Chapter 14: Problem 12
Why must a balanced chemical equation be used when determining \(K_{e q} ?\)
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Chapter 14: Problem 12
Why must a balanced chemical equation be used when determining \(K_{e q} ?\)
These are the key concepts you need to understand to accurately answer the question.
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Changes in the concentrations of the reactants and products at equilibrium have no effect on the value of the equilibrium constant. Explain this statement.
The ionic substance \(\mathrm{T}_{3} \mathrm{U}_{2}\) ionizes to form \(\mathrm{T}^{2+}\) and \(\mathrm{U}^{3-}\) ions. The solubility of \(\mathrm{T}_{3} \mathrm{U}_{2}\) is \(3.77 \times 10^{-20} \mathrm{mol} / \mathrm{L}\) . What is the value of the solubility-product constant?
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.
Describe and explain how the concentrations of \(A, B, C,\) and \(D\) change from the time when A and \(B\) first combine to the point at which equilibrium is established for the reaction $$A+B \rightleftarrows C+D$$
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$$
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