Kapitel 2

kapitel 5Kompletta lösningar (med videor) till utvalda övningsuppgifter i kapitel 2 i Ehinger: Kemi 2 (NA förlag).

Hoppa direkt till uppgift: 2.12, 2.13, 2.16, 2.22, 2.24, 2.28, 2.29

2.12

\[K= \frac {[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \hspace{100cm} \hspace{100cm} \]

\[1,4 \cdot 10^{-4} = \frac {x^2}{0,036\text{M} \cdot 0,0089\text{M}} \hspace{100cm} \]

\[0,036\text{M} \cdot 0,0089\text{M} \cdot 1,4 \cdot 10^{-4} = x^2 \hspace{100cm} \]

\[x = \sqrt{0,036\text{M} \cdot 0,0089\text{M} \cdot 1,4 \cdot 10^{-4}} = 2,117924 \cdot 10^{-4}\text{M} \approx 2,1 \cdot 10^{-4}\text{M} \hspace{100cm}\]

2.13

[Video kommer i augusti–september 2021]

2HI(g) ⇌ H2(g) + I2(g)

\[K = \frac {[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \hspace{100cm}\]

\[0,020 = \frac {0,0200\text{M} \cdot [\text{I}_2]}{(1,33\text{M})^2} \hspace{100cm}\]

\[[\text{I}_2] =0,020 \cdot \frac {(1,33\text{M})^2}{0,0200\text{M}} = 1,7689\text{M} \approx 1,8\text{M} \hspace{100cm}\]

2.16

a.

\[Q = \frac {[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} = \frac {1,00 \cdot 1,00}{1,00^2} = 1 > K = 1,36 \cdot 10^{-3} \hspace{100cm}\]

Eftersom \(Q > K\) kommer reaktionen att gå åt vänster.

b.

  [HI] [H2] [I2]  
f.r. 1,00 1,00 1,00 M
\(+2x\) \(-x\) \(-x\) M
v.j. \(1,00+2x\) \(1,00-x\) \(1,00-x\) M

 

\[K = \frac {[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \hspace{100cm} \]

\[1,36 \cdot 10^{-3} = \frac {(1,00-x)(1,00-x)}{(1,00+2x)^2} = \frac {(1,00-x)^2}{(1,00+2x)^2} \hspace{100cm} \]

\[\sqrt{1,36 \cdot 10^{-3}} = \frac {1,00-x}{1,00+2x} \hspace{100cm} \]

\[x = 0,896965 \hspace{100cm} \]

\[[\text{HI}] = (1,00-2x)\text{M} = (1,00 - 2 \cdot 0,896965)\text{M} = 2,79393\text{M} \approx 2,79\text{M} \hspace{100cm} \]

\[[\text{H}_2] = [\text{I}_2] = (1,00 - x)\text{M} = (1,00 - 0,896965)\text{M} = 0,103035\text{M} \approx 0,103\text{M} \hspace{100cm}\]

2.22

a.

O2(g) + N2(g) ⇌ 2NO(g)

\[Q = \frac {[\text{NO}]^2}{[\text{O}_2][\text{N}_2]} = \frac {(0,22\text{M})^2}{1,0\text{M} \cdot 1,0\text{M}} = 0,0484 < K = 0,097 \hspace{100cm}\]

Reaktionen kommer att gå åt höger.

  [O2] [N2] [NO]  
f.r. \[1,0\] \[1,0\] \[0,22\] M
\[-x\] \[-x\] \[+2x\] M
v.j. \[1,0-x\] \[1,0-x\] \[0,22+2x\] M


\[K = \frac {[\text{NO}]^2}{[\text{O}_2][\text{N}_2]} \hspace{100cm} \]

\[0,097 = \frac {(0,22 + 2x)^2}{(1,0-x)(1,0-x)} = \frac {(0,22 + 2x)^2}{(1,0-x)^2} \hspace{100cm} \]

\[\sqrt{0,097} = \sqrt{\frac {(0,22 + 2x)^2}{(1,0-x)^2}} = \frac {0,22 + 2x}{1,0-x} \hspace{100cm} \]

\[\sqrt{0,097} \cdot (1,0-x) = 0,22 + 2x \hspace{100cm} \]

\[\sqrt{0,097} - \sqrt{0,097} \cdot x = 0,22 + 2x \hspace{100cm} \]

\[\sqrt{0,097} - 0,22 = 2x + \sqrt{0,097} \cdot x = x(2 + \sqrt{0,097}) \hspace{100cm} \]

\[x = \frac {\sqrt{0,097} - 0,22}{2 + \sqrt{0,097}} = 0,03956317 \hspace{100cm} \]

\[[\text{O}_2] = [\text{N}_2] = (1,0 - x)\text{M} = (1,0 - 0,03956317)\text{M} = 0,96043683\text{M} \approx 0,96\text{M} \hspace{100cm} \]

\[[\text{NO}] = (0,22 + 2x)\text{M} = (0,22 + 2 \cdot 0,03956317)\text{M} = 0,29912635\text{M} \approx 0,30\text{M} \hspace{100cm}\]

