Tauira o ngā Pātai Kōrero mō te Kinetika Matū
Ko te kinetika matū he peka o te matū e ako ana i ngā tere o ngā tauhohenga matū me ngā āhuatanga e awe ana i a rātou. He mea nui te māramatanga hōhonu ki te kinetika matū mō ngā kaipūtaiao matū me ngā miihini kia whakawhanakehia ai ngā tukanga ahumahi whai hua, kia mārama ai hoki ki ngā tauhohenga matū koiora rerekē e puta ana i roto i ngā rauropi ora. Ka matapakihia e tēnei tuhinga ētahi tauira raruraru e pā ana ki te kinetika matū me ā rātou otinga hei whakarato i te māramatanga hōhonu ake mō tēnei kaupapa.
Tauira Pātai 1: Te Whakatau i te Raupapa Tauhohenga
Pātai:
Ko te whārite tere whānui o tētahi tauhohenga e whai ake nei:
\[ R = k[A]^m[B]^n \]
Kei hea:
– Ko te tere tauhohenga te \( R \),
– Ko te pūmau tere te \( k \),
– Ko \([A] \) me \([B]\) ngā kukū o ngā matū tauhohenga A me B,
– Ko \( m \) me \( n \) ngā ota tauhohenga e pā ana ki a A me B.
E mōhiotia ana i whakahaerehia te whakamātautau me ngā rerekētanga kukū e whai ake nei:
| Whakamātautau | \([A]\) (mol/L) | \([B]\) (mol/L) | Tere tauhohenga (mol/(Ls)) |
|—————–|——————-|——————–|————————|
| 1 | 0,10 | 0,20 | 0,030 |
| 2 | 0,10 | 0,40 | 0,060 |
| 3 | 0,20 | 0,20 | 0,120 |
Whakatauhia te raupapa tauhohenga e pā ana ki a A me B me te uara o te pūmau tere \( k \).
Kōrero:
Hei whakatau i te raupapa tauhohenga e pā ana ki a A me B, me whakarite e tātou ngā tere tauhohenga me ngā rerekētanga kukū rerekē.
Tuatahi, ka whakatauhia e tātou te raupapa o te tauhohenga e pā ana ki a B mā te whakataurite i ngā whakamātautau 1 me te 2:
\[ \frac{\text{R2}}{\text{R1}} = \frac{k[A]^m [B_2]^n}{k[A]^m [B_1]^n} \]
\[ \frac{0,060}{0,030} = \frac{[0,10]^m [0,40]^n}{[0,10]^m [0,20]^n} \]
\[ 2 = \left(\frac{0,40}{0,20}\right)^n \]
\[ 2 = 2^n \]
\[ n = 1 \]
Ko te raupapa tauhohenga e pā ana ki a B ko 1.
Muri iho, ka whakatauhia e mātou te raupapa tauhohenga e pā ana ki a A mā te whakataurite i ngā whakamātautau 1 me te 3:
\[ \frac{\text{R3}}{\text{R1}} = \frac{k[A_3]^m [B]^n}{k[A_1]^m [B]^n} \]
\[ \frac{0,120}{0,030} = \frac{[0,20]^m [0,20]^n}{[0,10]^m [0,20]^n} \]
\[ 4 = \left(\frac{0,20}{0,10}\right)^m \]
\[ 4 = 2^m \]
\[ m = 2 \]
Ko te raupapa tauhohenga e pā ana ki a A ko 2.
Nō reira, ko te whārite tere tauhohenga ko:
\[ R = k[A]^2[B] \]
Inaianei ka kitea e tātou te uara o te pūmau tere \( k \). Whakamahia ngā raraunga mai i te whakamātautau 1:
\[ 0,030 = k[0,10]^2[0,20] \]
\[ 0,030 = k \whakareatia ki te 0,01 \whakareatia ki te 0,20 \]
\[ 0,030 = k \whakareatia ki te 0,002 \]
\[ k = \frac{0,030}{0,002} \]
\[ k = 15 \ \kuputuhi{L}^2/(\kuputuhi{mol}^2 \cdot \kuputuhi{s}) \]
Nō reira, ko te pūmau tere \( k \) he 15 L²/(mol²·s).
