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Chapter 1: Problem 8

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) The value for \(K_{\mathrm{c}}\) at \(25^{\circ} \mathrm{C}\) is \(8.1 .\) What must happen in order for the reaction to reach equilibrium if the initial concentrations of all three species was 2.0 \(M\) ? (A) The rate of the forward reactions would increase, and \(\left[\mathrm{SO}_{3}\right]\) would decrease. (B) The rate of the reverse reaction would increase, and \(\left[\mathrm{SO}_{2}\right]\) would decrease. (C) Both the rate of the forward and reverse reactions would increase, and the value for the equilibrium constant would also increase. (D) No change would occur in either the rate of reaction or the concentrations of any of the species.

Short Answer

Expert verified
The correct answer is (A): The rate of the forward reaction would increase, and the concentration of \(SO_{3}\) would decrease.

Step by step solution

01

Write down the balanced equation and expression for Qc

The balanced reaction is: \(2 SO_{3}(g) \leftrightarrow O_{2}(g)+2 SO_{2}(g)\). The reaction quotient \(Q_{c}\) is given by: \[Q_{c} = \frac{[O_{2}][SO_{2}]^{2}}{[SO_{3}]^{2}}\]
02

Calculation of initial Qc

Based on the given initial concentrations, the reaction quotient Qc is calculated as follow: \[Q_{c} = \frac{(2.0)(2.0)^{2}}{(2.0)^{2}} = \frac{8}{4} = 2\]
03

Compare Qc with Kc

The value of Qc calculated in step 2 is less than the given Kc (2 < 8.1). This means the reaction will proceed to the right to achieve equilibrium.
04

Interpretation

Since the reaction moves to the right (forward direction), the concentration of the reactant \(SO_{3}\) will decrease, while the concentration of the products \(O_{2}\) and \(SO_{2}\) will increase. Hence, the forward reaction rate will increase.

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Most popular questions from this chapter

$$\begin{array}{|c|c|}\hline \text { Time (Hours) } & {[\mathrm{A}] M} \\\ \hline 0 & {0.40} \\ \hline 1 & {0.20} \\ \hline 2 & {0.10} \\ \hline 3 & {0.05} \\ \hline\end{array}$$ Reactant A underwent a decomposition reaction. The concentration of A was measured periodically and recorded in the chart above. Based on the data in the chart, which of the following is the rate law for the reaction? (A) Rate \(=k[\mathrm{A}]\) (B) Rate \(=k[\mathrm{A}]^{2}\) (C) Rate \(=2 k[\mathrm{A}]\) (D) Rate \(=\frac{1}{2} k[\mathrm{A}]\)

$$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \mapsto 2 \mathrm{NOCl}(g) \Delta G^{\circ}=132.6 \mathrm{kJ} / \mathrm{mol}$$ For the equilibrium above, what would happen to the value of \(\Delta G^{\circ}\) if the concentration of \(\mathrm{N}_{2}\) were to increase and why? (A) It would increase because the reaction would become more themodynamically favored. (B) It would increase because the reaction would shift right and create more products. (C) It would decrease because there are more reactants present. (D) It would stay the same because the value of \(K_{e q}\) would not change.

How many moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) must be added to 500 milliliters of water to create a solution that has a 2 -molar concentration of the Na' ion? (Assume the volume of the solution does not change). (A) 0.5 \(\mathrm{mol}\) (B) 1 \(\mathrm{mol}\) (C) 2 \(\mathrm{mol}\) (D) 5 \(\mathrm{mol}\)

Which compound, \(\mathrm{CaCl}_{2}\) or \(\mathrm{CaO}\) , would you expect to have a high melting point? Why? (A) \(\mathrm{CaCl}_{2}\) because there are more ions per lattice unit (B) \(\mathrm{CaCl}_{2}\) because a chlorine ion is smaller than an oxygen ion (C) Cao, because the charge of oxygen ion exceeds that of chlorine ion (D) CaO, because the common charges of calcium and oxygen ions are identical in magnitude

During a chemical reaction, NO(g) gets reduced and no nitrogen- containing compound is oxidized. Which of the following is a possible product of this reaction? (A) \(\mathrm{NO}_{2}(g)\) (B) \(\mathrm{N}_{2}(g)\) (C) \(\mathrm{NO}_{3}^{-}(a q)\) (D) \(\mathrm{NO}_{2}^{-}(a q)\)
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