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At \(1130^{\circ} \mathrm{C}\) the equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)$$ is \(2.25 \times 10^{-4}\). If \(\left[\mathrm{H}_{2} \mathrm{~S}\right]=4.84 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=\) \(1.50 \times 10^{-3} M,\) calculate \(\left[\mathrm{S}_{2}\right]\).

Short Answer

Expert verified
\([\mathrm{S}_2] = 2.34 \times 10^{-3} \mathrm{M}\).

Step by step solution

01

Understanding the Reaction

We are given the equilibrium reaction \(2 \mathrm{H}_2 \mathrm{S}(g) \rightleftharpoons 2 \mathrm{H}_2(g) + \mathrm{S}_2(g)\) and need to determine the concentration of \(\mathrm{S}_2\). The equilibrium constant \(K_c\) is provided as \(2.25 \times 10^{-4}\).
02

Writing the Expression for the Equilibrium Constant

The equilibrium constant expression \(K_c\) for the reaction is given by \[ K_c = \frac{[\mathrm{H}_2]^2[\mathrm{S}_2]}{[\mathrm{H}_2 \mathrm{S}]^2} \] where the concentrations are evaluated at equilibrium.
03

Inserting Known Values

Substitute the known concentration values into the equilibrium constant expression: \[ 2.25 \times 10^{-4} = \frac{(1.50 \times 10^{-3})^2 [\mathrm{S}_2]}{(4.84 \times 10^{-3})^2} \]
04

Solving for \( [\mathrm{S}_2] \)

Solve the equation for \( [\mathrm{S}_2] \):1. Calculate \( (1.50 \times 10^{-3})^2 = 2.25 \times 10^{-6} \).2. Calculate \( (4.84 \times 10^{-3})^2 = 2.34 \times 10^{-5} \).3. Rearrange the equation to find \( [\mathrm{S}_2] \) as follows:\[ [\mathrm{S}_2] = \frac{2.25 \times 10^{-4} \times 2.34 \times 10^{-5}}{2.25 \times 10^{-6}} \]
05

Calculating \( [\mathrm{S}_2] \)

Compute \( [\mathrm{S}_2] \) using the rearranged equation:1. Multiply numerator terms: \( 2.25 \times 10^{-4} \times 2.34 \times 10^{-5} = 5.265 \times 10^{-9} \).2. Divide by the denominator: \( \frac{5.265 \times 10^{-9}}{2.25 \times 10^{-6}} = 2.34 \times 10^{-3} \).3. So, \( [\mathrm{S}_2] = 2.34 \times 10^{-3} \mathrm{M} \).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant over time, even though both reactions are still occurring. It's important to note that equilibrium does not mean the reactants and products are equal in concentration, but rather stable in their proportions.

In the context of the given exercise, the chemical equilibrium is represented by the reaction between hydrogen sulfide (\(2 \mathrm{H}_2 \mathrm{S}(g) \rightleftharpoons 2 \mathrm{H}_2(g) + \mathrm{S}_2(g)\)). At equilibrium, the concentrations of hydrogen sulfide, hydrogen, and sulfur are stable when the system is maintained at a constant temperature of 1130°C.
Le Chatelier's Principle
Le Chatelier's Principle is a helpful guideline for predicting how a chemical system at equilibrium responds to disturbances, such as changes in concentration, temperature, or pressure. According to this principle, if any of these conditions change, the equilibrium will shift in the direction that tends to reduce the impact of the change.

In the provided reaction, suppose we increased the concentration of hydrogen sulfide, \(\mathrm{H}_2 \mathrm{S}\). According to Le Chatelier's Principle, the system would adjust to counteract this change by favoring the forward reaction to produce more hydrogen and sulfur. Conversely, if hydrogen or sulfur were increased, the equilibrium would shift toward the formation of more \(\mathrm{H}_2 \mathrm{S}\), partially alleviating the increase in \(\mathrm{H}_2\) or \(\mathrm{S}_2\) concentrations.

Through recognizing these shifts, chemists can manipulate conditions to increase the yield of desired products.
Reaction Quotient
The reaction quotient, sometimes denoted as \(Q_c\), serves as a valuable indicator in determining the current state of a reaction system compared to its equilibrium state. It is calculated using the same expression as the equilibrium constant \(K_c\) but with current concentrations instead of equilibrium concentrations.

For the reaction specified in the exercise, the expression for the reaction quotient \(Q_c\) is:
  • \(Q_c = \frac{[\mathrm{H}_2]^2[\mathrm{S}_2]}{[\mathrm{H}_2 \mathrm{S}]^2}\)
Comparing \(Q_c\) with \(K_c\) can tell us whether the system is at equilibrium:
  • If \(Q_c = K_c\), the system is at equilibrium.
  • If \(Q_c < K_c\), the forward reaction will be favored to reach equilibrium.
  • If \(Q_c > K_c\), the backward reaction will be favored.
In this exercise, although the actual concentrations at a given moment were not used to calculate \(Q_c\), understanding this relationship is crucial for predicting how changes affect the position of equilibrium.

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Most popular questions from this chapter

The decomposition of ammonium hydrogen sulfide $$\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ is an endothermic process. A 6.1589 -g sample of the solid is placed in an evacuated 4.000 - \(L\) vessel at exactly \(24^{\circ} \mathrm{C}\). After equilibrium has been established, the total pressure inside is 0.709 atm. Some solid \(\mathrm{NH}_{4}\) HS remains in the vessel. (a) What is the \(K_{P}\) for the reaction? (b) What percentage of the solid has decomposed? (c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel?

The equilibrium constant \(K_{P}\) for the reaction$$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$is 1.05 at \(250^{\circ} \mathrm{C}\). The reaction starts with a mixture of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) at pressures of \(0.177 \mathrm{~atm}, 0.223\)atm, and 0.111 atm, respectively, at \(250^{\circ} \mathrm{C}\). When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.

Consider the equilibrium system \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\). Sketch the change in concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) with time for these situations: (a) Initially only A is present; (b) initially only B is present; (c) initially both A and \(\mathrm{B}\) are present (with \(\mathrm{A}\) in higher concentration). In each case, assume that the concentration of \(\mathrm{B}\) is higher than that of \(\mathrm{A}\) at equilibrium.

Consider the gas-phase reaction $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)$$ Predict the shift in the equilibrium position when helium gas is added to the equilibrium mixture (a) at constant pressure and (b) at constant volume.

One mole of \(\mathrm{N}_{2}\) and 3 moles of \(\mathrm{H}_{2}\) are placed in a flask at \(397^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is found to be 0.21. The \(K_{P}\) for the reaction is \(4.31 \times 10^{-4}\).

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