Tauira Pātai 2: Te Haurua-Ora o te Tauhohenga Tuarua-Ota
Pātai:
I runga i te tauhohenga tuarua me te whārite tere:
\[ R = k[A]^2 \]
Ko te pūmau tere (\( k \)) o te tauhohenga he 0,5 L/(mol·s). Mena ko te kukū tīmatanga o te matū tauhohenga \( [A]_0 \) he 1 mol/L, kimihia te haurua-ora o te tauhohenga.
Kōrero:
Mō ngā tauhohenga tuarua-raupapa, ka taea te tatau i te haurua-ora (\( t_{1/2} \)) mā te whakamahi i te whārite:
\[ t_{1/2} = \frac{1}{k[A]_0} \]
Whakakapia ngā uara e mōhiotia ana:
\[ t_{1/2} = \frac{1}{0,5 \times 1} \]
\[ t_{1/2} = \frac{1}{0,5} \]
\[ t_{1/2} = 2 \ \text{s} \]
Nō reira, ko te haurua-ora o tētahi tauhohenga tuarua me te pūmau tere o te 0,5 L/(mol·s) me te kukū tīmatanga o te matū tauhohenga o te 1 mol/L he 2 hēkona.
Tauira Pātai 3: Pūngao Whakahohe mā te Whārite Arrhenius
Pātai:
E rua ngā pūmau tere rerekē o tētahi tauhohenga i ngā pāmahana rerekē e rua:
– I te 300 K, ko te pūmau tere (\( k_1 \)) he 0,2 L/(mol·s)
– I te 350 K, ko te pūmau tere (\( k_2 \)) he 0,4 L/(mol·s)
Tātaihia te pūngao whakahohe (\( E_a \)) o te tauhohenga mā te whakamahi i te whārite Arrhenius:
\[ k = A e^{-E_a/(RT)} \]
Kōrero:
Ka taea te tuhi i te whārite Arrhenius i roto i te āhua logarithmic penei:
\[ \ln k = \ln A – \frac{E_a}{RT} \]
Ka taea e tātou te whakamahi i ngā raraunga pūmau tere e rua i ngā pāmahana rerekē e rua hei whakatau i \( E_a \):
Me tuhi e tātou ētahi whārite e rua mō ēnei āhuatanga e rua:
\[ \ln k_1 = \ln A – \frac{E_a}{R \cdot T_1} \]
\[ \ln k_2 = \ln A – \frac{E_a}{R \cdot T_2} \]
Mā te tango i ēnei whārite e rua:
\[ \ln k_2 – \ln k_1 = \left(\ln A – \frac{E_a}{R \cdot T_2}\right) – \left(\ln A – \frac{E_a}{R \cdot T_1}\right) \]
\[ \ln \left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} – \frac{1}{T_1}\right) \]
Whakakapia ngā uara o \( k_1 \), \( k_2 \), \( T_1 \), me \( T_2 \):
\[ \ln \left(\frac{0,4}{0,2}\right) = -\frac{E_a}{8,314} \left(\frac{1}{350} – \frac{1}{300}\right) \]
\[ \ln (2) = -\frac{E_a}{8,314} \left(\frac{1}{350} – \frac{1}{300}\right) \]
\[ 0,693 = -\frac{E_a}{8,314} \left(\frac{300 – 350}{350 \cdot 300}\right) \]
\[ 0,693 = -\frac{E_a}{8,314} \left(\frac{-50}{105000}\right) \]
\[ 0,693 = \frac{E_a}{8,314} \left(\frac{1}{2100}\right) \]
\[ 0,693 = \frac{E_a}{17462850/2100} \]
\[ 0,693 = \frac{E_a}{8314} \]
\[ E_a = 0,693 \whakareatia e 8314 \]
\[ E_a = 5761,842 \ \kuputuhi{J/mol} \]
Nō reira, ko te pūngao whakahohe (\( E_a \)) mō te tauhohenga he tata ki te 5761,842 J/mol, tata rānei ki te 5,76 kJ/mol.
-
He mea nui te mōhio ki te nekeneke matū me te mārama ki te matapaki i ngā raruraru pēnei i ēnei i roto i ngā momo mara, inā koa te umanga matū me te rangahau pūtaiao. Mā te tauira raruraru i runga ake nei ka mārama ki ngā tikanga mō te whakatau i te raupapa tauhohenga, te haurua-ora, me te pūngao whakahohe, he mea nui ēnei mō te whanaketanga hangarau me te māramatanga hohonu ake ki ngā tikanga tauhohenga matū.