2.24

[Video kommer i augusti–september 2021.]

\[K = \frac {[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} \hspace{100cm}\]

\[[\mathrm{NH_3}] = 0,070\mathrm{M} \hspace{100cm} \\ [\mathrm{N_2}] = 0,30\mathrm{M} \hspace{100cm} \\ [\mathrm{H_2}] = 0,20\mathrm{M} \hspace{100cm} \]

\[K = \frac {(0,070\mathrm{M})^2}{0,30\mathrm{M} \cdot (0,20\mathrm{M})^3} = 2,04166667\frac {\mathrm{M}^2}{\mathrm{M} \cdot \mathrm{M}^3} \approx 2,0\mathrm{M}^{-2} \hspace{100cm}\]

2.28

[Video kommer i augusti–september 2021.]

2BrCl ⇌ Br2 + Cl2

\[Q = \frac {[\text{Br}_2][\text{Cl}_2]}{[\text{BrCl}]^2} = \frac {0,15\text{M} \cdot 0,15\text{M}}{(0,30\text{M})^2} = 0,25 < K = 0,46 \hspace{100cm}\]

Reaktionen kommer att gå åt höger.

  [BrCl] [Br2] [Cl2]  
f.r. \[0,30\] \[0,15\] \[0,15\] M
\[-2x\] \[+x\] \[+x\] M
v.j. \[0,30 - 2x\] \[0,15 + x\] \[0,15 + x\] M


\[K = \frac {[\text{Br}_2][\text{Cl}_2]}{[\text{BrCl}]^2} \hspace{100cm} \]

\[0,46 = \frac {(0,15+x)(0,15+x)}{(0,30-2x)^2} = \frac {(0,15+x)^2}{(0,30-2x)^2} \hspace{100cm} \]

\[\sqrt{0,46} = \sqrt{\frac {(0,15+x)^2}{(0,30-2x)^2}} = \frac {0,15+x}{0,30-2x} \hspace{100cm} \]

\[\sqrt{0,46} \cdot (0,30 - 2x) = 0,15 + x \hspace{100cm} \]

\[\sqrt{0,46} \cdot 0,30 - \sqrt{0,46} \cdot 2x = 0,15 + x \hspace{100cm} \]

\[\sqrt{0,46} \cdot 0,30 - 0,15 = x + \sqrt{0,46} \cdot 2x \hspace{100cm} \]

\[\sqrt{0,46} \cdot 0,30 - 0,15 = x\left(1 + \sqrt{0,46} \cdot 2 \right) \hspace{100cm} \]

\[\frac {\sqrt{0,46} \cdot 0,30 - 0,15}{1 + \sqrt{0,46} \cdot 2} = x = 0,02269072 \hspace{100cm} \]

\[[\text{BrCl]} = (0,30 - 2x)\text{M} = (0,30 - 2 \cdot 0,02269072)\text{M} = \hspace{100cm} \]

\[= 0,25461857\text{M} \approx 0,25\text{M} \hspace{100cm}\]

2.29

[Video kommer i augusti–september 2021.]

AgCl(s) ⇌ Ag+(aq) + Cl(aq)

\[K_{\text{s, AgCl}} = [\text{Ag}^+][\text{Cl}^-] = 1,1 \cdot 10^{-10}\text{M}^2 \hspace{100cm} \]

\[[\text{Ag}^+] = [\text{Cl}^-] = c_{\text{AgCl}} \hspace{100cm} \]

\[(c_{\text{AgCl}})^2 = 1,1 \cdot 10^{-10}\text{M}^2 \hspace{100cm} \]

\[c_{\text{AgCl}} = \sqrt{1,1 \cdot 10^{-10}\text{M}^2} = 1,0488088 \cdot 10^{-5}\text{M} \hspace{100cm} \]

\[\begin{aligned} n_{\text{AgCl}} &= c_{\text{AgCl}} \cdot V = 1,0488088 \cdot 10^{-5}\text{mol/dm}^3 \cdot 1,0\text{dm}^3 = \hspace{100cm} \\ &= 1,0488088 \cdot 10^{-5}\text{mol} \end{aligned}\]

\[\begin{aligned} m_{\text{AgCl}} &= n_{\text{AgCl}} \cdot M_{\text{AgCl}} = \hspace{100cm} \\ &= 1,0488088 \cdot 10^{-5}\text{mol} \cdot (107,9 + 35,45)\text{g/mol} = \\ &= 1,50346748 \cdot 10^{-3}\text{g} \approx 1,5\text{mg} \end{aligned}\]