Ngā Tauira Pātai e Matapaki ana i te Taurite Matū i te Ao Ahumahi
He ariā nui te taurite matū i roto i te matū, ā, he whānui te whakamahinga i roto i ngā momo mara ahumahi. I roto i te tauhohenga matū, ka puta te taurite ina ōrite te tere o te tauhohenga whakamua ki te tere o te tauhohenga whakamuri, kia mau tonu ai te kukū o ngā matū tauhohenga me ngā hua i roto i te wā. He maha ngā ahumahi, pērā i te rongoā, te matū hinu, me te tukatuka kai, e whakawhirinaki nui ana ki te mārama me te whakahaere i te taurite matū hei whakapai ake i te whakaputanga me te whai huatanga. Ka matapakihia e tēnei tuhinga ētahi tauira o ngā raruraru e pā ana ki te taurite matū i roto i te horopaki ahumahi me pēhea te whakaoti i aua raruraru.
Tauira Pātai 1: Ahumahi Haukini (Tukanga Haber-Bosch)
Pātai:
Ka puta te haukini (NH4) i te tukanga Haber-Bosch3) mai i te hauota (N2) me te hauwai (H2) e ai ki te tauhohenga:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]
I te 500 K, ko te pūmau taurite (K c ) mō tēnei tauhohenga ko 6.0 x 10^-2. Mēnā ka tīmata tātou me te 1.00 mol N 2 me te 3.00 mol H 2 i roto i tētahi tauhohenga he 1.00 L te rōrahi, tatauhia te kukū o ia wāhanga i te taurite.
Kōrero:
1. Tāutuhia te huringa o te kukū mō ia wāhanga o te pūnaha.
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]
Kia waiho a x hei ngā mole o NH3 e hangaia ana i te taurite, ko te huringa o te kukū penei:
- N2: -x mol/L
- H2-3x mol/L
– NH3: +2x mol/L
2. Whakaritehia te whārite taurite i runga i te pūmau taurite (Kc):
\[
K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = 6.0 \times 10^{-2}
\]
Te kukū tuatahi me te huringa o te kukū:
– [N]2] = 1.00 – x
– [H2] = 3.00 – 3x
– [NH]3] = 2x
3. Monohia ēnei uara ki te whārite taurite:
\[
6.0 \times 10^{-2} = \frac{(2x)^2}{(1.00 – x)(3.00 – 3x)^3}
\]
4. Tātaihia te uara o te x mā te whakamahi i te whakamātautau me te hapa, i ētahi atu tikanga tau rānei hei whakaoti i te whārite.
I muri i te tatau, ka whiwhi tātou i te x = 0.46. Nō reira:
– [N]2] = 1.00 – 0.46 = 0.54 mol/L
– [H2] = 3.00 – 3(0.46) = 1.62 mol/L
– [NH]3] = 2(0.46) = 0.92 mol/L
Tauira Pātai 2: Ahumahi Waikawa Whāwhā (Tukanga Whakapā)
Pātai:
I roto i te tukanga whakapā, ka hurihia te hauhā whanariki (SO2) ki roto i te whāwhā toruwaikura (SO3) mā te tauhohenga:
\[ 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \]
Ko te pūmau taurite (K c ) mō tēnei tauhohenga i te 600 K he 350. Mena kei roto i te tauhohenga te 0.50 mol SO 2 , 0.25 mol O 2 , me te 0.10 mol SO 3 , tatauhia ngā kukū o ngā wāhanga i te taurite i roto i te rōrahi o te 2.00 L.
Kōrero:
1. Whakatauhia te kukū tuatahi:
– [NĀ2]tīmatanga = 0.50 mol / 2.00 L = 0.25 M
– [O2]tīmatanga = 0.25 mol / 2.00 L = 0.125 M
– [NĀ3]tīmatanga = 0.10 mol / 2.00 L = 0.05 M
2. Me waiho ko x te huringa o te kukū o te SO2.3 e hangaia ana i te taurite:
– [NĀ2]: 0.25 – x
– [O2]: 0.125 – \(\frac{x}{2}\)
– [NĀ3]: 0.05 + x
3. Mono ki te whārite taurite:
\[
350 = \frac{(0.05 + x)^2}{(0.25 – x)^2 \cdot (0.125 – \frac{x}{2})}
\]
4. Mā te whakaoti i tēnei whārite (mā te whakamahi i tētahi tikanga tau, i tētahi tātaitai hōtaka rānei), ka kitea ko x = 0.165. Kātahi:
– [NĀ2] = 0.25 – 0.165 = 0.085 M
– [O2] = 0.125 – \(\frac{0.165}{2}\) = 0.0425 M
– [NĀ3] = 0.05 + 0.165 = 0.215 M
Tauira Pātai 3: Te Hanganga Ethylbenzene
Pātai:
I te hanga o te ethylbenzene, ka puta te styrene mā te tango i te hauwai o te ethylbenzene (C6H5CH2CH3):
\[ \kuputuhi{C}_6\kuputuhi{H}_5\kuputuhi{CH}_2\kuputuhi{CH}_3(g) \rightleftharpoons \kuputuhi{C}_6\kuputuhi{H}_5\kuputuhi{CH=CH}_2(g) + \kuputuhi{H}_2(g) \]
Mena ko te pūmau taurite (K c ) mō tēnei tauhohenga i te 700 K he 2.5, ā, i te tīmatanga he 1.0 mol o te ethylbenzene i roto i te rōrahi o te 1.0 L, tatauhia te kukū i te taurite.
Kōrero:
1. Whakatauhia te kukū tuatahi:
– [C]6H5CH2CH3] = 1.0 M
– [C]6H5CH=CH2] = 0 M (nā te mea kāore anō kia pirau)
– [H2] = 0 M
2. Me waiho a x hei huringa o te kukū o te C6H5CH=CH2 e hangaia ana i te taurite:
– [C]6H5CH2CH3]: 1.0 – x
– [C]6H5CH=CH2]: x
– [H2]: x
3. Mono ki te whārite taurite:
\[
2.5 = \frac{x \cdot x}{1.0 – x} = \frac{x^2}{1.0 – x}
\]
4. Mā te whakaoti i tēnei whārite tapawhā, ka kitea ko x = 0.62. Kātahi:
– [C]6H5CH2CH3] = 1.0 – 0.62 = 0.38 M
– [C]6H5CH=CH2] = 0.62 M
– [H2] = 0.62 M
I roto i ēnei tauira e toru, kua kite tātou i te whakamahinga o te ariā o te taurite matū i roto i ngā horopaki ahumahi rerekē. He kaupapa matua, he kaupapa nui hoki te taurite matū i roto i ngā tukanga ahumahi, nā te mea ka taea e te whakahaere tika i te taurite matū te whakapai ake i te whai huatanga o te whakaputa me te kounga o te hua. Mā te māramatanga hohonu ki te taurite matū ka taea e te miihini, e te kaimahi ahumahi rānei te hoahoa me te whakahaere i ngā tukanga kia tino pai.
