm{SF}_{4}\) (B) \(\mathrm{PCl}_{5}\) (C) \(\mathrm{BF}_{3}\) (D) \(\mathrm{CO}_{2}\)

A strong acid/strong base titration is completed using an indicator which changes color at the exact equivalence point of the titration. The protonated form of the indicator is HIn, and the deprotonated form is In. At the equivalence point of the reaction: (A) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]\) (B) \([\mathrm{HIn}]=1 /\left[\mathrm{In}^{-}\right]\) (C) \([\mathrm{HIn}]=2\left[\mathrm{In}^{-}\right]\) (D) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]^{2}\)

Which of the following expressions is equal to the \(K_{\mathrm{sp}}\) of \(\mathrm{Ag}_{2} \mathrm{CO}_{3} ?\) (A) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CO}_{3}^{2-}\right]\) (B) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CO}_{3}^{2-}\right]^{2}\) (C) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CO}_{3}^{2-}\right]\) (D) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CO}_{3}^{2-}\right]^{2}\)
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