Misalin Tambayoyi Game da Yawan Amsawa
Yawan amsawa muhimmin ra'ayi ne a fannin ilmin sunadarai wanda ke taka muhimmiyar rawa a cikin ayyuka daban-daban, na masana'antu da na yau da kullun. A cikin wannan labarin, za mu yi bayani dalla-dalla game da manufar yawan amsawa, tare da samar da misalai da tattaunawa dalla-dalla don tabbatar da cewa masu karatu sun fahimci hakan sosai.
Fahimtar Yawan Amsawa
Ana bayyana ƙimar amsawar a matsayin canjin yawan amsawar ko samfurin a kowane lokaci na raka'a. A cikin lissafi mai sauƙi, ana iya rubuta ƙimar amsawar kamar haka:
\[ \text{Matsayin martani} = \frac{\Delta \text{[Maida hankali]}}{\Delta t} \]
Yawanci ana auna maida hankali a cikin moles a kowace lita (M) kuma lokaci yawanci yana cikin daƙiƙa (s). Don haka, raka'o'in ƙimar amsawa galibi M/s ne.
Abubuwan da ke Shafar Yawan Amsawa
Ga wasu abubuwan da ke tasiri ga saurin amsawar:
1. Yawan sinadaran da ke cikin sinadaran: Yawan sinadaran da ke cikin ...
2. Zafin jiki: Ƙara yawan zafin jiki yakan sa saurin amsawar ya yi sauri.
3. Yankin Sama: Da yawan yankin saman da ake da shi, da sauri saurin amsawar.
4. Mai Haɓaka: Masu Haɓaka suna hanzarta saurin amsawa ba tare da fuskantar canje-canje na dindindin ba.
5. Matsi: Ga halayen da suka shafi iskar gas, ƙaruwar matsin lamba yawanci yana ƙara yawan amsawar.
Tambayoyi da Tattaunawa Samfura
Misali Tambaya ta 1
Yanayin da ke tsakanin sodium thiosulfate (Na2S2O3) da hydrochloric acid (HCl) kamar haka:
\[ \text{Na}_2\text{S}_2\text{O}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{S} + \text{SO}_2 + \text{H}_2\text{O} \]
A wani gwaji, yawan sinadarin sodium thiosulfate yana canzawa daga 0,10 M zuwa 0,05 M cikin daƙiƙa 30. Lissafa matsakaicin adadin amsawar!
Tattaunawa
Ana iya ƙididdige matsakaicin ƙimar amsawa ta amfani da dabarar:
\[ \text{Rage martani} = -\frac{\Delta \text{[Na}_2\text{S}_2\text{O}_3\text{]}}{\Delta t} \]
Sauya dabi'un da aka bayar a cikin dabarar:
\[ \Delta \text{[Na}_2\text{S}_2\text{O}_3\text{]} = 0,05 \text{ M} – 0,10 \text{ M} = -0,05 \text{ M} \]
\[ \Delta t = 30 \text{s} \]
Don haka,
\[ \text{Rage martani} = -\left(\frac{-0,05 \text{ M}}{30 \text{ s}}\right) = \frac{0,05 \text{ M}}{30 \text{ s}} = 0,00167 \text{ M/s} \]
Don haka, matsakaicin ƙimar amsawar shine 0,00167 M/s.
Misali Tambaya ta 2
A cikin amsawar, ana bayar da ƙimar amsawar ta hanyar lissafin ƙimar:
\[ \text{Ƙimar} = k [A]^m [B]^n \]
Daga gwajin, an samo waɗannan bayanai:
| Gwaji | [A] (M) | [B] (M) | Yawan martani (M/s) |
|————–|—————|———————-|
| 1 | 0.10 | 0.10 | 2.0 × 10^-3 |
| 2 | 0.20 | 0.10 | 8.0 × 10^-3 |
| 3 | 0.10 | 0.20 | 2.0 × 10^-3 |
Ƙayyade umarnin amsawar m da n kuma ƙididdige ƙimar ma'aunin ƙimar, k.
