3.1 Tasapainovakion ratkaiseminen ja reaktion suunnan ennustaminen

3.7

$c\left(N_2O_4\right)=\frac{24{,}4mmol}{0{,}372l}=0{,}06559139...$
$c\left(NO_2\right)=0{,}0016136$

$\begin{array}{l|l} &N_2O_4&2\ NO_2\\ \hline c_{alku}&0{,}065591&0\\ muutos&-\frac{1}{2}\cdot0{,}016136&+0{,}016136\\ c_{tasap}&0{,}057523&0{,}16136 \end{array}$

$K_c=\frac{\left(NO_2\right)^2}{N_2O_4}=0{,}04526371...\approx0{,}453\ \frac{mol}{dm^3}$

$K_c=\frac{\left(\frac{mol}{dm^3}\right)^2}{\frac{mol}{dm^3}}=\frac{mol}{dm^3}$

3.3

$n\left(NO\right)=0{,}100mol$

$n\left(H_2\right)=0{,}0500mol$

$n\left(H_2O\right)=0{,}100mol$
$c\left(NO\right)=0{,}062mol\text{/}l$

$\begin{array}{l|l} &NO&H_2&N_2&H_2O\\ \hline c_{alku}&0{,}100&0{,}05&0&0{,}1\\ muutos&-0{,}038&-0{,}038&0{,}038&+0{,}038\\ c_{tasap.}&0{,}062&0{,}012&0{,}038&0{,}138 \end{array}$
typen tasapainokonsentraatio 0,038 mol/l
vedyn tasapainokonsentraatio 0,012mol/l
veden tasapainokonsentraatio 0,138mol/l

$K_c=\frac{0{,}038\cdot0{,}138}{0{,}062\cdot0{,}012}=7{,}04838...\approx7{,}05$

$K_c=\frac{\text{mol/dm}\cdot\text{mol/dm}^3}{\text{mol/dm}^3\cdot\text{mol/dm}^3}=1$
tasapainovakio on 7,05

3.2

$n\left(N_2\right)=2{,}80\cdot10^{-3}mol$
$n\left(O_2\right)=2{,}5\cdot10^{-5}mol$

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

$K_c=\frac{\left(2{,}00\cdot10^{-2}\right)^2}{\left(2{,}80\cdot10^{-3}\right)^2\cdot2{,}5\cdot10^{-5}}=2{,}04\cdot10^6\ dm^3\text{/}mol$

3.1

a)
$2\ HI\left(g\right)\xrightleftharpoons[]{}H_2\left(g\right)+I_2\left(g\right)$
b)
$K_c=\frac{\left[H_2\right]\ \cdot\left[I_2\right]}{\left[HI\right]^2}=\frac{\left(4{,}79\cdot10^{-4}mol\right)^2}{\left(3{,}53\cdot10^{-4}mol\right)^2}=1{,}8412875...\approx1{,}84$
$K_c=\frac{\text{mol/dm}\cdot\text{mol/dm}^3}{\left(\text{mol/dm}^3\right)^2}=1$