Tambayoyi Misali Game da Daidaiton Sinadarai
Daidaiton sinadarai muhimmin ra'ayi ne a fannin ilmin sunadarai wanda ke bayyana yanayin da yawan martanin gaba da na baya a cikin martanin sinadarai suke daidai. A ƙarƙashin wannan yanayi, yawan amsawa da samfura suna nan daram. Wannan labarin zai samar da misalai da dama na matsaloli da mafita don taimakawa fahimtar manufar daidaiton sinadarai.
Ka'idoji na Asali na Daidaiton Sinadarai
Daidaiton sinadarai yana faruwa ne lokacin da amsawar sinadarai ta iya ci gaba ta hanyar juyawa ko kuma ta hanyoyi biyu. Ana iya nuna amsawar da za a iya juyawa kamar haka:
\[ \text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD} \]
Ina:
- A da B su ne sinadaran amsawa,
- C da D samfura ne,
– a, b, c, da d su ne ma'aunin stoichiometric na kowane abu.
Idan aka cimma daidaito, ƙimar amsawar gaba (samar da samfura) daidai take da ƙimar amsawar baya (samar da amsawa). A wannan lokacin, kodayake amsawar ta ci gaba da aiki da sauri, yawan duk abubuwan da ke cikinta ba ya canzawa.
Daidaito Mai Daidaito (K)
Ana iya bayyana ma'aunin daidaito \(K_c\) na amsawar da ke sama kamar haka:
\[ K_c = \frac{{[\text{C}]^c [\text{D}]^d}}{{[\text{A}]^a [\text{B}]^b}} \]
Inda [X] shine yawan sinadarin X. Lokacin amfani da matsin lamba na ɗan lokaci a cikin daidaiton iskar gas, ana bayyana daidaiton ma'auni kamar haka \(K_p\).
Tambayoyi Misali Game da Daidaito a Sinadarai
Tambaya ta 1: Martani da Bayanan Tattara Hankali
Ana sanya adadin mole 1 na N₂ da moles 3 na H₂ a cikin akwati mai lita 1 a wani zafin jiki. Amsar tana gudana kamar haka:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]
Bayan an cimma daidaito, an sami 0,8 mol na N₂. A ƙididdige daidaiton daidaiton \(K_c\).
Tattaunawa:
1. Ƙayyade canjin yawan aiki:
Da farko, adadin moles shine:
– \([\text{N}_2]_{initial} = 1 \, \text{mol/L}\]
– \([\text{H}_2]_{initial} = 3 \, \text{mol/L}\]
– \([\text{NH}_3]_{initial} = 0 \, \text{mol/L}\]
A ma'auni, adadin moles:
- \([\rubutu {N}_2] = 0,8 \, \rubutu {mol/L}\]
Canji a cikin N₂ = 1 – 0,8 = 0,2 mol/L
2. Tsarin canje-canje:
\[
\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)
\]
Don haka, canje-canje ga H₂ da NH₃:
– \([\text{H}_2] = 3 \cdot 0,2 = 0,6 \, \text{mol/L}\]
– \([\text{NH}_3] = 2 \cdot 0,2 = 0,4 \, \text{mol/L}\]
Daidaiton mole:
– \([\text{H}_2] = 3 – 0,6 = 2,4 \, \text{mol/L}\]
– \([\text{NH}_3] = 0 + 0,4 = 0,4 \, \text{mol/L}\]
3. Lissafi \(K_c\):
\[
K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}
\]
Maye gurbin dabi'un:
– \([\text{NH}_3] = 0,4 \, \text{mol/L}\]
- \([\rubutu {N}_2] = 0,8 \, \rubutu {mol/L}\]
– \([\text{H}_2] = 2,4 \, \text{mol/L}\]
\[
K_c = \frac{(0,4)^2}{(0,8)(2,4)^3}
\]
\[
= \frac{0,16}{0,8 \cdot 13,824}
\]
\[
= \frac{0,16}{11,0592}
\kimanin 0,0145
\]
Tambaya ta 2: Tasirin Canje-canje a Hankali
Adadin N₂O₄(g) yana rikidewa zuwa 2NO₂(g) a cikin akwati mai rufewa. A wani zafin jiki, ma'aunin daidaiton \(K_c\) shine 0,36. Idan ma'aunin farko na N₂O₄(g) shine 1,0 M kuma babu NO₂(g) a farko, ƙididdige yawan NO₂(g) a ma'auni.
Tattaunawa:
1. Amfani da Teburin ICE:
\[
\begin{daidai}
\text{Amsa:} & \ \ \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \\
\text{Farko:} & \ \ [\text{N}_2\text{O}_4]_{0} = 1.0 \, \text{M}, \ [\text{NO}_2]_{0} = 0 \\
\text{Canza:} & \ \ [\text{N}_2\text{O}_4]_{eq} = 1.0 – x, \ [\text{NO}_2]_{eq} = 2x \\
\end{daidai}
\]
2. Haɗawa da \(K_c\):
\[
K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}
= \frac{(2x)^2}{1.0 – x}
= \frac{4x^2}{1 – x}
\]
3. Kayyade x:
\[
K_c = 0,36
\]
Don haka, canji:
\[
0,36 = \frac{4x^2}{1 – x}
\]
ninka giciye:
\[
0,36(1 – x) = 4x^2
\]
\[
0,36 – 0,36x = 4x^2
\]
Matsar da komai zuwa gefe ɗaya:
\[
4x^2 + 0,36x - 0,36 = 0
\]
4. Magance Daidaito na Huɗu:
Yi amfani da dabarar quadratic:
\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
Inda a = 4, b = 0,36 da c = -0,36:
\[
x = \frac{-0,36 \pm \sqrt{(0,36)^2 – 4(4)(-0,36)}}{2(4)}
\]
\[
x = \frac{-0,36 \pm \sqrt{0,1296 + 5,76}}{8}
\]
\[
x = \frac{-0,36 \pm \sqrt{5,8896}}{8}
\]
Tunda maida hankali ba zai iya zama mara kyau ba, za mu zaɓi tushen mai kyau:
\[
x \kimanin 0,36
\]
5. NO₂ maida hankali:
\[
[\text{NO}_2]_{eq} = 2x = 2 \cdot 0,36 = 0,72 \, \text{M}
\]
Kammalawa
Fahimtar daidaiton sinadarai shine mabuɗin yin hasashen yadda tsarin sinadarai zai yi aiki a ƙarƙashin wasu yanayi. Tare da aiki da fahimta mai zurfi, za mu iya magance matsalolin da suka shafi wannan ra'ayi, bayyana yanayin ƙarshe na tsarin, da kuma fahimtar yanayin halayen sinadarai a cikin tsarin rufewa. Ganewa da amfani da daidaitaccen ma'auni (K_c) zai sauƙaƙa hasashen yawan dukkan nau'ikan halittu a cikin tsarin a daidaito.
