Teksti

Esim. Lämpötilassa 448°C reaktion
$H_2\left(g\right)+I_2\left(g\right)\xrightleftharpoons[]{}2HI\left(g\right)$

Tasapainovakio o 51. Ennusta, mihin suuntaan reaktio etenee, kun 2,0 litran astian suljetaan 0,020 mol vetyjodidia, 0,010 mol vetyä sekä 0,030 ol jodia ja seos kuumennetaan 448°C.

$C\left(HI\right)=\frac{0{,}020mol}{2{,}0l}=0{,}010\frac{mol}{l}$

$C\left(H_2\right)=\frac{0{,}010mol}{2{,}0l}=0{,}0050\frac{mol}{l}$
$C\left(I_2\right)=\frac{0{,}030mol}{2{,}0l}=0{,}015\frac{mol}{l}$

Sijoitetaan nämä konsentraatiot reaktion tasapainovakion lausekkeeseen:

$K_c=\frac{\left[HI\right]^2}{\left[H_2\right]\left[I_2\right]}=\frac{\left(0{,}010\frac{mol}{l}\right)^2}{\left(0{,}0050\frac{mol}{l}\right)\left(0{,}015\frac{mol}{l}\right)}=1{,}333...\approx1{,}3<51$

Reaktio etenee reaktiotuotteiden suuntaan eli vetyjodidia muodostuu lisää, kunnes tasapainotila saavutetaan.

3.1

$2HI\left(g\right)\xrightleftharpoons[]{}H_2\left(g\right)+I_2\left(g\right)$

$C\left(HI\right)=3{,}53\cdot10^{-3}\frac{mol}{l}$
$C\left(H_2\right)=4{,}79\cdot10^{-4}\frac{mol}{l}$
$C\left(I_2\right)=4{,}79\cdot10^{-4}\frac{mol}{l}$

$K_c=\frac{\left[H_2\right]\left[I_2\right]}{\left[HI\right]^2}=\frac{\left(4{,}79\cdot10^{-4}\frac{mol}{l}\right)\cdot\left(4{,}79\cdot10^{-4}\ \frac{mol}{l}\right)}{\left(3{,}53\cdot10^{-3}\frac{mol}{l}\right)^2}=0{,}01841...\approx0{,}0184$

3.2

$n\left(N_2\right)=2{,}80\cdot10^{-3}mol$

$n\left(O_2\right)=2{,}50\cdot10^{-5}mol$

$n\left(N_2O\right)=2{,}00\cdot10^{-2}mol$

$C\left(N_2\right)=\frac{2{,}80\cdot10^{-3}mol}{2{,}0l}=1{,}4\cdot10^{-3}\frac{mol}{l}$
$C\left(O_2\right)=\frac{2{,}50\cdot10^{-5}mol}{2{,}0l}=1{,}25\cdot10^{-5}\frac{mol}{l}$
$C\left(N_2O\right)=\frac{2{,}00\cdot10^{-2}mol}{2{,}0l}=0{,}01\frac{mol}{l}$

$K_c=\frac{\left[N_2O\right]^2}{\left[N_2\right]^2\left[O_2\right]}=\frac{\left(0{,}01\frac{mol}{l}\right)^2}{\left(1{,}4\cdot10^{-3}\frac{mol}{l}\right)^2\left(1{,}25\cdot10^{-5}\frac{mol}{l}\right)}=4081632{,}658\approx4\ 081\ 600$

$=4{,}08\cdot10^6\ \left(\frac{mol}{l}\right)^{-1}$

3.3

$\ 2NO\left(g\right)+2H_2\left(g\right)\xrightleftharpoons[]{}N_2\left(g\right)+2H_2O\left(g\right)$

$\begin{array}{l|l} C_{alku}\left(\frac{mol}{l}\right)&0{,}100&0{,}0500&0&0{,}100\\ \hline Muutos\left(\frac{mol}{l}\right)&-0{,}038&-0{,}038&+0{,}019&+0{,}038\\ C_{tasap}\left(\frac{mol}{l}\right)&0{,}062&0{,}012&0{,}019&0{,}138 \end{array}$

$K_c=\frac{\left[N_2\right]\left[H_2O\right]^2}{\left[NO\right]^2\left[H_2\right]^2}=\frac{0{,}019\cdot0{,}138^2}{0{,}062^2\cdot0{,}012^2}=653{,}68106...\approx650\ \left(\frac{mol}{l}\right)^{-1}$

3.4

3.5

3.6

3.7

$N_2O_4\left(g\right)\xrightleftharpoons[]{}2NO_2\left(g\right)$

$n_{alku}\left(N_2O_4\right)=24{,}4mmol=24{,}4\cdot10^{-3}mol$

$V=0{,}372l$
$T=25°C+273{,}15K=298{,}15K$

$p=0{,}400\ bar$

$R=0{,}08314510\ \frac{bar\cdot dm^3}{mol\cdot K}$

$c\left(N_2O_4\right)_{alku}=\frac{n}{V}=\frac{24{,}4\cdot10^{-3}mol}{0{,}372\ l}=\frac{61}{930}\frac{mol}{l}$

$pV=nRT<=>n=\frac{pV}{RT}$
$n\left(NO_2\right)=\frac{pV}{RT}=\frac{0{,}400bar\cdot0{,}372l}{0{,}08314510\ \frac{bar\cdot dm^3}{mol\cdot K}\cdot298{,}15K}=6{,}002490171\cdot10^{-3}mol$

