10. for the reaction at equilibrium: $2\text{no}_{\text{(g)}} + \text{cl}_{2\text{(g)}} \ ightleftharpoons 2\text{nocl}_{\text{(g)}}$ if $\text{no} = 0.30\text{ m}$, $\text{cl}_2 = 0.20\text{ m}$, $\text{nocl} = 0.80\text{ m}$, the value of $k_{\text{eq}}$ is: (a) 36 (b) 0.028 (c) 13 (d) 0.075 11. for the following equilibrium: $\text{pbr}_{3\text{(g)}} + \text{br}_{2\text{(g)}} \ ightleftharpoons \text{pbr}_{5\text{(g)}}$ $k_{\text{eq}} = 0.090$ at $250^\circ\text{c}$ and $k_{\text{eq}} = 0.17$ at $100^\circ\text{c}$ the reverse reaction at $100^\circ\text{c}$ is: (a) exothermic and $k_{\text{eq}} = 5.9$ (b) exothermic and $k_{\text{eq}} = 11.1$ (c) endothermic and $k_{\text{eq}} = 11.0$ (d) endothermic and $k_{\text{eq}} = 5.9$ 12. for the following reaction at equilibrium: $\text{h}_{2\text{(g)}} + \text{s}_{\text{(s)}} \ ightleftharpoons \text{h}_2\text{s}_{\text{(g)}}$ $k_{\text{eq}} = 16.0$ it could be said that: (a) $\text{h}_2 = \text{h}_2\text{s}$ (b) $\text{h}_2 \text{h}_2\text{s}$ (d) $\text{h}_2 = \text{s}$ 13. according to the table of solubilities, what will result when $0.10\text{ m } \text{sr(oh)}_2$ is mixed with $0.10\text{ m } \text{mgso}_4$? (a) no precipitate forms (b) only $\text{srso}_4$ will precipitate (c) only $\text{mg(oh)}_2$ will precipitate (d) both $\text{srso}_4$ and $\text{mg(oh)}_2$ will precipitate 14. calculate the maximum $\text{ag}^+_{\text{(aq)}}$ that can exist in a solution containing $0.10\text{ m } \text{na}_2\text{cro}_{4\text{(aq)}}$ without precipitating $\text{ag}_2\text{cro}_{4\text{(s)}}$? $k_{\text{sp}}\text{ for } \text{ag}_2\text{cro}_4 = 1.1 \times 10^{-12}$ (a) $1.1 \times 10^{-11}\text{ m}$ (b) $1.1 \times 10^{-6}\text{ m}$ (c) $3.3 \times 10^{-6}\text{ m}$ (d) $7.4 \times 10^{-3}\text{ m}$ 15. in a $50.0\text{ ml}$ sample of saturated $\text{al(oh)}_3$ solution the $\text{oh}^-_{\text{(aq)}} = 2.6 \times 10^{-3}\text{ m}$. the solubility of $\text{al(oh)}_3$ is: (a) $8.7 \times 10^{-4}\text{ m}$ (b) $7.8 \times 10^{-3}\text{ m}$ (c) $1.7 \times 10^{-2}\text{ m}$ (d) $2.6 \times 10^{-3}\text{ m}$

Question

10. for the reaction at equilibrium: $2\text{no}_{\text{(g)}} + \text{cl}_{2\text{(g)}} \
ightleftharpoons 2\text{nocl}_{\text{(g)}}$
if $\text{no} = 0.30\text{ m}$, $\text{cl}_2 = 0.20\text{ m}$, $\text{nocl} = 0.80\text{ m}$, the value of $k_{\text{eq}}$ is:
(a) 36
(b) 0.028
(c) 13
(d) 0.075

11. for the following equilibrium: $\text{pbr}_{3\text{(g)}} + \text{br}_{2\text{(g)}} \
ightleftharpoons \text{pbr}_{5\text{(g)}}$
$k_{\text{eq}} = 0.090$ at $250^\circ\text{c}$ and $k_{\text{eq}} = 0.17$ at $100^\circ\text{c}$
the reverse reaction at $100^\circ\text{c}$ is:
(a) exothermic and $k_{\text{eq}} = 5.9$
(b) exothermic and $k_{\text{eq}} = 11.1$
(c) endothermic and $k_{\text{eq}} = 11.0$
(d) endothermic and $k_{\text{eq}} = 5.9$

12. for the following reaction at equilibrium: $\text{h}_{2\text{(g)}} + \text{s}_{\text{(s)}} \
ightleftharpoons \text{h}_2\text{s}_{\text{(g)}}$ $k_{\text{eq}} = 16.0$
it could be said that:
(a) $\text{h}_2 = \text{h}_2\text{s}$
(b) $\text{h}_2 < \text{h}_2\text{s}$
(c) $\text{h}_2 > \text{h}_2\text{s}$
(d) $\text{h}_2 = \text{s}$

13. according to the table of solubilities, what will result when $0.10\text{ m } \text{sr(oh)}_2$ is mixed with $0.10\text{ m } \text{mgso}_4$?
(a) no precipitate forms
(b) only $\text{srso}_4$ will precipitate
(c) only $\text{mg(oh)}_2$ will precipitate
(d) both $\text{srso}_4$ and $\text{mg(oh)}_2$ will precipitate

14. calculate the maximum $\text{ag}^+_{\text{(aq)}}$ that can exist in a solution containing $0.10\text{ m } \text{na}_2\text{cro}_{4\text{(aq)}}$ without precipitating $\text{ag}_2\text{cro}_{4\text{(s)}}$? $k_{\text{sp}}\text{ for } \text{ag}_2\text{cro}_4 = 1.1 \times 10^{-12}$
(a) $1.1 \times 10^{-11}\text{ m}$
(b) $1.1 \times 10^{-6}\text{ m}$
(c) $3.3 \times 10^{-6}\text{ m}$
(d) $7.4 \times 10^{-3}\text{ m}$

