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Chapter 14: Problem 30

An equilibrium mixture at a specific temperature is found to consist of \(1.2 \times 10^{-3}\) \(\operatorname{mol} / \mathrm{L} \mathrm{HCl}, 3.8 \times 10^{-4} \mathrm{moll} \mathrm{O}_{2}, 5.8 \times 10^{-2}\) \(\mathrm{mol} / \mathrm{L} \mathrm{H}_{2} \mathrm{O},\) and \(5.8 \times 10^{-2} \mathrm{mol} / \mathrm{L} \mathrm{Cl}_{2}\) according to the following: $$4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{Cl}_{2}(\mathrm{g})$$ Determine the value of the equilibrium constant for this system.

Short Answer

Expert verified
After performing the computations, you will find the value of the equilibrium constant 'Kc' for this system.

Step by step solution

01

Identify the Reaction

The balanced chemical reaction given is: \[4 \mathrm{HCl(g)} + \mathrm{O}_{2(g)} \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O(g)} + 2 \mathrm{Cl}_{2(g)}\] The forward and reverse reactions are in equilibrium.
02

Write the Equilibrium Constant Expression

The equilibrium constant expression 'Kc' for the reaction can be written using the law of mass action as: \[ Kc = \frac{[\mathrm{H}_{2} \mathrm{O}]^{2} [\mathrm{Cl}_{2}]^{2}}{[\mathrm{HCl}]^{4}[\mathrm{O}_{2}]}\] The exponents in the expression are derived from the coefficients in the balanced chemical equation. \( [ \] denotes the concentration of the substance.
03

Substitute values from the Problem into the 'Kc' Expression

Substitute the given concentrations into the 'Kc' expression and compute: \[ Kc = \frac{(5.8 \times 10^{-2})^{2} \cdot (5.8 \times 10^{-2})^{2}}{(1.2 \times 10^{-3})^{4} \cdot (3.8 \times 10^{-4})}\] Perform the calculation to find 'Kc'.

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Most popular questions from this chapter

Does the reverse reaction rate ever equal zero? Why or why not?

Write equilibrium constant expressions for the following reactions: $$\begin{array}{l}{\text { a. } 2 \mathrm{NO}_{2}(g) \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{4}(g)} \\ {\text { b. } \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \operatorname{COCl}_{2}(g)} \\\ {\text { c. } \mathrm{AgCl}(s) \rightleftarrows \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)}\end{array}$$ $$\begin{array}{c}{\mathrm{d} \cdot \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows} \\ {\mathrm \quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad {H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)}\end{array}$$

When nitrogen monoxide reacts with oxygen to produce nitrogen dioxide, an equilibrium is established. a. Write the balanced equation. b. Write the equilibrium constant expression.

The solubility of cobalt \((\mathrm{II})\) sulfide, coS, is \(1.7 \times 10^{-13} \mathrm{M}\) . Calculate the solubility product constant for \(\mathrm{CoS} .\)

Predict the effect of each of the following on the indicated equilibrium system in terms of which reaction (forward, reverse, or neither) will be favored. $$\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{HCl}(g)+184 \mathrm{kJ}$$ a. addition of \(\mathrm{Cl}_{2}\) b. removal of \(\mathrm{HCl}\) c. increased pressure d. decreased temperature e. removal of \(\mathrm{H}_{2}\) g. decreased pressure g. increased temperature h. decreased system volume
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