$c\left(NO_2\right)_{alku}=\frac{6{,}002490171\cdot10^{-3}mol}{0{,}372l}=0{,}01613572...\approx0{,}016136\ \frac{mol}{l}$

$N_2O_4\left(g\right)\xrightleftharpoons[]{}2NO_2\left(g\right)$

$\begin{array}{l|l} c_{alku}\left(\frac{mol}{l}\right)&\frac{61}{930}\frac{mol}{l}&0\\ \hline Muutos\left(\frac{mol}{l}\right)&-\frac{1}{2}\cdot0{,}016136&+0{,}016136\\ c_{tasap.}\left(\frac{mol}{l}\right)&0{,}057523&0{,}016136 \end{array}$

$K_c=\frac{\left[NO_2\right]^2}{\left[N_2O_4\right]}=\frac{\left(0{,}016136\ \frac{mol}{l}\right)^2}{0{,}057523\frac{mol}{l}}=4{,}52637199\cdot10^{-3}\frac{mol}{l}\approx4{,}53'10^{-3}\frac{mol}{l}$

3.9

$N_2+3H_2\xrightleftharpoons[]{}2NH_3$

$T=472°C+273{,}15K=745{,}15K$

$K_c=0{,}104\ \left(\frac{mol}{l}\right)^{-2}$
$V=1{,}00l$
$n\left(H_2\right)=2{,}00\ mol$
$n\left(N_2\right)=1{,}00\ mol$

$n\left(NH_3\right)=2{,}00\ mol$

$c\left(H_2\right)_{alku}=\frac{2{,}00mol}{1{,}00l}=2{,}00\ \frac{mol}{dm^3}$
$c\left(N_2\right)_{alku}=\frac{1{,}00mol}{1{,}00l}=1{,}00\ \frac{mol}{dm^3}$
$c\left(NH_3\right)_{alku}=\frac{2{,}00mol}{1{,}00l}=2{,}00\ \frac{mol}{dm^3}$

$K_c=\frac{\left[NH_3\right]^2}{\left[N_2\right]\left[H_2\right]^3}=\frac{\left(2{,}00\ \frac{mol}{dm^3}\right)^2}{\left(1{,}00\ \frac{mol}{dm^3}\right)\left(2{,}00\ \frac{mol}{dm^3}\right)^3}=0{,}5\ \left(\frac{mol}{dm^3}\right)^{-2}$

$0{,}5\ \left(\frac{mol}{dm^3}\right)^{-2}>0{,}104\left(\frac{mol}{dm^3}\right)^{-2}$

v: Lähtöaineiden suuntaan.

b) Palautuvan reaktion nopeus on suurempi, sillä ammoniakin konsentraation tulee pienentyä, jotta tasapainotila muodostuisi.

3.12

$PCl_5\left(g\right)\xrightleftharpoons[]{}PCl_3\left(g\right)+Cl_2\left(g\right)$

3.13

$p_iV=n_iRT$

$p_i=\frac{n_i}{V}RT\ \ \ \ \ \left|\frac{n_i}{V}=c\right|$

$p_i=c_iRT$
$c_i=\frac{p_i}{RT}$

$p_i=40\ kPa=40\ 000\ Pa$

$T=\left(273{,}15+60\right)K=333{,}15K$

$c_i=\frac{p_i}{RT}=\frac{0{,}4\ bar}{0{,}0831451\ \frac{bar\cdot dm^3}{mol\cdot K}\cdot333{,}15K}=0{,}01444...\frac{mol}{dm^3}\approx0{,}014\ \frac{mol}{dm^3}$

3.16

Tehtävässä annetut alkuarvot:

$t=107°C\rightarrow T=\left(107+273{,}15\right)K=380{,}15K$

$V\left(astia\right)=1{,}04\ dm^3=1{,}04\cdot10^{-3}m^3$
$p=1{,}59\ MPa=1{,}59\cdot10^6\ Pa$
$n\left(CO\right)_{tasap.}=0{,}122mol$
$n\left(H_2\right)$

a)
$n\left(kaasut\right)_{tasap.}=\frac{pV}{RT}=\frac{1{,}59\cdot10^6\ Pa\cdot1{,}04\cdot10^{-5}m^3}{8{,}31451\ \frac{Pa\cdot m^3}{mol\cdot K}\cdot380{,}15K}=0{,}52317\ mol$

$n\left(CH_3OH\right)_{tasap.}=n\left(kaasut\right)_{tasap.}-n\left(CO\right)_{tasap.}-n\left(H_2\right)_{tasap.}=\left(0{,}52317-0{,}122-0{,}298\right)mol=0{,}10317\ mol\approx0{,}103\ mol$

Lasketaan tasapainokonsentraatiot

$\left[CO\right]=\frac{0{,}122mol}{1{,}04dm^3}=0{,}11731\ \frac{mol}{dm^3}$

$\left[H_2\right]=\frac{0{,}298mol}{1{,}04dm^3}=0{,}28654\ \frac{mol}{dm^3}$
$\left[CH_3OH\right]=\frac{0{,}103mol}{1{,}04dm^3}=0{,}099038\ \frac{mol}{dm^3}$

$K_c=\frac{\left[Ch_3OH\right]}{\left[CO\right]\left[H_2\right]^2}=\frac{0{,}099038\ \frac{mol}{dm^3}}{\left(0{,}11731\ \frac{mol}{dm^3}\right)\cdot\left(0{,}29654\ \frac{mol}{dm^3}\right)^2}$

$=10{,}282\ \left(\frac{mol}{dm^3}\right)^{-2}\approx10{,}3\ \left(\frac{mol}{dm^3}\right)^{-2}$