15. in a $50.0\text{ ml}$ sample of saturated $\text{al(oh)}_3$ solution the $\text{oh}^-_{\text{(aq)}} = 2.6 \times 10^{-3}\text{ m}$. the solubility of $\text{al(oh)}_3$ is:
(a) $8.7 \times 10^{-4}\text{ m}$
(b) $7.8 \times 10^{-3}\text{ m}$
(c) $1.7 \times 10^{-2}\text{ m}$
(d) $2.6 \times 10^{-3}\text{ m}$

Accepted Answer

{ "quality": "clear", "question_count": 6, "skills_matched": [ "step_cot", "table_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Equilibrium Constant Calculations", "Le Chateliers Principle", "Precipitation Reactions" ], "new_concepts": [ "Solubility Product Constant", "Heterogeneous Equilibrium", "Molar Solubility" ], "current_concepts": [ "Equilibrium Constant Calculations", "Le Chateliers Principle", "Precipitation Reactions", "Solubility Product Constant", "Heterogeneous Equilibrium", "Molar Solubility" ] } ## Calculate the equilibrium constant for Question 10 Using the *Equilibrium Constant Calculations* knowledge point $$ K_{eq} = \frac{[\text{NOCl}]^2}{[\text{NO}]^2[\text{Cl}_2]} = \frac{(0.80)^2}{(0.30)^2(0.20)} = \frac{0.64}{0.09 \times 0.20} = \frac{0.64}{0.018} \approx 35.56 \approx 36 $$ ## Determine the reverse reaction constant for Question 11 Using the *Equilibrium Constant Calculations* knowledge point $$ K_{eq,\text{reverse}} = \frac{1}{K_{eq,\text{forward}}} = \frac{1}{0.17} \approx 5.88 \approx 5.9 $$ Using the *Le Chateliers Principle* knowledge point, since $K_{eq}$ decreases from $0.17$ at $100^\circ\text{C}$ to $0.090$ at $250^\circ\text{C}$ as temperature increases, the forward reaction is exothermic. Thus, the reverse reaction is endothermic. ## Analyze heterogeneous equilibrium for Question 12 Using the *Le Chateliers Principle* knowledge point $$ K_{eq} = \frac{[\text{H}_2\text{S}]}{[\text{H}_2]} = 16.0 \implies [\text{H}_2\text{S}] = 16.0 [\text{H}_2] \implies [\text{H}_2] < [\text{H}_2\text{S}] $$ ## Predict precipitation for Question 13 Using the *Precipitation Reactions* knowledge point Mixing $0.10\text{ M }\text{Sr(OH)}_2$ and $0.10\text{ M }\text{MgSO}_4$ yields $\text{SrSO}_4$ and $\text{Mg(OH)}_2$. According to standard solubility rules, $\text{SrSO}_4$ is insoluble (precipitates) and $\text{Mg(OH)}_2$ is insoluble (precipitates). Thus, both $\text{SrSO}_4$ and $\text{Mg(OH)}_2$ will precipitate. ## Calculate maximum silver ion concentration for Question 14 To prevent precipitation of $\text{Ag}_2\text{CrO}_4$: $$ Q_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] \le K_{sp} $$ $$ [\text{Ag}^+]^2(0.10) \le 1.1 \times 10^{-12} \implies [\text{Ag}^+]^2 \le 1.1 \times 10^{-11} \implies [\text{Ag}^+] \le 3.3 \times 10^{-6}\text{ M} $$ ## Calculate molar solubility of aluminum hydroxide for Question 15 The dissolution of $\text{Al(OH)}_3$ is: $$ \text{Al(OH)}_3(s) ightleftharpoons \text{Al}^{3+}(aq) + 3\text{OH}^-(aq) $$ The molar solubility $S$ of $\text{Al(OH)}_3$ is equal to the concentration of $\text{Al}^{3+}$ at equilibrium: $$ S = [\text{Al}^{3+}] = \frac{[\text{OH}^-]}{3} = \frac{2.6 \times 10^{-3}\text{ M}}{3} \approx 8.7 \times 10^{-4}\text{ M} $$ | No. | Answer | |---|---| | 10 | (a) 36 | | 11 | (d) Endothermic and $K_{eq} = 5.9$ | | 12 | (b) $[\text{H}_2] < [\text{H}_2\text{S}]$ | | 13 | (d) Both $\text{SrSO}_4$ and $\text{Mg(OH)}_2$ will precipitate | | 14 | (c) $3.3 \times 10^{-6}\text{ M}$ | | 15 | (a) $8.7 \times 10^{-4}\text{ M}$ | { "subject": "Natural Science", "question_type": "Multi-part", "knowledge_point": [ "Natural Science", "Chemistry", "Chemical Equilibrium" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate the equilibrium constant for Question 10

Determine the reverse reaction constant for Question 11

Analyze heterogeneous equilibrium for Question 12

Predict precipitation for Question 13

Calculate maximum silver ion concentration for Question 14

Calculate molar solubility of aluminum hydroxide for Question 15

Answer:

No.	Answer
11	(d) Endothermic and \(K_{eq} = 5.9\)
12	(b) \([\text{H}_2] < [\text{H}_2\text{S}]\)
13	(d) Both \(\text{SrSO}_4\) and \(\text{Mg(OH)}_2\) will precipitate
14	(c) \(3.3 \times 10^{-6}\text{ M}\)
15	(a) \(8.7 \times 10^{-4}\text{ M}\)