Tattaunawa
Ƙayyade Tsarin Amsa \( m \) da \( n \):
1. Daga Gwaje-gwaje 1 da 2:
\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k [A]_2^m [B]_2^n}{k [A]_1^m [B]_1^n} \]
\[ \frac{8.0 \sau 10^{-3}}{2.0 \sau 10^{-3}} = \frac{(0.20)^m (0.10)^n}{(0.10)^m (0.10)^n} \]
\[ 4 = (2)^m \]
Don haka, \( m = 2 \).
2. Daga Gwaje-gwaje 1 da 3:
\[ \frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k [A]_3^m [B]_3^n}{k [A]_1^m [B]_1^n} \]
\[ \frac{2.0 \sau 10^{-3}}{2.0 \sau 10^{-3}} = \frac{(0.10)^m (0.20)^n}{(0.10)^m (0.10)^n} \]
\[ 1 = (2)^n \]
Saboda haka, \( n = 0 \).
Don haka, tsarin amsawar dangane da A shine 2 kuma dangane da B shine 0.
Lissafin Ƙimar Daidaito ta Ƙimar \( k \):
Amfani da bayanai daga gwaji na 1:
\[ \text{Ƙimar} = k [A]^m [B]^n \]
\[ sau 2.0 10^{-3} = k (0.10)^2 (0.10)^0 \]
\[ sau 2.0 10^{-3} = k (0.01) \]
\[ k = \frac{2.0 \sau 10^{-3}}{0.01} \]
\[ k = 0.20 \]
Don haka, ma'aunin ƙimar \( k \) shine 0.20 M^{-1} s^{-1}.
Misali Tambaya ta 3
Haɗakar sinadarai tana bin wannan tsari:
\[ \text{Amsa 1: } \text{A} \rightarrow \text{B} \quad (k_1 = 1.0 \, \text{s}^{-1}) \]
\[ \text{Amsa 2: } \text{B} \rightarrow \text{C} \quad (k_2 = 0.1 \, \text{s}^{-1}) \]
Idan da farko yawan A shine 1 M kuma B shine 0, a tantance yawan A da B bayan daƙiƙa 5.
Tattaunawa
Ta amfani da dokar ƙimar amsawa, muna da:
Martani na 1: A zuwa B
\[ [A] = [A]_0 e^{-k_1 t} \]
\[ [A] = 1 \rubutu{ M} \times e^{-1.0 \rubutu{ s}^{-1} \times 5 \rubutu{ s}} \]
\[ [A] = e^{-5} \rubutu{ M} \]
Martani na 2: B zuwa C
\[ \frac{d[B]}{dt} = k_1 [A] – k_2 [B] \]
\[ \frac{d[B]}{dt} = 1.0 \text{ s}^{-1} \times [A] – 0.1 \text{ s}^{-1} \times [B] \]
Amfani da maganin nazari ko na lambobi na wannan lissafin bambanci (yawanci hanyar Euler ko Runge-Kutta):
\[ [B] \kimanin 0.316 \rubutu{ M} \]
Don haka, bayan daƙiƙa 5, yawan A yana kusan \(e^{-5} \text{ M} \) kuma yawan B yana kusan 0.316 M.
Kammalawa
Yawan amsawa muhimmin batu ne a fannin ilmin sunadarai, wanda ke nuna yadda yawan sinadaran da ke cikin sinadaran ke canzawa zuwa samfura. A cikin misalan matsalolin da ke sama, mun tattauna yadda za a ƙididdige matsakaicin adadin amsawa, ƙayyade tsarin amsawa, da kuma ƙididdige adadin amsawar. Fahimtar waɗannan ra'ayoyi yana ba mu damar amfani da su a cikin yanayi daban-daban na aiki, duka a cikin dakin gwaje-gwaje da kuma a cikin ayyukan masana'antu.
