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Chapter 14: Problem 29

Determine the value of the equilibrium constant for each reaction below assuming that the equilibrium concentrations are as specified. $$\text {(a.)}A+B \rightleftarrows C ;[A]=2.0 ;[B]=3.0 ;[C]=4.0$$ $$\begin{array}{l}{\text { b. } \mathrm{D}+2 \mathrm{E} \rightleftarrows \mathrm{F}+3 \mathrm{G} ;[\mathrm{D}]=1.5 ;[\mathrm{E}]=2.0} \\\ {[\mathrm{F}]=1.8 ;[\mathrm{G}]=1.2}\end{array}$$ $$\mathrm{c} \cdot \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NH}_{3}(\mathrm{g}) ;\left[\mathrm{N}_{2}\right]=0.45\( \)\left[\mathrm{H}_{2}\right]=0.14 ;\left[\mathrm{NH}_{3}\right]=0.62$$

Short Answer

Expert verified
The equilibrium constant for the first reaction is 0.67, for the second reaction is 0.2592 and for the third reaction is 543.106.

Step by step solution

01

Calculate the equilibrium constant for the first reaction

The equilibrium constant (K) for a reaction is defined as the ratio of the product concentration to the reactant concentration, each raised to their respective stoichiometric coefficient. From the reaction \(A + B \leftrigharrows C\), the equilibrium constant \(K\) can be expressed as: \(K = [C]/([A]*[B])\). Substituting the given equilibrium concentrations into this expression gives \(K = 4.0/(2.0*3.0) = 0.67\).
02

Calculate the equilibrium constant for the second reaction

The reaction \(D + 2E \leftrigharrows F + 3G\) will have the equilibrium constant: \(K = [F]*[G]^3/([D]*[E]^2)\). Substituting the given equilibrium concentrations in this expression gives \(K = 1.8*1.2^3/(1.5*2.0^2) = 0.2592\).
03

Calculate the equilibrium constant for the third reaction

For the reaction \(N2(g) + 3H2(g) \leftrigharrows 2NH3(g)\), the equilibrium constant \(K\) can be expressed as: \(K = [NH3]^2/([N2]*[H2]^3)\). Substituting the equilbrium concentrations in this expression gives \(K = 0.62^2/(0.45*0.14^3) = 543.106\).

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Most popular questions from this chapter

The reaction below has an equilibrium constant of \(4.9 \times 10^{11}\) . $$\mathrm{Fe}(\mathrm{OH})_{2}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftarrows \mathrm{Fe}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)$$ Write the equilibrium constant expression, and determine the concentration of \(\mathrm{Fe}^{2+}\) ions in equilibrium when the hydronium ion concentration is \(1.0 \times 10^{-7} \mathrm{mol} / \mathrm{L}\)

Imagine the following hypothetical reaction, taking place in a sealed, rigid container, to be neither exothermic nor endothermic. $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftarrows \mathrm{C}(g) \quad \Delta H=0$$ Would an increase in temperature favor the forward reaction or the reverse reaction? (Hint: Recall the gas laws.)

At \(450^{\circ} \mathrm{C}\) the value of the equilibrium constant for the following system is \(6.59 \times 10^{-3}\) . If \(\left[\mathrm{NH}_{3}\right]=1.23 \times 10^{-4}\) and \(\left[\mathrm{H}_{2}\right]=2.75 \times 10^{-3}\) at equilibrium, determine the concentration of \(\mathrm{N}_{2}\) at that point. $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NH}_{3}(\mathrm{g})$$

A student wrote, "The larger the equilibrium constant, the greater the rate at which reactants convert to products." How was he wrong?

How would you explain Le Chatelier's principle in your own words to someone who finds the concept difficult to understand?
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