A sample of \(\mathrm{H}_{2} \mathrm{S}\) gas is placed in an evacuated, sealed container and heated until the following decomposition reaction occurs at \(1000 \mathrm{K} :\) $$2 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \qquad K_{\mathrm{c}}=1.0 \times 10^{-6}$$ If, at a given point in the reaction, the value for the reaction quotient \(Q\) is determined to be \(2.5 \times 10^{-8},\) which of the following is occurring? (A) The concentration of the reactant is decreasing while the concentration of the products is increasing. (B) The concentration of the reactant is increasing while the concentration of the products is decreasing. (C) The system has passed the equilibrium point, and the concentration of all species involved in the reaction will remain constant. (D) The concentrations of all species involved are changing at the same rate.

Short Answer

Expert verified
(A) The concentration of the reactant is decreasing while the concentration of the products is increasing.

Step by step solution

01

Understanding the Reaction Quotient and Equilibrium Constant

The reaction quotient (Q) gives the same ratio as the equilibrium constant (\(K_c\)), but Q can be calculated at any time, while \(K_c\) is only true at equilibrium. If Q < \(K_c\), the reaction would shift in the forward direction to reach equilibrium. Conversely, if Q > \(K_c\), the reaction would shift in the reverse direction to reach equilibrium.
02

Comparing Q and \(K_c\)

The given value for Q is \(2.5 \times 10^{-8}\), and the given equilibrium constant \(K_c\) is \(1.0 \times 10^{-6}\). So the condition Q < \(K_c\) is met as \(2.5 \times 10^{-8}\) < \(1.0 \times 10^{-6}\).
03

Deciding the Direction of Reaction

Because Q < \(K_c\), the reaction will need to shift in the forward direction to reach equilibrium. This means that more reactants will be transformed into products, thus the concentration of the reactants will decrease while that of the products will increase.

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Most popular questions from this chapter

\(2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) The data below was gathered for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at 310 \(\mathrm{K}\) via the equation above. \(\begin{array}{|c|c|}\hline \text { Time(s) } & {\left[\mathrm{N}_{2} \mathbf{O}_{5}\right](M)} \\ \hline 0 & {0.250} \\ \hline 500 . & {0.190} \\\ \hline 1000 . & {0.145} \\ \hline 2000 . & {0.085} \\ \hline\end{array}\) (a) How does the rate of appearance of NO_{2} \text { compare to the rate of } disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}\) ? Justify your answer. (b) The reaction is determined to be first order overall. On the axes below, create a graph of some function of concentration vs. time that will produce a straight line. Label and scale your axes appropriately. (c) (i) What is the rate constant for this reaction? Include units. (ii) What would the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) be at \(t=1500 \mathrm{s} ?\) (iii) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) ? (d) Would the addition of a catalyst increase, decrease, or have no effect on the following variables? Justify your answers. (i) Rate of disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (ii) Magnitude of the rate constant (iii) Half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\)

20.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is placed in a beaker and titrated with a solution of \(1.0 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) resulting in the creation of a precipitate. What will happen to the conductivity of the solution after additional \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is added past the equivalence point? (A) The conductivity will increase as additional ions are being added to the solution. (B) The conductivity will stay constant as the precipitation reaction has gone to completion. (C) The conductivity will decrease as the solution will be diluted with the addition of additional \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\). (D) The conductivity will stay constant as equilibrium has been established.

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\(2 \mathrm{Ag}^{+}(a q)+\mathrm{Fe}(s) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Fe}^{2+}(a q)\) Which of the following would cause an increase in potential in the voltaic cell described by the above reaction? (A) Increasing \(\left[\mathrm{Fe}^{2+}\right]\) (B) Adding more \(\mathrm{Fe}(s)\) (C) Decreasing \(\left[\mathrm{Fe}^{2+}\right]\) (D) Removing some \(\mathrm{Fe}(s)\)

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